Shell Programming and Scripting

Can somebody please help me to remove the last character of a string.??

I have a string variable, in which I dynamically put values in a for loop.I want to remove the last character from the string.

But, the problem is I will not know which character can come in the string (Its inside for loop). So I tried using substr. substr does not work here.

Is there any alternative to do this?? Here my code goes.

....
for files in $rawSourceFiles
loop
MID_INFILE=`cat “${files}” | grep “Message-Id:"`
len=`expr “${MID_INFILE}” : ‘.*'`
len=`expr $len - 1`
trimmedMID_INFILE=`expr substr “${MID_INFILE}” 1 ${len}`

.....
...
done

But, it does not seem working. It throws an error “expr: syntax error” ..

What all I want here is in trimmedMID_INFILE should be the string MID_INFILE with last character trimmed. (last char could be anything like "\n", "/", etc.)


Can anybody please enlighten me?? I would be so grateful. Thanks in advance.

Considering the ''expr' error that you are getting is from the code where you are trying to 'trim', than use the following:

Code:

trimmedMID_INFILE=`echo $MID_INFILE | cut -b 1-$len`

- MID_INFILE is you string.
- cut -b (cut by byte)
- 1-$len (from 1st byte till $len - considering $len is already length - 1)

Thanks Hemangjan.. !!! Thanks a lot.. It worked.. Yay...!!!



One more question,

if [ $a = $b ]
then
echo SUCCESS
fi

If I want to case-insensitive-compare these two variables, what should I use?/

Sorry to ask these dumb questions. I'm a newbie to SHELL SCRIPTING..

:-(

This link may help
https://www.unix.com/shell-programmin...ve-checks.html