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When to wrap quotes around a shell variable?
(5 answers)
Closed 4 years ago.
In this script I was just wondering why the double quotes are necessary around the variable $line
$cat script1
#!/bin/bash
exec 3<$1
exec 4<$2
exec 5>$3
while read line <&3
do
echo "$line" >&5
done
while read line <&4
do
echo "$line" >&5
done
You need quotes around line if you don't want spaces (IFS characters, default spaces, tabs and newlines) to expand. Observe:
printf "a b\n" |
while read line; do
echo $line
done
this will output:
a b
With qoutes:
printf "a b\n" |
while read line; do
echo "$line"
done
will output:
a b
Quotes prohibit shell expansion. Grab a good read on quotes. Also read about IFS and be aware of read delimiter -d option and -r option.
If you whish to read from file, preserving leading and trailing whitespaces and the whole line, see here and use:
while IFS= read -r line; do
echo "$line"
done
Also note that using while read loops to parse files is very slow on bash. Try to use bash commands and moreutils and standard unix commands to parse files. If you really need to parse file line by line, xargs and parallel are good programs for that. The script you presented may be just substituted by cat "$1" "$2" >"$5" (or more like cat "$1" "$2" | sed 's/^[ \t]*//;s/[ \t]*$//' > "$5", because leading and trailing whilespaces will not be preserved).
Also quote your variables. The exec 3<$1 will fail if someone cals your function with file with spaces in the name, like: ./script1 "filename with spaces.txt". Use exec 3<"$1".
I have a bash script which asks for two arguments with a space between them. Now I would like to automate filling out the prompt in the command line with reading from a text file. The text file contains a list with the argument combinations.
So something like this in the command line I think;
for line in 'cat text.file' ; do script.sh ; done
Can this be done? What am I missing/doing wrong?
Thanks for the help.
A while loop is probably what you need. Put the space separated strings in the file text.file :
cat text.file
bingo yankee
bravo delta
Then write the script in question like below.
#!/bin/bash
while read -r arg1 arg2
do
/path/to/your/script.sh "$arg1" "$arg2"
done<text.file
Don't use for to read files line by line
Try something like this:
#!/bin/bash
ARGS=
while IFS= read -r line; do
ARGS="${ARGS} ${line}"
done < ./text.file
script.sh "$ARGS"
This would add each line to a variable which then is used as the arguments of your script.
'cat text.file' is a string literal, $(cat text.file) would expand to output of command however cat is useless because bash can read file using redirection, also with quotes it will be treated as a single argument and without it will split at space tab and newlines.
Bash syntax to read a file line by line, but will be slow for big files
while IFS= read -r line; do ... "$line"; done < text.file
unsetting IFS for read command preserves leading spaces
-r option preserves \
another way, to read whole file is content=$(<file), note the < inside the command substitution. so a creative way to read a file to array, each element a non-empty line:
read_to_array () {
local oldsetf=${-//[^f]} oldifs=$IFS
set -f
IFS=$'\n' array_content=($(<"$1")) IFS=$oldifs
[[ $oldsetf ]]||set +f
}
read_to_array "file"
for element in "${array_content[#]}"; do ...; done
oldsetf used to store current set -f or set +f setting
oldifs used to store current IFS
IFS=$'\n' to split on newlines (multiple newlines will be treated as one)
set -f avoid glob expansion for example in case line contains single *
note () around $() to store the result of splitting to an array
If I were to create a solution determined by the literal of what you ask for (using a for loop and parsing lines from a file) I would use iterations determined by the number of lines in the file (if it isn't too large).
Assuming each line has two strings separated by a single space (to be used as positional parameters in your script:
file="$1"
f_count="$(wc -l < $file)"
for line in $(seq 1 $f_count)
do
script.sh $(head -n $line $file | tail -n1) && wait
done
You may have a much better time using sjsam's solution however.
I want to add the string %%% to the beginning of some specific lines in a text file.
This is my script:
#!/bin/bash
a="c:\Temp"
sed "s/$a/%%%$a/g" <File.txt
And this is my File.txt content:
d:\Temp
c:\Temp
e:\Temp
But nothing changes when I execute it.
I think the 'sed' command is not finding the pattern, possibly due to the \ backslashes in the variable a.
I can find the c:\Temp line if I use grep with -F option (to not interpret strings):
cat File.txt | grep -F "$a"
But sed seems not to implement such '-F` option.
Not working neither:
sed 's/$a/%%%$a/g' <File.txt
sed 's/"$a"/%%%"$a"/g' <File.txt
I have found similar threads about replacing with sed, but they don't refer to variables.
How can I replace the desired lines by using a variable adding them the %%% char string?
EDIT: It would be fine that the $a variable could be entered via parameter when calling the script, so it will be assigned like:
a=$1
Try it like this:
#!/bin/sh
a='c:\\Temp' # single quotes
sed "s/$a/%%%$a/g" <File.txt # double quotes
Output:
Johns-MacBook-Pro:sed jcreasey$ sh x.sh
d:\Temp
e:\Temp
%%%c:\Temp
You need the double slash '\' to escape the '\'.
The single quotes won't expand the variables.
