how to print variables and comment separately? - linux

How to display comments from variables file?
Content of variables.conf
VAR1=/destination1 # comment
VAR2=/destination2 # comment
Script
VAR1=./variables.conf
for file in $(cat $VAR1 | sed 's/^/$/');
do
echo "${file%=*}"
done
Output
$VAR1
$VAR2
I need output to be like that and cannot figured out
$VAR1 # comment
$VAR2 # comment

I don't know why people are overcomplicating this so badly. Simply loop line by line and extract the comment the way you extract the name:
VAR1=./variables.conf
sed 's/^/$/' "$VAR1" | while read -r file
do
echo "${file%=*} #${file##*#}"
done
or just do it all with sed in the first place:
sed -e 's/^/$/; s/=[^#]*/ /' "$VAR1"

I've scripted a little function for you. I expect that the var isn't defined twice, otherwise you'll have to make sure that grep only gets the first. Also, this will not work on really old bash versions, I use version 5.0.3.
#! /bin/bash
SCRIPTNAME=$0
VAR1=/destination1 # comment 1
VAR2=/destination2 # comment 2
function print_comment {
local var_name=$1
local -n var_content=$1
local comment=""
comment=$(grep -E "^${var_name}=.*" $SCRIPTNAME | sed -r 's/^[^#].*(#.*)$/\1/')
echo "$var_content $comment"
}
print_comment VAR1
print_comment VAR2
exit 0
This outputs:
$ ./tst.sh
/destination1 # comment 1
/destination2 # comment 2

A way with mapping variable name as key and comment as value in a Bash 4+ associative array:
#!/usr/bin/env bash
varfile='variables.conf'
# shellcheck source=./variables.conf
. "$varfile"
# shellcheck disable=SC2155 # safe assignement
declare -A comments="($(
while IFS=$'\37' read -r k v; do
printf '[%q]=%q\n' "$k" "$v"
done< <(sed 's/^\([^=]*\)=[^#]\+#\ \?\(.*\)/\1\x1F\2/' "$varfile")
))"
for varname in "${!comments[#]}"; do
comment="${comments[$varname]}"
value="${!varname}"
printf '%-20s %-30s %s\n' "$varname" "$value" "$comment"
done

Comments are by definition invisible outside the script, they cannot be exported or used (well, kinda, but that's another story).
It would be possible to do what you mean by creating two string variables with the intended comments and later use an echo command to dislay them, something like :
#! /bin/bash
comment1="# Hey-ho"
comment2="# Let's go"
VAR1=/destination1
VAR2=/destination2
echo -e "$VAR1 $comment1\n$VAR2 $comment2" #why waste a good echo ;)
This should do the job. You can BTW skip de '#' as waht will be shown is obviosuly a string, but just to keep your syntax. And of course, my last commend won't be printed.
Of course, you will have to duplicate work, but as I said, comments are comments and not meant to be shown unless you are reading the code.

Related

Why am I getting command not found error on numeric comparison?

