take input from a variable as pattern to awk

Applications Programming
1

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

Try

log=`awk -vpat1="$c" '$0 ~ pat1' in.txt`;
3

From where you want to take your contents? From the shell?

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

or from awk itself?

4

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

Oh, I see. Just remove the backslash before the $c variable:

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

and it will do the work.

6

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

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 $c�s contents.

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

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

this way awk will not complain any more.

8

Thanks pfylnn, I got it. It was because of newline in $c.