Variable passed as argument


 
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# 1  
Old 03-03-2013
Variable passed as argument

I have a script.

Code:
#!/bin/sh

cur_$1_modify_time=Hello
echo "cur_$1_modify_time"

When I run like

sh /root/script1 jj

I expect value "Hello" being assigned to variable "cur_jj_modify_time" and output being "Hello" ie echoing $cur_jj_modify_time

But the output comes as
Code:
# sh /root/script1 jj
/root/script1: line 3: cur_jj_modify_time=Hello: command not found
cur_jj_modify_time

Please help.
# 2  
Old 03-03-2013
Code:
#!/bin/sh

eval cur_$1_modify_time=Hello
echo "cur_$1_modify_time"

https://www.unix.com/man-page/posix/1posix/eval/
This User Gave Thanks to anbu23 For This Post:
# 3  
Old 03-03-2013
anbu23's answer does not produce the desired output, i.e. hello.

Try this:
Code:
#!/bin/bash

eval cur_$1_modify_time=Hello
eval fpm=$(echo \$cur_$1_modify_time)
echo $fpm

This User Gave Thanks to fpmurphy For This Post:
# 4  
Old 03-03-2013
fpmurphy is correct. Your code worked perfectly.
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