I am trying to parse each line of a file and look for a particular string. The script seems to be doing its intended job, however, in parallel it tries to execute the if command on line 6:
#!/bin/bash
for line in $(cat $1)
do
echo $line | grep -e "Oct/2015"
if($?==0); then
echo "current line is: $line"
fi
done
and I get the following (my script is readlines.sh)
./readlines.sh: line 6: 0==0: command not found
First: As Mr. Llama says, you need more spaces. Right now your script tries to look for a file named something like /usr/bin/0==0 to run. Instead:
[ "$?" -eq 0 ] # POSIX-compliant numeric comparison
[ "$?" = 0 ] # POSIX-compliant string comparison
(( $? == 0 )) # bash-extended numeric comparison
Second: Don't test $? at all in this case. In fact, you don't even have good cause to use grep; the following is both more efficient (because it uses only functionality built into bash and requires no invocation of external commands) and more readable:
if [[ $line = *"Oct/2015"* ]]; then
echo "Current line is: $line"
fi
If you really do need to use grep, write it like so:
if echo "$line" | grep -q "Oct/2015"; then
echo "Current line is: $line"
fi
That way if operates directly on the pipeline's exit status, rather than running a second command testing $? and operating on that command's exit status.
#Charles Duffy has a good answer which I have up-voted as correct (and it is), but here's a detailed, line by line breakdown of your script and the correct thing to do for each part of it.
for line in $(cat $1)
As I noted in my comment elsewhere this should be done as a while read construct instead of a for cat construct.
This construct will wordsplit each line making spaces in the file separate "lines" in the output.
All empty lines will be skipped.
In addition when you cat $1 the variable should be quoted. If it is not quoted spaces and other less-usual characters appearing in the file name will cause the cat to fail and the loop will not process the file.
The complete line would read:
while IFS= read -r line
An illustrative example of the tradeoffs can be found here. The linked test script follows. I tried to include an indication of why IFS= and -r are important.
#!/bin/bash
mkdir -p /tmp/testcase
pushd /tmp/testcase >/dev/null
printf '%s\n' '' two 'three three' '' ' five with leading spaces' 'c:\some\dos\path' '' > testfile
printf '\nwc -l testfile:\n'
wc -l testfile
printf '\n\nfor line in $(cat) ... \n\n'
let n=1
for line in $(cat testfile) ; do
echo line $n: "$line"
let n++
done
printf '\n\nfor line in "$(cat)" ... \n\n'
let n=1
for line in "$(cat testfile)" ; do
echo line $n: "$line"
let n++
done
let n=1
printf '\n\nwhile read ... \n\n'
while read line ; do
echo line $n: "$line"
let n++
done < testfile
printf '\n\nwhile IFS= read ... \n\n'
let n=1
while IFS= read line ; do
echo line $n: "$line"
let n++
done < testfile
printf '\n\nwhile IFS= read -r ... \n\n'
let n=1
while IFS= read -r line ; do
echo line $n: "$line"
let n++
done < testfile
rm -- testfile
popd >/dev/null
rmdir /tmp/testcase
Note that this is a bash-heavy example. Other shells do not tend to support -r for read, for example, nor is let portable. On to the next line of your script.
do
As a matter of style I prefer do on the same line as the for or while declaration, but there's no convention on this.
echo $line | grep -e "Oct/2015"
The variable $line should be quoted here. In general, meaning always unless you specifically know better, you should double-quote all expansion--and that means subshells as well as variables. This insulates you from most unexpected shell weirdness.
You decclared your shell as bash which means you will have there "Here string" operator <<< available to you. When available it can be used to avoid the pipe; each element of a pipeline executes in a subshell, which incurs extra overhead and can lead to unexpected behavior if you try to modify variables. This would be written as
grep -e "Oct/2015" <<<"$line"
Note that I have quoted the line expansion.
You have called grep with -e, which is not incorrect but is needless since your pattern does not begin with -. In addition you have full-quoted a string in shell but you don't attempt to expand a variable or use other shell interpolation inside of it. When you don't expect and don't want the contents of a quoted string to be treated as special by the shell you should single quote them. Furthermore, your use of grep is inefficient: because your pattern is a fixed string and not a regular expression you could have used fgrep or grep -F, which does string contains rather than regular expression matching (and is far faster because of this). So this could be
grep -F 'Oct/2015' <<<"$line"
Without altering the behavior.
if($?==0); then
This is the source of your original problem. In shell scripts commands are separated by whitespace; when you say if($?==0) the $? expands, probably to 0, and bash will try to execute a command called if(0==0) which is a legal command name. What you wanted to do was invoke the if command and give it some parameters, which requires more whitespace. I believe others have covered this sufficiently.
You should never need to test the value of $? in a shell script. The if command exists for branching behavior based on the return code of whatever command you pass to it, so you can inline your grep call and have if check its return code directly, thus:
if grep -F 'Oct/2015` <<<"$line" ; then
Note the generous whitespace around the ; delimiter. I do this because in shell whitespace is usually required and can only sometiems be omitted. Rather than try to remember when you can do which I recommend an extra one space padding around everything. It's never wrong and can make other mistakes easier to notice.
As others have noted this grep will print matched lines to stdout, which is probably not something you want. If you are using GNU grep, which is standard on Linux, you will have the -q switch available to you. This will suppress the output from grep
if grep -q -F 'Oct/2015' <<<"$line" ; then
If you are trying to be strictly standards compliant or are in any environment with a grep that doesn't know -q the standard way to achieve this effect is to redirect stdout to /dev/null/
if printf "$line" | grep -F 'Oct/2015' >/dev/null ; then
In this example I also removed the here string bashism just to show a portable version of this line.
echo "current line is: $line"
There is nothing wrong with this line of your script, except that although echo is standard implementations vary to such an extent that it's not possible to absolutely rely on its behavior. You can use printf anywhere you would use echo and you can be fairly confident of what it will print. Even printf has some caveats: Some uncommon escape sequences are not evenly supported. See mascheck for details.
printf 'current line is: %s\n' "$line"
Note the explicit newline at the end; printf doesn't add one automatically.
fi
No comment on this line.
done
In the case where you did as I recommended and replaced the for line with a while read construct this line would change to:
done < "$1"
This directs the contents of the file in the $1 variable to the stdin of the while loop, which in turn passes the data to read.
In the interests of clarity I recommend copying the value from $1 into another variable first. That way when you read this line the purpose is more clear.
I hope no one takes great offense at the stylistic choices made above, which I have attempted to note; there are many ways to do this (but not a great many correct) ways.
Be sure to always run interesting snippets through the excellent shellcheck and explain shell when you run into difficulties like this in the future.
And finally, here's everything put together:
#!/bin/bash
input_file="$1"
while IFS= read -r line ; do
if grep -q -F 'Oct/2015' <<<"$line" ; then
printf 'current line is %s\n' "$line"
fi
done < "$input_file"
If you like one-liners, you may use AND operator (&&), for example:
echo "$line" | grep -e "Oct/2015" && echo "current line is: $line"
or:
grep -qe "Oct/2015" <<<"$line" && echo "current line is: $line"
Spacing is important in shell scripting.
Also, double-parens is for numerical comparison, not single-parens.
if (( $? == 0 )); then

