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How to pass shell variables to awk's pattern?

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# 1  
Old 09-02-2010
How to pass shell variables to awk's pattern?
How would I get folders owned by specific users.. I want to pass users as a shell variable to awk.

Code:
 
drwxr-x--x   3 user1  allusers     512 Oct 14  2006 946157019/
drwxr-x--x   3 user2  allusers     512 Mar  9  2008 94825883/
drwxr-x--x   3 user3  allusers     512 Mar  9  2008 948390501/

For some reason when I pass search expression as argument, its not working. But when I replace $var1 with hardcoded string it works..

Code:
 
$>ls -lt | /usr/xpg4/bin/awk -v var1='user1|user2' '$3 ~ /"'$var1'"/'

How do we pass pattern as argument to awk?
Same approach works if its awk's statement.
# 2  
Old 09-02-2010
Code:
ls -lt | /usr/xpg4/bin/awk -v var1='user1|user2' '$3 ~ var1'

# 3  
Old 09-02-2010
That works,, why this version of syntax does not use !!?

Quote:
'$3 ~ /var1/'
instead it uses
Quote:
'$3 ~ var1'
When I replace var1 with its value without // wrapping it does not work..
# 4  
Old 09-02-2010
Hi.

From the awk man page:

Quote:
/re/ is a constant regular expression
Here, the regular expression would be re (the literal string re), not the value of the variable re.
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