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[Solved] Assigning Shell variable

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# 1  
Old 03-10-2014
[Solved] Assigning Shell variable
Hello,

Need a small help to execute below script.

Code:
#!/bin/bash
. new.txt
for no in 3 4
do
echo $((uname_$no))
done

new.txt contains
Code:
uname_1="XXXXXX"
uname_2="YYYYY"
uname_3="ZZZZZ"
.........

I am getting errors if i run the above script. I doubt the syntax is wrong in "echo $((uname_$no))" line. Correct me if i am wrong.
# 2  
Old 03-10-2014
Hello,

Following may help.

Either use the following.
Code:
uname_$no=echo $no

OR try following.
Code:
echo "uname_"$no

Please let us know if you have any queries for same.


Thanks,
R. Singh
# 3  
Old 03-10-2014
That is the wrong syntax. $(( .. )) is used for arithmetic expansion. Try:
Code:
eval echo "\"\$uname_$no\""

If you have control over the source file format, since you are using bash, you could use arrays instead:
Code:
uname[1]="XXXXXX"
uname[2]="YYYYY"
uname[3]="ZZZZZ"

and
Code:
echo "${uname[no]}"

Last edited by Scrutinizer; 03-10-2014 at 06:55 AM..
# 4  
Old 03-10-2014
This will work in modern versions of Bash:
Code:
#!/bin/bash

. ./new.txt

for no in 1 2 3
do
    ref="uname_$no"
    echo ${!ref}
done

nameref provides similar functionality in ksh93
These 2 Users Gave Thanks to fpmurphy For This Post:
prasanna2166 rbatte1
# 5  
Old 03-10-2014
Hello Ravinder,

Thanks for your reply.

Code:
echo "uname_"$no

By using your above code, it just displays 2 or 3 ($no) not the actual one which needs to be displayed.


Expected output:
YYYYY
ZZZZZ

My objective is to display the values present in new.txt file.
# 6  
Old 03-10-2014
Hello,

Could you please try the following.

Code:
for no in  3 4
do
echo "uname_"$no
done < "check_out"


output will be as follow.

Code:
uname_3
uname_4


NOTE: Where check_out is the input file as same provided by you.


Thanks.
R. Singh
# 7  
Old 03-10-2014
Thanks fpmurphy, i tried your code and it is working fine.
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