Make SED command work for any variable

deploy.sh
USERNAME="Tom"
PASSWORD="abc123"
FILE="config.conf"
sed -i "s/\PLACEHOLDER_USERNAME/$USERNAME/g" $FILE
sed -i "s/\PLACEHOLDER_PASSWORD/$PASSWORD/g" $FILE
config.conf
deloy="PLACEHOLDER_USERNAME"
pass="PLACEHOLDER_PASSWORD"
This file puts my variables defined in deploy into my config file. I can't source the file so I want put my variables in this way.
Question
I want a command that is generic to work for all placeholder variables using some sort of while loop rather than needing one command per variable. This means any term starting with placeholder_ in the file will try to be replaced with the value of the variable defined already in deploy.sh
All variables should be set and not empty. I guess if there is the ability to print a warning if it can't find the variable that would be good but it isn't mandatory for this.
Basically, use shell code to write a sed script and then use sed -i .bak -f sed.script config.conf to apply it:
trap "rm -f sed.script; exit 1" 0 1 2 3 13 15
for var in USERNAME PASSWORD
do
echo "s/PLACEHOLDER_$var/${!var}/"
done > sed.script
sed -i .bak -f sed.script config.conf
rm -f sed.script
trap 0
The main 'tricks' here are:
knowing that ${!var} expands to the value of the variable named by $var, and
knowing that sed will take a script full of commands via -f sed.script, and
knowing how to use trap to ensure temporary files are cleaned up.
You could also use sed -e "s/.../.../" -e "s/.../.../" -i .bak config.conf too, but the script file is easier, I think, especially if you have more than 2 values to substitute. If you want to go down this route, use a bash array to hold the arguments to sed. A more careful script would use at least $$ in the script file name, or use mktemp to create the temporary file.
Revised answer
The trouble is, although much closer to being generic, it is still not generic since I have to manually put in what variables I want to change. Can it not be more like "for each placeholder_, find the variable in deploy.sh and add that variable, so it can work for any number of variables.
So, find what the variables are in the configuration file, then apply the techniques of the previous answer to solve that problem:
trap "rm -f $tmp; exit 1" 0 1 2 3 13 15
for file in "$#"
do
for var in $(sed 's/.*PLACEHOLDER_\([A-Z0-9_]*\).*/\1/' "$file")
do
value="${!var}"
[ -z "$value" ] && { echo "$0: variable $var not set for $file" >&2; exit 1; }
echo "s/PLACEHOLDER_$var/$value/"
done > $tmp
sed -i .bak -f $tmp "$file"
rm -f $tmp
done
trap 0
This code still pulls the values from the environment. You need to clarify what is required if you want to extract the settings from the shell script, but it can be done — the script will have to be sufficiently self-aware to find its source so it can search it for the names. But the basics are in this answer; the rest is a question of tinkering until it does what you need.
#!/bin/ksh
TemplateFile=$1
SourceData=$2
(sed 's/.*/#V0r:PLACEHOLDER_&:r0V#/' ${SourceData}; cat ${TemplateFile}) | sed -n "
s/$/²/
H
$ {
x
s/^\(\n *\)*//
# also reset t flag
t varxs
:varxs
s/^#V0r:\([a-zA-Z0-9_]\{1,\}\)=\([^²]*\):r0V#²\(\n.*\)\"\1\"/#V0r:\1=\2:r0V#²\3\2/
t varxs
# clean the line when no more occurance in text
s/^[^²]*:r0V#²\n//
# and next
t varxs
# clean the marker
s/²\(\n\)/\1/g
s/²$//
# display the result
p
}
"
call like this: YourScript.ksh YourTemplateFile YourDataSourceFile where:
YourTemplateFile is the file that contain the structure with generic value like deloy="PLACEHOLDER_USERNAME"
YourDataSourceFile is the file that contain all the peer Generic value = specific value like USERNAME="Tom"

