Bash: How to remove the last character of a string?


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# 1  
Question Bash: How to remove the last character of a string?

In bash, how can one remove the last character of a string? In perl, the chop function would remove the last character. However, I do not know how to do the same job in bash.

Many thanks in advance.
# 2  
There's a nice trimString() function in blib (the bash library link-> //bit.ly/i7tZWH) that does this. Of course it's still under development so your mileage may vary.

Essentially you can use something like... ${1:0:$((${#1}-1))}.

Note the subshell call may no be necessary...

Cheers,
Mikel King
# 3  
No math needed:

Code:
foo="hello world"
echo ${foo%?}
hello worl

This User Gave Thanks to agama For This Post:
# 4  
You can pipe the line into awk like this...

script.sh | awk -F "until: " '{print $2}'

It's crude but works. I'm sure there are a dozen other ways but that's the first that comes to mind.

Cheers,
Mikel
# 5  
carldamon

general use sed for only search and destroy missions. So I would not want to steer you in the wrong direction. But I believe you can through multiple pipings, or in combination with cut. I just find awk to be my go to tool for these sorts of manipulations.

Cheers,
m
# 6  
Hello carldamon,

You seem to have barged into my thread by posting a question that is not related to my original post.

Remove your post from this thread, and create a new thread of your own, please.

LessNux (OP)

----------

Update:

Hello mikelking,

The question that carldamon posted in this thread has nothing to do with my original post. Therefore, your replies to carldamon have nothing to do with my original post. So, remove your replies to carldamon, please. However, keep your reply to my original post. Thanks.

LessNux (OP)

Last edited by LessNux; 02-18-2012 at 01:11 AM..
# 7  
We can achieve that easily with traditional POSIX filters .

Quote:
echo $foo | cut -c 1-$(expr `echo "$foo" | wc -c` - 2)
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