I have a file named testfile.It contains some information like
methun:x:500:500:comment:/home/methun:bin/bash
salahuddin:x:501:500:comment:/home/methun:bin/bash
Now implemented a following shell program:
echo "Enter a Name:"
read username
users='cat /mypractice/myfiles/testfile | awk -F ':' '{print $1}''
for user in $users
do
if [ "$user" == "$username" ]; then
echo "Name found and Enter a new name to change."
read newUsername
#need code to change text on my file --->testfile
fi
done
Now suppose I need to change methun to Moin. Comment to newcomment. I used
sed -i 's/"$user"/"$newuser"/g' /mypractice/myfiles/testfile
But it not working here. I test with it in my testfile singly it change and replace all.But i need to change only that position i want .
I also tried with usermod but it will not works here..
Can anyone give me the solution or correct my code...Thanks
You are using g flag in your sed command which means global substitution (will change all occurrences). Also, the variables although quoted are wrapped inside single quotes and hence are not interpolated.
Try this:
sed -i "s/^$user/$newuser/" /mypractice/myfiles/testfile
I have placed a ^ anchor in the substitution part which means only substitute if the word is at the beginning of the line. This protects you from making a change if the name of the user is not at the start but somewhere in the middle.
Related
I need to extract path from a string. I found examples in another post, but missing additional steps.
I have a string as below:
title="test test good dskgkdh hdfyr /rlsmodules/svnrepo/SOURCE/CBL/MQ/BASELINE/MQO000.CBL kdlkfg nsfgf trhrnrt"
cobsrc=$(awk '{match($0,/\/[^"]*/,a);print a[0]}' <<< $title)
echo $cobsrc
Output is
/rlsmodules/svnrepo/SOURCE/CBL/MQ/BASELINE/MQO000.CBL kdlkfg nsfgf trhrnrt
I need only
/rlsmodules/svnrepo/SOURCE/CBL/MQ/BASELINE/MQO000.CBL
What modification is required?
An existing post on similar query:
how to extract path from string in shell script
Four solutions, in order of my own preference.
First option would be simple parameter expansion, in two steps:
$ title="/${title#*/}"
$ title="${title%% *}"
$ echo "$title"
/rlsmodules/svnrepo/SOURCE/CBL/MQ/BASELINE/MQO000.CBL
The first line removes everything up to the first slash (while prepending a slash to replace the one that's stripped", the second line removes everything from the first bit of whitespace that remains.
Or, if you prefer, use a regex:
$ [[ $title =~ ^[^/]*(/[^ ]+)\ ]]
$ echo ${BASH_REMATCH[1]}
/rlsmodules/svnrepo/SOURCE/CBL/MQ/BASELINE/MQO000.CBL
The regex translates as:
null at the beginning of the line,
a run of zero or more non-slashes,
an atom:
a slash followed by non-space characters
a space, to end the previous atom.
The $BASH_REMATCH array contains the content of the bracketed atom.
Next option might be grep -o:
$ grep -o '/[^ ]*' <<<"$title"
(Result redacted -- you know what it'll be.)
You could of course assign this output to a variable using command substitution, which you already know about.
Last option is another external tool...
$ sed 's:^[^/]*::;s/ .*//' <<<"$title"
This is the same functionality as is handled by the parameter expansion (at the top of the answer) only in a sed script, which requires a call to an external program. Included only for pedantry. :)
Could you please try following.
echo "$title" | awk 'match($0,/\/.*\/[^ ]*/){print substr($0,RSTART,RLENGTH)}'
Output will be as follows.
/rlsmodules/svnrepo/SOURCE/CBL/MQ/BASELINE/MQO000.CBL
Solution 2nd: Considering that your variable don't have space in between its value then following may help you too.
echo "$title" | awk '{sub(/[^/]* /,"");sub(/ .*/,"")} 1'
I have a for loop that takes a CIDR that contains an illegal character for file names ("/").
part of the command is to save the output with the name of the CIDR.
for i in $(more subnets.lst);do shodan download $i-shodan net:$i;done
the download argument is followed by the result of the more command, which are the CIDR (192.168.21.0/24).
is there a way in bash to rename a variable while the loop is running ?
I remember doing this years back in batch files by subtracting from the str length, but that won't help me as I just need to replace the "/" with a "-"(or any other compliant char.