How to check if sed has changed a file

I am trying to find a clever way to figure out if the file passed to sed has been altered successfully or not.
Basically, I want to know if the file has been changed or not without having to look at the file modification date.
The reason why I need this is because I need to do some extra stuff if sed has successfully replaced a pattern.
I currently have:
grep -q $pattern $filename
if [ $? -eq 0 ]
then
sed -i s:$pattern:$new_pattern: $filename
# DO SOME OTHER STUFF HERE
else
# DO SOME OTHER STUFF HERE
fi
The above code is a bit expensive and I would love to be able to use some hacks here.
A bit late to the party but for the benefit of others, I found the 'w' flag to be exactly what I was looking for.
sed -i "s/$pattern/$new_pattern/w changelog.txt" "$filename"
if [ -s changelog.txt ]; then
# CHANGES MADE, DO SOME STUFF HERE
else
# NO CHANGES MADE, DO SOME OTHER STUFF HERE
fi
changelog.txt will contain each change (ie the changed text) on it's own line. If there were no changes, changelog.txt will be zero bytes.
A really helpful sed resource (and where I found this info) is http://www.grymoire.com/Unix/Sed.html.
I believe you may find these GNU sed extensions useful
t label
If a s/// has done a successful substitution since the last input line
was read and since the last t or T command, then branch to label; if
label is omitted, branch to end of script.
and
q [exit-code]
Immediately quit the sed script without processing any more input, except
that if auto-print is not disabled the current pattern space will be printed.
The exit code argument is a GNU extension.
It seems like exactly what are you looking for.
This might work for you (GNU sed):
sed -i.bak '/'"$old_pattern"'/{s//'"$new_pattern"'/;h};${x;/./{x;q1};x}' file || echo changed
Explanation:
/'"$old_pattern"'/{s//'"$new_pattern"'/;h} if the pattern space (PS) contains the old pattern, replace it by the new pattern and copy the PS to the hold space (HS).
${x;/./{x;q1};x} on encountering the last line, swap to the HS and test it for the presence of any string. If a string is found in the HS (i.e. a substitution has taken place) swap back to the original PS and exit using the exit code of 1, otherwise swap back to the original PS and exit with the exit code of 0 (the default).
You can diff the original file with the sed output to see if it changed:
sed -i.bak s:$pattern:$new_pattern: "$filename"
if ! diff "$filename" "$filename.bak" &> /dev/null; then
echo "changed"
else
echo "not changed"
fi
rm "$filename.bak"
You could use awk instead:
awk '$0 ~ p { gsub(p, r); t=1} 1 END{ exit (!t) }' p="$pattern" r="$repl"
I'm ignoring the -i feature: you can use the shell do do redirections as necessary.
Sigh. Many comments below asking for basic tutorial on the shell. You can use the above command as follows:
if awk '$0 ~ p { gsub(p, r); t=1} 1 END{ exit (!t) }' \
p="$pattern" r="$repl" "$filename" > "${filename}.new"; then
cat "${filename}.new" > "${filename}"
# DO SOME OTHER STUFF HERE
else
# DO SOME OTHER STUFF HERE
fi
It is not clear to me if "DO SOME OTHER STUFF HERE" is the same in each case. Any similar code in the two blocks should be refactored accordingly.
In macos I just do it as follows:
changes=""
changes+=$(sed -i '' "s/$to_replace/$replacement/g w /dev/stdout" "$f")
if [ "$changes" != "" ]; then
echo "CHANGED!"
fi
I checked, and this is faster than md5, cksum and sha comparisons
I know it is a old question and using awk instead of sed is perhaps the best idea, but if one wants to stick with sed, an idea is to use the -w flag. The file argument to the w flag only contains the lines with a match. So, we only need to check that it is not empty.
perl -sple '$replaced++ if s/$from/$to/g;
END{if($replaced != 0){ print "[Info]: $replaced replacement done in $ARGV(from/to)($from/$to)"}
else {print "[Warning]: 0 replacement done in $ARGV(from/to)($from/$to)"}}' -- -from="FROM_STRING" -to="$DESIRED_STRING" </file/name>
Example:
The command will produce the following output, stating the number of changes made/file.
perl -sple '$replaced++ if s/$from/$to/g;
END{if($replaced != 0){ print "[Info]: $replaced replacement done in $ARGV(from/to)($from/$to)"}
else {print "[Warning]: 0 replacement done in $ARGV(from/to)($from/$to)"}}' -- -from="timeout" -to="TIMEOUT" *
[Info]: 5 replacement done in main.yml(from/to)(timeout/TIMEOUT)
[Info]: 1 replacement done in task/main.yml(from/to)(timeout/TIMEOUT)
[Info]: 4 replacement done in defaults/main.yml(from/to)(timeout/TIMEOUT)
[Warning]: 0 replacement done in vars/main.yml(from/to)(timeout/TIMEOUT)
Note: I have removed -i from the above command , so it will not update the files for the people who are just trying out the command. If you want to enable in-place replacements in the file add -i after perl in above command.
check if sed has changed MANY files
recursive replace of all files in one directory
produce a list of all modified files
workaround with two stages: match + replace
g='hello.*world'
s='s/hello.*world/bye world/g;'
d='./' # directory of input files
o='modified-files.txt'
grep -r -l -Z -E "$g" "$d" | tee "$o" | xargs -0 sed -i "$s"
the file paths in $o are zero-delimited
$ echo hi > abc.txt
$ sed "s/hi/bye/g; t; q1;" -i abc.txt && (echo "Changed") || (echo "Failed")
Changed
$ sed "s/hi/bye/g; t; q1;" -i abc.txt && (echo "Changed") || (echo "Failed")
Failed
https://askubuntu.com/questions/1036912/how-do-i-get-the-exit-status-when-using-the-sed-command/1036918#1036918
Don't use sed to tell if it has changed a file; instead, use grep to tell if it is going to change a file, then use sed to actually change the file. Notice the single line of sed usage at the very end of the Bash function below:
# Usage: `gs_replace_str "regex_search_pattern" "replacement_string" "file_path"`
gs_replace_str() {
REGEX_SEARCH="$1"
REPLACEMENT_STR="$2"
FILENAME="$3"
num_lines_matched=$(grep -c -E "$REGEX_SEARCH" "$FILENAME")
# Count number of matches, NOT lines (`grep -c` counts lines),
# in case there are multiple matches per line; see:
# https://superuser.com/questions/339522/counting-total-number-of-matches-with-grep-instead-of-just-how-many-lines-match/339523#339523
num_matches=$(grep -o -E "$REGEX_SEARCH" "$FILENAME" | wc -l)
# If num_matches > 0
if [ "$num_matches" -gt 0 ]; then
echo -e "\n${num_matches} matches found on ${num_lines_matched} lines in file"\
"\"${FILENAME}\":"
# Now show these exact matches with their corresponding line 'n'umbers in the file
grep -n --color=always -E "$REGEX_SEARCH" "$FILENAME"
# Now actually DO the string replacing on the files 'i'n place using the `sed`
# 's'tream 'ed'itor!
sed -i "s|${REGEX_SEARCH}|${REPLACEMENT_STR}|g" "$FILENAME"
fi
}
Place that in your ~/.bashrc file, for instance. Close and reopen your terminal and then use it.
Usage:
gs_replace_str "regex_search_pattern" "replacement_string" "file_path"
Example: replace do with bo so that "doing" becomes "boing" (I know, we should be fixing spelling errors not creating them :) ):
$ gs_replace_str "do" "bo" test_folder/test2.txt
9 matches found on 6 lines in file "test_folder/test2.txt":
1:hey how are you doing today
2:hey how are you doing today
3:hey how are you doing today
4:hey how are you doing today hey how are you doing today hey how are you doing today hey how are you doing today
5:hey how are you doing today
6:hey how are you doing today?
$SHLVL:3
Screenshot of the output:
References:
https://superuser.com/questions/339522/counting-total-number-of-matches-with-grep-instead-of-just-how-many-lines-match/339523#339523
https://unix.stackexchange.com/questions/112023/how-can-i-replace-a-string-in-a-files/580328#580328

