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for loop substitution error

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# 1  
Old 09-20-2011
for loop substitution error
PHP Code:
for i in ${ls -al |grep prod | awk '{ print $NF }'}
do
$i,  bootinfo -s $i
done 
I am trying to get $i and the numeric value from bootinfo -s $i, but getting a substitution error.

Please advise.

Thank you
# 2  
Old 09-20-2011
I believe you want parenthesis, not curly braces around your commandline:
Code:
$(ls -al |grep prod | awk '{ print $NF }')

Last edited by Franklin52; 09-21-2011 at 03:30 AM.. Reason: Please use code tags, thank you
This User Gave Thanks to gary_w For This Post:
Daniel Gate
# 3  
Old 09-20-2011
Code:
for i in $( ls -al |grep prod | awk '{ print $NF }' )
...
done

--ahamed
This User Gave Thanks to ahamed101 For This Post:
Daniel Gate
# 4  
Old 09-20-2011
Ahhh! Yes, it should have been ( )
PHP Code:
$i,  bootinfo -s $i 
gives me permission errors due to $i

PHP Code:
echo $i,  bootinfo -s $i 
doesn't give me the numeric value.

How do you put those two values next to each other?

Please advise.
# 5  
Old 09-20-2011
Perhaps:
Code:
printf "%s " "$i"
bootinfo -s "$i"

?

Also, is the ls, grep and awk necessary?
Code:
for i in *prod* ; do

This User Gave Thanks to Scott For This Post:
Daniel Gate
# 6  
Old 09-20-2011
YES, you are right!!!! Thank you so much
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