So you escape the slash in single quotes and pass it into the double quotes.
Of course you could also just do this:
#!/bin/sh
sed 's/\(.*Temp\)/%%%&/' <File.txt
If you want to get input from the command line you have to allow for the fact that \ is an escape character there too. So the user needs to type 'c:\\' or the interpreter will just wait for another character. Then once you get it, you will need to escape it again. (printf %q).
#!/bin/sh
b=`printf "%q" $1`
sed "s/\($b\)/%%% &/" < File.txt
The issue you are having has to do with substitution of your variable providing a regular expression looking for a literal c:Temp with the \ interpreted as an escape by the shell. There are a number of workarounds. Seeing the comments and having worked through the possibilities, the following will allow the unquoted entry of the search term:
#!/bin/bash
## validate that needed input is given on the command line
[ -n "$1" -a "$2" ] || {
printf "Error: insufficient input. Usage: %s <term> <file>\n" "${0//*\//}" >&2
exit 1
}
## validate that the filename given is readable
[ -r "$2" ] || {
printf "Error: file not readable '%s'\n" "$2" >&2
exit 1
}
a="$1" # assign a
filenm="$2" # assign filename
## test and fix the search term entered
[[ "$a" =~ '/' ]] || a="${a/:/:\\}" # test if \ removed by shell, if so replace
a="${a/\\/\\\\}" # add second \
sed -e "s/$a/%%%$a/g" "$filenm" # call sed with output to stdout
Usage:
$ bash sedwinpath.sh c:\Temp dat/winpath.txt
d:\Temp
%%%c:\Temp
e:\Temp
Note: This allows both single-quoted or unquoted entry of the dos path search term. To edit in place use sed -i. Additionally, the [[ operator and =~ operator are limited to bash.
I could have sworn the original question said replace, but to append, just as you suggest in the comments. I have updated the code with:
sed -e "s/$a/%%%$a/g" "$filenm"
Which provides the new output:
$ bash sedwinpath.sh c:\Temp dat/winpath.txt
d:\Temp
%%%c:\Temp
e:\Temp
Remember: If you want to edit the file in place use sed -i or sed -i.bak which will edit the actual file (and if -i.bak is given create a backup of the original in originalname.bak). Let me know if that is not what you intended and I'm happy to edit again.
Creating your script with a positional parameter of $1
#!/bin/bash
a="$1"
cat <file path>|sed "s/"$1"/%%%"$1"/g" > "temporary file"
Now whenever you want sed to find "c:\Temp" you need to use your script command line as follows
bash <my executing script> c:\\\\Temp
The first backslash will make bash interpret any backslashes that follows therefore what will be save in variable "a" in your executing script is "c:\\Temp". Now substituting this variable in sed will cause sed to interpret 1 backlash since the first backslash in this variable will cause sed to start interpreting the other backlash.
when you Open your temporary file you will see:
d:\Temp
%%%c:\Temp
e:\Temp
I am trying to use a bash shell variable as a command parameter but can't
Here is what works:
sed -n '2p' <file>
gives me line 2 of file
What I want to do:
sed -n '$variable p' <file>
Of course, this does not work. I have tried every possible syntax combination without success. How can I incorporate a variable in place of a constant?
Variables are expanded inside doublequotes, not inside singlequotes:
sed -n "$variable p" <file>
#Barmar has the right answer to your question.
I fear you are going to use this as a technique to iterate over the lines of a file.
This will be very inefficient:
for linenum in $(seq $(wc -l < filename)); do
line=$(sed -n "$linenum p" filename)
# do something with $line
done
The idiomatic way to iterate over the lines of a file is:
while IFS= read -r line; do
# do something with "$line"
done < filename
Put the variable outside the string:
sed -n $variable'p'
Say, I have a file foo.txt specifying N arguments
arg1
arg2
...
argN
which I need to pass to the command my_command
How do I use the lines of a file as arguments of a command?
If your shell is bash (amongst others), a shortcut for $(cat afile) is $(< afile), so you'd write:
mycommand "$(< file.txt)"
Documented in the bash man page in the 'Command Substitution' section.
Alterately, have your command read from stdin, so: mycommand < file.txt
As already mentioned, you can use the backticks or $(cat filename).
What was not mentioned, and I think is important to note, is that you must remember that the shell will break apart the contents of that file according to whitespace, giving each "word" it finds to your command as an argument. And while you may be able to enclose a command-line argument in quotes so that it can contain whitespace, escape sequences, etc., reading from the file will not do the same thing. For example, if your file contains:
a "b c" d
the arguments you will get are:
a
"b
c"
d
If you want to pull each line as an argument, use the while/read/do construct:
while read i ; do command_name $i ; done < filename
command `< file`
will pass file contents to the command on stdin, but will strip newlines, meaning you couldn't iterate over each line individually. For that you could write a script with a 'for' loop:
for line in `cat input_file`; do some_command "$line"; done
Or (the multi-line variant):
for line in `cat input_file`
do
some_command "$line"
done
Or (multi-line variant with $() instead of ``):
for line in $(cat input_file)
do
some_command "$line"
done
References:
For loop syntax: https://www.cyberciti.biz/faq/bash-for-loop/
You do that using backticks:
echo World > file.txt
echo Hello `cat file.txt`
If you want to do this in a robust way that works for every possible command line argument (values with spaces, values with newlines, values with literal quote characters, non-printable values, values with glob characters, etc), it gets a bit more interesting.