Simplify a BASH scripting design

I have need to execute a command in a script an arbitrary number of times with associated arbitrary parameters.
I've decided the script will take its cue from a parameter file (parameter.txt) where lines are of the form:
label param1 param2
For each line in parameter.txt, I'll call the command with the specified parameters.
So far, my tinkering is moving along the lines of the following, but it's looking messy:
while read line; do
echo $line | sed -r 's/[^ ]+ ([^ ]+).+/\1/' &&
echo $line | sed -r 's/[^ ]+ [^ ]+ ([^ ]+)/\1/'
done < parameter.txt
My command is of the form:
mycmd -a param1 -b param2 > label
Could I get some suggestions how I might simplify this?
I'm doing this for a small embedded system whose 'helper' commands are in short supply (xargs for example isn't available, and things like awk are hobbled busybox implementations), and I'm using version 2 (2.04g I think) of BASH.
while read label param1 param2; do
mycmd -a "$param1" -b "$param2" > "$label"
done < parameter.txt
I'd suggest a function, as long as there aren't any embedded spaces.
function x()
{
mycmd -a $2 -b $3 >$1
}
while read line; do x $line ; done <parameter.txt
Try this:
while read line ; do
set -- $line
dest="$1"
shift
mycmd "$#" > "$dest"
done < parameter.txt
should work. If the parameters in the file have spaces, you will have to quote them properly.
I suggest to add the -a, -b to the file parameter.txt because generating them on the fly is probably brittle.
If you don't like this solution, then I suggest to create a new script from this one which contains the actual commands. That way, you can easily debug any problems.
When the script looks okay, you can source it with source ./generated.sh (yes, you have to specify the path).

