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Why awk perform differently when using variable?

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Top Forums Shell Programming and Scripting Why awk perform differently when using variable?
# 1  
Old 01-20-2015
Why awk perform differently when using variable?
Hi Gurus,

I hit a hard block in my script.
when using awk command with variable, I got different result.
Please see below:
my test file as below:
Code:
$ cat demofile.txt
filename-yyyy-abcd
filename-xxx-week-pass
filename-xxx-week-run

for testing purpose, I put 3 awk command in one script.
Code:
$cat vartest.ksh
#!/bin/ksh

var=abcd
awk  '$0~/a/{print "ac", $0}!$0~/a/{print "su" ,$0}' a="$var" demofile.txt

echo "using -v "
/usr/xpg4/bin/awk -v a=$var '$0~/a/{print "ac", $0}!$0~/a/{print "su" ,$0}'  demofile.txt

echo "without variable"
awk  '$0~/abcd/{print "ac", $0}!$0~/abcd/{print "su" ,$0}' demofile.txt

the output as below:
Code:
$./vartest.ksh
ac filename-yyyy-abcd
ac filename-xxx-week-pass
ac filename-xxx-week-run
using -v
ac filename-yyyy-abcd
ac filename-xxx-week-pass
ac filename-xxx-week-run
without variable
ac filename-yyyy-abcd
su filename-xxx-week-pass
su filename-xxx-week-run

only 3rd command give me correct result.
my server is
Code:
uname -a
SunOS  5.10 Generic_150400-04 sun4v sparc sun4v

Thanks in advance.
# 2  
Old 01-20-2015
If /abcd/ is a literal string abcd,
then /a/ is a literal string a.
$0~a is the variable a.
This User Gave Thanks to MadeInGermany For This Post:
ken6503
# 3  
Old 01-20-2015
Quote:
Originally Posted by MadeInGermany
If /abcd/ is a literal string abcd,
then /a/ is a literal string a.
$0~a is the variable a.
Thanks for your quick response.

I tried the script
I got 2nd and 3rd result. why 1st one gave me error?

Code:
$ cat vartest.ksh
#!/bin/ksh
var=abcd
awk  '$0~a {print "ac", $0} $0!~a {print "su" ,$0}' a="$var" demofile.txt
echo "using -v "
/usr/xpg4/bin/awk -v a=$var '$0~a{print "ac", $0} $0!~a{print "su" ,$0}'  demofile.txt
echo "without variable"
awk  '$0~/abcd/{print "ac", $0} !$0~/abcd/{print "su" ,$0}' demofile.txt

$ ./vartest.ksh
awk: syntax error near line 1
awk: bailing out near line 1
using -v
ac filename-yyyy-abcd
su filename-xxx-week-pass
su filename-xxx-week-run
without variable
ac filename-yyyy-abcd
su filename-xxx-week-pass
su filename-xxx-week-run

# 4  
Old 01-21-2015
Hello ken6503,

Could you please use change awk to /usr/xpg4/bin/awk or /usr/xpg6/bin/awk on a Solaris/SunOS system. This should fix the error:
Quote:
awk: syntax error near line 1
awk: bailing out near line 1

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
ken6503
# 5  
Old 01-21-2015
Quote:
Originally Posted by RavinderSingh13
Hello ken6503,

Could you please use change awk to /usr/xpg4/bin/awk or /usr/xpg6/bin/awk on a Solaris/SunOS system. This should fix the error:



Thanks,
R. Singh
Hi R.Singh,

you are right. I need to use /usr/xpg4/bin/awk or nawk

thank you very much.
# 6  
Old 01-21-2015
One more thing, you apply the condition twice, one time positive, one time negative.
Consider changing to either
Code:
awk  '$0~a {print "ac", $0; next} {print "su" ,$0}' a="$var" demofile.txt

or
Code:
awk  '{if ($0~a) {print "ac", $0} else {print "su" ,$0}}' a="$var" demofile.txt

# 7  
Old 01-21-2015
Or
Code:
awk  'BEGIN{split("su ac", CM)} {print CM[($0~a)+1], $0}' a="$var" file
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