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take input from a variable as pattern to awk

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# 1  
Old 12-13-2011
take input from a variable as pattern to awk
Hi everyone,

Can anyone tell me how to take contents of a variable as a pattern for awk command. Am doing as below, but doesnt get any output:
$c = "Tue Dec";
$log = ` awk '/ \$c /' in.txt`;
print $log;
# 2  
Old 12-13-2011
Try
Code:
log=`awk -vpat1="$c" '$0 ~ pat1' in.txt`;

# 3  
Old 12-13-2011
From where you want to take your contents? From the shell?

Code:
 C="Tue Dec"
 LOG=`awk '/'"$C"'/' in.txt`
 echo $LOG

or from awk itself?
# 4  
Old 12-13-2011
Thanks pfylnn for your reply.

Its a perl script. Actually my code is as below:

system("lastlog -u user > last.txt");
`sed -n '2p' last.txt > last1.txt`;
$c = `cut -c44-62 last1.txt`;
print $c;

$log = ` awk '/\$c/' in.txt`;
print $log;
# 5  
Old 12-13-2011
Oh, I see. Just remove the backslash before the $c variable:

Code:
$log = ` awk '/$c/' in.txt`;

and it will do the work.
# 6  
Old 12-13-2011
hey Pfylnn,

if i reeive backslash it gives me this error:

awk: /Tue
awk: ^ unterminated regexp
sh: line 1: /: is a directory

---------- Post updated at 07:12 AM ---------- Previous update was at 07:08 AM ----------

Hey carloM,

Thanks for ur reply. But the code u suggest gives me this error:

awk: ./OSS20.pl ~ pat1
awk: ^ syntax error
# 7  
Old 12-13-2011
Quote:
Originally Posted by anandrec
hey Pfylnn,

if i reeive backslash it gives me this error:

awk: /Tue
awk: ^ unterminated regexp
sh: line 1: /: is a directory
I guess this is because there is a line break in the $c variable (or maybe somewhere else). Have you checked it? If you just asign directly the value "Tue Dec" to $c it will work. So probably there is something different in $cīs contents.

[EDIT] You can 'chomp' $c before using it to remove the line break:

Code:
$c = `cut -c44-62 last1.txt`;
chomp $c;
$log = ` awk '/$c/' in.txt`;
print $log;

this way awk will not complain any more.
Last edited by pflynn; 12-13-2011 at 08:41 AM..
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