Appending a line to a file only if it does not already exist

I need to add the following line to the end of a config file:
include "/configs/projectname.conf"
to a file called lighttpd.conf
I am looking into using sed to do this, but I can't work out how.
How would I only insert it if the line doesn't already exist?
Just keep it simple :)
grep + echo should suffice:
grep -qxF 'include "/configs/projectname.conf"' foo.bar || echo 'include "/configs/projectname.conf"' >> foo.bar
-q be quiet
-x match the whole line
-F pattern is a plain string
https://linux.die.net/man/1/grep
Edit:
incorporated #cerin and #thijs-wouters suggestions.
This would be a clean, readable and reusable solution using grep and echo to add a line to a file only if it doesn't already exist:
LINE='include "/configs/projectname.conf"'
FILE='lighttpd.conf'
grep -qF -- "$LINE" "$FILE" || echo "$LINE" >> "$FILE"
If you need to match the whole line use grep -xqF
Add -s to ignore errors when the file does not exist, creating a new file with just that line.
Try this:
grep -q '^option' file && sed -i 's/^option.*/option=value/' file || echo 'option=value' >> file
Using sed, the simplest syntax:
sed \
-e '/^\(option=\).*/{s//\1value/;:a;n;ba;q}' \
-e '$aoption=value' filename
This would replace the parameter if it exists, else would add it to the bottom of the file.
Use the -i option if you want to edit the file in-place.
If you want to accept and keep white spaces, and in addition to remove the comment, if the line already exists, but is commented out, write:
sed -i \
-e '/^#\?\(\s*option\s*=\s*\).*/{s//\1value/;:a;n;ba;q}' \
-e '$aoption=value' filename
Please note that neither option nor value must contain a slash /, or you will have to escape it to \/.
To use bash-variables $option and $value, you could write:
sed -i \
-e '/^#\?\(\s*'${option//\//\\/}'\s*=\s*\).*/{s//\1'${value//\//\\/}'/;:a;n;ba;q}' \
-e '$a'${option//\//\\/}'='${value//\//\\/} filename
The bash expression ${option//\//\\/} quotes slashes, it replaces all / with \/.
Note: Just trapped into a problem. In bash you may quote "${option//\//\\/}", but in the sh of busybox, this does not work, so you should avoid the quotes, at least in non-bourne-shells.
All combined in a bash function:
# call option with parameters: $1=name $2=value $3=file
function option() {
name=${1//\//\\/}
value=${2//\//\\/}
sed -i \
-e '/^#\?\(\s*'"${name}"'\s*=\s*\).*/{s//\1'"${value}"'/;:a;n;ba;q}' \
-e '$a'"${name}"'='"${value}" $3
}
Explanation:
/^\(option=\).*/: Match lines that start with option= and (.*) ignore everything after the =. The \(…\) encloses the part we will reuse as \1later.
/^#?(\s*'"${option//////}"'\s*=\s*).*/: Ignore commented out code with # at the begin of line. \? means «optional». The comment will be removed, because it is outside of the copied part in \(…\). \s* means «any number of white spaces» (space, tabulator). White spaces are copied, since they are within \(…\), so you do not lose formatting.
/^\(option=\).*/{…}: If matches a line /…/, then execute the next command. Command to execute is not a single command, but a block {…}.
s//…/: Search and replace. Since the search term is empty //, it applies to the last match, which was /^\(option=\).*/.
s//\1value/: Replace the last match with everything in (…), referenced by \1and the textvalue`
:a;n;ba;q: Set label a, then read next line n, then branch b (or goto) back to label a, that means: read all lines up to the end of file, so after the first match, just fetch all following lines without further processing. Then q quit and therefore ignore everything else.
$aoption=value: At the end of file $, append a the text option=value
More information on sed and a command overview is on my blog:
https://marc.wäckerlin.ch/computer/stream-editor-sed-overview-and-reference
If writing to a protected file, #drAlberT and #rubo77 's answers might not work for you since one can't sudo >>. A similarly simple solution, then, would be to use tee --append (or, on MacOS, tee -a):
LINE='include "/configs/projectname.conf"'
FILE=lighttpd.conf
grep -qF "$LINE" "$FILE" || echo "$LINE" | sudo tee --append "$FILE"
Here's a sed version:
sed -e '\|include "/configs/projectname.conf"|h; ${x;s/incl//;{g;t};a\' -e 'include "/configs/projectname.conf"' -e '}' file