You can do this with using bash Parameter Expansion:
$ echo $var
192.168.21.0/24
$ echo "${var//\//-}"
192.168.21.0-24
So in your command, just use "${i//\//-}" whenever needed without changing the original value of $i. If you want to set variable i to the new value:
i="${i//\//-}"
On a side note, use while loop to read lines from a file, not cat, more or brothers, like:
while IFS= read -r line; do ....; done
I worked it out for myself:
for i in $(more subnets-test);do shodan download $(echo $i | tr "/" "-") net:$i;done
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
How can I use grep to find an exact word inside of a file entered by the user as string?
For example I need to select the word I want to find and the file I want to find it in. I've been told I am really close but something is not working as it should be. I'm using bash shell under Linux.
Here's what I've done so far:
#!/bin/bash
echo "Find the file you want to search the word in?"
read filename
echo "Enter the word you want to find."
read word1
grep $word1 $filename
How can I use grep to find an exact word inside of a file entered by
the user as string.
Try using -F option. Type man grep on shell for more details.
grep -F "$word1" "$filename"
It's recommended to enclose search string and file name with quotes to avoid unexpected output because of white spaces.
Not sure why you have fi on the last line. It's not needed.
Try:
grep -R WORD ./ to search the entire current directory, or grep WORD ./path/to/file.ext to search inside a specific file.
#!/bin/bash/
echo "Find the file you want to search the word in?"
read filename
echo "Enter the word you want to find."
read word
cat $filename | grep "$word"
This works fine to find the exact word match in a file.
If you want an exact word match within lines use
grep "\b$word\b"
I would like to copy the contents of a variable (here called var) into a file.
The name of the file is stored in another variable destfile.
I'm having problems doing this. Here's what I've tried:
cp $var $destfile
I've also tried the same thing with the dd command... Obviously the shell thought that $var was referring to a directory and so told me that the directory could not be found.
How do I get around this?
Use the echo command:
var="text to append";
destdir=/some/directory/path/filename
if [ -f "$destdir" ]
then
echo "$var" > "$destdir"
fi
The if tests that $destdir represents a file.
The > appends the text after truncating the file. If you only want to append the text in $var to the file existing contents, then use >> instead:
echo "$var" >> "$destdir"
The cp command is used for copying files (to files), not for writing text to a file.
echo has the problem that if var contains something like -e, it will be interpreted as a flag. Another option is printf, but printf "$var" > "$destdir" will expand any escaped characters in the variable, so if the variable contains backslashes the file contents won't match. However, because printf only interprets backslashes as escapes in the format string, you can use the %s format specifier to store the exact variable contents to the destination file:
printf "%s" "$var" > "$destdir"
None of the answers above work if your variable:
starts with -e
starts with -n
starts with -E
contains a \ followed by an n
should not have an extra newline appended after it
and so they cannot be relied upon for arbitrary string contents.
In bash, you can use "here strings" as:
cat <<< "$var" > "$destdir"
As noted in the comment by Ash below, #Trebawa's answer (formulated in the same room as mine!) using printf is a better approach than cat.
All of the above work, but also have to work around a problem (escapes and special characters) that doesn't need to occur in the first place: Special characters when the variable is expanded by the shell. Just don't do that (variable expansion) in the first place. Use the variable directly, without expansion.
Also, if your variable contains a secret and you want to copy that secret into a file, you might want to not have expansion in the command line as tracing/command echo of the shell commands might reveal the secret. Means, all answers which use $var in the command line may have a potential security risk by exposing the variable contents to tracing and logging of the shell.
For variables that are already exported, use this:
printenv var >file
That means, in case of the OP question:
printenv var >"$destfile"
Note: variable names are case sensitive.
Warning: It is not a good idea to export a variable just for the sake of printing it with printenv. If you have a non-exported script variable that contains a secret, exporting it will expose it to all future child processes (unless unexported, for example using export -n).
If I understood you right, you want to copy $var in a file (if it's a string).
echo $var > $destdir
When you say "copy the contents of a variable", does that variable contain a file name, or does it contain a name of a file?
I'm assuming by your question that $var contains the contents you want to copy into the file:
$ echo "$var" > "$destdir"
This will echo the value of $var into a file called $destdir. Note the quotes. Very important to have "$var" enclosed in quotes. Also for "$destdir" if there's a space in the name. To append it:
$ echo "$var" >> "$destdir"
you may need to edit a conf file in a build process:
echo "db-url-host=$POSTGRESQL_HOST" >> my-service.conf
You can test this solution with running before export POSTGRESQL_HOST="localhost"