To write to a file, given an array of arguments:
printf '%s\0' "${arguments[#]}" >file
...replace with "argument one", "argument two", etc. as appropriate.
To read from that file and use its contents (in bash, ksh93, or another recent shell with arrays):
declare -a args=()
while IFS='' read -r -d '' item; do
args+=( "$item" )
done <file
run_your_command "${args[#]}"
To read from that file and use its contents (in a shell without arrays; note that this will overwrite your local command-line argument list, and is thus best done inside of a function, such that you're overwriting the function's arguments and not the global list):
set --
while IFS='' read -r -d '' item; do
set -- "$#" "$item"
done <file
run_your_command "$#"
Note that -d (allowing a different end-of-line delimiter to be used) is a non-POSIX extension, and a shell without arrays may also not support it. Should that be the case, you may need to use a non-shell language to transform the NUL-delimited content into an eval-safe form:
quoted_list() {
## Works with either Python 2.x or 3.x
python -c '
import sys, pipes, shlex
quote = pipes.quote if hasattr(pipes, "quote") else shlex.quote
print(" ".join([quote(s) for s in sys.stdin.read().split("\0")][:-1]))
'
}
eval "set -- $(quoted_list <file)"
run_your_command "$#"
If all you need to do is to turn file arguments.txt with contents
arg1
arg2
argN
into my_command arg1 arg2 argN then you can simply use xargs:
xargs -a arguments.txt my_command
You can put additional static arguments in the xargs call, like xargs -a arguments.txt my_command staticArg which will call my_command staticArg arg1 arg2 argN
Here's how I pass contents of a file as an argument to a command:
./foo --bar "$(cat ./bar.txt)"
None of the answers seemed to work for me or were too complicated. Luckily, it's not complicated with xargs (Tested on Ubuntu 20.04).
This works with each arg on a separate line in the file as the OP mentions and was what I needed as well.
cat foo.txt | xargs my_command
One thing to note is that it doesn't seem to work with aliased commands.
The accepted answer works if the command accepts multiple args wrapped in a string. In my case using (Neo)Vim it does not and the args are all stuck together.
xargs does it properly and actually gives you separate arguments supplied to the command.
I suggest using:
command $(echo $(tr '\n' ' ' < parameters.cfg))
Simply trim the end-line characters and replace them with spaces, and then push the resulting string as possible separate arguments with echo.
In my bash shell the following worked like a charm:
cat input_file | xargs -I % sh -c 'command1 %; command2 %; command3 %;'
where input_file is
arg1
arg2
arg3
As evident, this allows you to execute multiple commands with each line from input_file, a nice little trick I learned here.
Both solutions work even when lines have spaces:
readarray -t my_args < foo.txt
my_command "${my_args[#]}"
if readarray doesn't work, replace it with mapfile, they're synonyms.
I formerly tried this one below, but had problems when my_command was a script:
xargs -d '\n' -a foo.txt my_command
After editing #Wesley Rice's answer a couple times, I decided my changes were just getting too big to continue changing his answer instead of writing my own. So, I decided I need to write my own!
Read each line of a file in and operate on it line-by-line like this:
#!/bin/bash
input="/path/to/txt/file"
while IFS= read -r line
do
echo "$line"
done < "$input"
This comes directly from author Vivek Gite here: https://www.cyberciti.biz/faq/unix-howto-read-line-by-line-from-file/. He gets the credit!
Syntax: Read file line by line on a Bash Unix & Linux shell:
1. The syntax is as follows for bash, ksh, zsh, and all other shells to read a file line by line
2. while read -r line; do COMMAND; done < input.file
3. The -r option passed to read command prevents backslash escapes from being interpreted.
4. Add IFS= option before read command to prevent leading/trailing whitespace from being trimmed -
5. while IFS= read -r line; do COMMAND_on $line; done < input.file
And now to answer this now-closed question which I also had: Is it possible to `git add` a list of files from a file? - here's my answer:
Note that FILES_STAGED is a variable containing the absolute path to a file which contains a bunch of lines where each line is a relative path to a file I'd like to do git add on. This code snippet is about to become part of the "eRCaGuy_dotfiles/useful_scripts/sync_git_repo_to_build_machine.sh" file in this project, to enable easy syncing of files in development from one PC (ex: a computer I code on) to another (ex: a more powerful computer I build on): https://github.com/ElectricRCAircraftGuy/eRCaGuy_dotfiles.
while IFS= read -r line
do
echo " git add \"$line\""
git add "$line"
done < "$FILES_STAGED"
References:
Where I copied my answer from: https://www.cyberciti.biz/faq/unix-howto-read-line-by-line-from-file/
For loop syntax: https://www.cyberciti.biz/faq/bash-for-loop/
Related:
How to read contents of file line-by-line and do git add on it: Is it possible to `git add` a list of files from a file?