If your string is in a variable:
string='include "/configs/projectname.conf"'
sed -e "\|$string|h; \${x;s|$string||;{g;t};a\\" -e "$string" -e "}" file
If, one day, someone else have to deal with this code as "legacy code", then that person will be grateful if you write a less exoteric code, such as
grep -q -F 'include "/configs/projectname.conf"' lighttpd.conf
if [ $? -ne 0 ]; then
echo 'include "/configs/projectname.conf"' >> lighttpd.conf
fi
another sed solution is to always append it on the last line and delete a pre existing one.
sed -e '$a\' -e '<your-entry>' -e "/<your-entry-properly-escaped>/d"
"properly-escaped" means to put a regex that matches your entry, i.e. to escape all regex controls from your actual entry, i.e. to put a backslash in front of ^$/*?+().
this might fail on the last line of your file or if there's no dangling newline, I'm not sure, but that could be dealt with by some nifty branching...
Here is a one-liner sed which does the job inline. Note that it preserves the location of the variable and its indentation in the file when it exists. This is often important for the context, like when there are comments around or when the variable is in an indented block. Any solution based on "delete-then-append" paradigm fails badly at this.
sed -i '/^[ \t]*option=/{h;s/=.*/=value/};${x;/^$/{s//option=value/;H};x}' test.conf
With a generic pair of variable/value you can write it this way:
var=c
val='12 34' # it handles spaces nicely btw
sed -i '/^[ \t]*'"$var"'=/{h;s/=.*/='"$val"'/};${x;/^$/{s//c='"$val"'/;H};x}' test.conf
Finally, if you want also to keep inline comments, you can do it with a catch group. E.g. if test.conf contains the following:
a=123
# Here is "c":
c=999 # with its own comment and indent
b=234
d=567
Then running this
var='c'
val='"yay"'
sed -i '/^[ \t]*'"$var"'=/{h;s/=[^#]*\(.*\)/='"$val"'\1/;s/'"$val"'#/'"$val"' #/};${x;/^$/{s//'"$var"'='"$val"'/;H};x}' test.conf
Produces that:
a=123
# Here is "c":
c="yay" # with its own comment and indent
b=234
d=567
As an awk-only one-liner:
awk -v s=option=value '/^option=/{$0=s;f=1} {a[++n]=$0} END{if(!f)a[++n]=s;for(i=1;i<=n;i++)print a[i]>ARGV[1]}' file
ARGV[1] is your input file. It is opened and written to in the for loop of theEND block. Opening file for output in the END block replaces the need for utilities like sponge or writing to a temporary file and then mving the temporary file to file.
The two assignments to array a[] accumulate all output lines into a. if(!f)a[++n]=s appends the new option=value if the main awk loop couldn't find option in file.
I have added some spaces (not many) for readability, but you really need just one space in the whole awk program, the space after print.
If file includes # comments they will be preserved.
Here's an awk implementation
/^option *=/ {
print "option=value"; # print this instead of the original line
done=1; # set a flag, that the line was found
next # all done for this line
}
{print} # all other lines -> print them
END { # end of file
if(done != 1) # haven't found /option=/ -> add it at the end of output
print "option=value"
}
Run it using
awk -f update.awk < /etc/fdm_monitor.conf > /etc/fdm_monitor.conf.tmp && \
mv /etc/fdm_monitor.conf.tmp /etc/fdm_monitor.conf
or
awk -f update.awk < /etc/fdm_monitor.conf | sponge /etc/fdm_monitor.conf
EDIT:
As a one-liner:
awk '/^option *=/ {print "option=value";d=1;next}{print}END{if(d!=1)print "option=value"}' /etc/fdm_monitor.conf | sponge /etc/fdm_monitor.conf
use awk
awk 'FNR==NR && /configs.*projectname\.conf/{f=1;next}f==0;END{ if(!f) { print "your line"}} ' file file
sed -i 's/^option.*/option=value/g' /etc/fdm_monitor.conf
grep -q "option=value" /etc/fdm_monitor.conf || echo "option=value" >> /etc/fdm_monitor.conf
here is an awk one-liner:
awk -v s="option=value" '/^option/{f=1;$0=s}7;END{if(!f)print s}' file
this doesn't do in-place change on the file, you can however :
awk '...' file > tmpfile && mv tmpfile file
Using sed, you could say:
sed -e '/option=/{s/.*/option=value/;:a;n;:ba;q}' -e 'aoption=value' filename
This would replace the parameter if it exists, else would add it to the bottom of the file.
Use the -i option if you want to edit the file in-place:
