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cannot pass a echo output to a variable in bash

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# 1  
Old 07-17-2012
cannot pass a echo output to a variable in bash
Hi,

I have a problem with passing a echo output into a variable in bash

Code:
file='1990.tar'
NAME='echo $file | cut -d '.' -f1';
echo $NAME

the result is
Code:
echo $file | cut -d . -f1

however with this one,
Code:
#!/bin/bash
file='1990.tar'
echo $file | cut -d '.' -f1

the result is what I need:
Code:
1990

How could I pass the right filename 1990 to the variable "NAME", thanks!
# 2  
Old 07-17-2012
Code:
file='1990.tar'
NAME=$(echo "${file}" | cut -d '.' -f1)
echo "${NAME}"

The $( ) notation is preferred.


The problem in your original script was that you used single quotes not backticks to enclose the executable pipeline.
Code:
file='1990.tar'
NAME=`echo $file | cut -d '.' -f1`
echo $NAME

This User Gave Thanks to methyl For This Post:
1988PF
# 3  
Old 07-17-2012
It's your quoting. You probably intended to use back ticks, but this is better.

Code:
NAME=$( echo $file | cut -d '.' -f1 )

You could also do it without having to invoke another process (best):

Code:
NAME="${file%.*}"

The syntax %.* causes the expansion of the variable to be truncated starting with the right most dot (.). Thus abc.def becomes just abc.
This User Gave Thanks to agama For This Post:
1988PF
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