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Full Discussion: if within nawk command
Top Forums UNIX for Dummies Questions & Answers if within nawk command Post 302098009 by kshelluser on Thursday 30th of November 2006 06:00:00 AM
Old 11-30-2006
if within nawk command

I am trying to nawk through a file and if the 24.25.26 charachters in the file are "000" then print a line to a new file in one format, otherwise print a line in a different format.

So far I have the following:

nawk '/^1/ {
if(substr($0,24,3) != 000) printf("%.5d%s\n",NR-1,"0"substr($0,14,13)"XXX"substr($0,2,6)substr($0,32,7)"+"" XXXXXXXX""XXXX "" ""N")
else printf("%.5d%s\n",NR-1,"0"substr($0,14,10)"801""XXXX"substr($0,2,6)substr($0,32,7)"+"" XXXXXXXX""XXXX "" ""N")
fi
}' filename >> newfile

however this is reulting in a syntax error as follows, due to the if statement:

nawk: syntax error at source line 1
context is
/^1/ {if(substr($0,24,3) != 000) printf("%.5d%s\n",NR-1,"0"substr($0,14,
13)"XXXX"substr($0,2,6)substr($0,32,7)"+"" XXXXXXXX""XXXX ""
""N") >>> else <<< printf("%.5d%s\n",NR-1,"0"substr($0,14,10)"801""XXXX"su
bstr($0,2,6)substr($0,32,7)"+"" XXXXXXXX""XXXX "" ""N") f
i}
nawk: illegal statement at source line 1

Can anyone confirm the correct syntax for an if/else statement within a nawk command.

Thanks.
 

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PERLKO(1)						 Perl Programmers Reference Guide						 PERLKO(1)

NAME
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Encode, Encode::KR, encoding, perluniintro, perlunicode AUTHORS
Jarkko Hietaniemi <jhi@iki.fi> XXX <jshin@mailaps.org> perl v5.16.2 2012-10-11 PERLKO(1)
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