sed -i -e '/option=/{s/.*/option=value/;:a;n;:ba;q}' -e 'aoption=value' filename
sed -i '1 h
1 !H
$ {
x
s/^option.*/option=value/g
t
s/$/\
option=value/
}' /etc/fdm_monitor.conf
Load all the file in buffer, at the end, change all occurence and if no change occur, add to the end
The answers using grep are wrong. You need to add an -x option to match the entire line otherwise lines like #text to add will still match when looking to add exactly text to add.
So the correct solution is something like:
grep -qxF 'include "/configs/projectname.conf"' foo.bar || echo 'include "/configs/projectname.conf"' >> foo.bar
Using sed: It will insert at the end of line. You can also pass in variables as usual of course.
grep -qxF "port=9033" $light.conf
if [ $? -ne 0 ]; then
sed -i "$ a port=9033" $light.conf
else
echo "port=9033 already added"
fi
Using oneliner sed
grep -qxF "port=9033" $lightconf || sed -i "$ a port=9033" $lightconf
Using echo may not work under root, but will work like this. But it will not let you automate things if you are looking to do it since it might ask for password.
I had a problem when I was trying to edit from the root for a particular user. Just adding the $username before was a fix for me.
grep -qxF "port=9033" light.conf
if [ $? -ne 0 ]; then
sudo -u $user_name echo "port=9033" >> light.conf
else
echo "already there"
fi
I elaborated on kev's grep/sed solution by setting variables in order to reduce duplication.
Set the variables in the first line (hint: $_option shall match everything on the line up until the value [including any seperator like = or :]).
_file="/etc/ssmtp/ssmtp.conf" _option="mailhub=" _value="my.domain.tld" \
sh -c '\
grep -q "^$_option" "$_file" \
&& sed -i "s/^$_option.*/$_option$_value/" "$_file" \
|| echo "$_option$_value" >> "$_file"\
'
Mind that the sh -c '...' just has the effect of widening the scope of the variables without the need for an export. (See Setting an environment variable before a command in bash not working for second command in a pipe)
You can use this function to find and search config changes:
#!/bin/bash
#Find and Replace config values
find_and_replace_config () {
file=$1
var=$2
new_value=$3
awk -v var="$var" -v new_val="$new_value" 'BEGIN{FS=OFS="="}match($1, "^\\s*" var "\\s*") {$2=" " new_val}1' "$file" > output.tmp && sudo mv output.tmp $file
}
find_and_replace_config /etc/php5/apache2/php.ini max_execution_time 60
If you want to run this command using a python script within a Linux terminal...
import os,sys
LINE = 'include '+ <insert_line_STRING>
FILE = <insert_file_path_STRING>
os.system('grep -qxF $"'+LINE+'" '+FILE+' || echo $"'+LINE+'" >> '+FILE)
The $ and double quotations had me in a jungle, but this worked.
Thanks everyone
Try:
LINE='include "/configs/projectname.conf"'
sed -n "\|$LINE|q;\$a $LINE" lighttpd.conf >> lighttpd.conf
Use the pipe as separator and quit if $LINE has been found. Otherwise, append $LINE at the end.
Since we only read the file in sed command, I suppose we have no clobber issue in general (it depends on your shell settings).
Using only sed I'd suggest the following solution:
sed -i \
-e 's#^include "/configs/projectname.conf"#include "/configs/projectname.conf"#' \
-e t \
-e '$ainclude "/configs/projectname.conf"' lighttpd.conf
s replace the line include "/configs/projectname.conf with itself (using # as delimiter here)
t if the replacement was successful skip the rest of the commands
$a otherwise jump to the last line and append include "/configs/projectname.conf after it
Almost all of the answers work but not in all scenarios or OS as per my experience. Only thing that worked on older systems and new and different flavours of OS is the following.
I needed to append KUBECONFIG path to bashrc file if it doesnt exist. So, what I did is
I assume that it exists and delete it.
with sed I append the string I want.
sed -i '/KUBECONFIG=/d' ~/.bashrc
echo 'export KUBECONFIG=/etc/rancher/rke2/rke2.yaml' >> ~/.bashrc
I needed to edit a file with restricted write permissions so needed sudo. working from ghostdog74's answer and using a temp file:
awk 'FNR==NR && /configs.*projectname\.conf/{f=1;next}f==0;END{ if(!f) { print "your line"}} ' file > /tmp/file
sudo mv /tmp/file file

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