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Top Forums Shell Programming and Scripting How to use if/else if with substr? Post 302895962 by Akshay Hegde on Thursday 3rd of April 2014 01:08:46 PM
Old 04-03-2014
Quote:
Originally Posted by newbie2010
I have a command like this:

Code:
listdb ID923 -l |gawk '{if (substr($0,37,1)==1 && NR == 3)print "YES" else if (substr ($0,37,1)==0 && NR == 3)  print "NO"}'

This syntax doesn't work. But I was able to get this to work:

Code:
listdb ID923 -l |gawk '{if (substr($0,37,1)==1 && NR == 3)print "YES"}'

What is the error I am making?
Any help is appreciated; I am relatively new to awk
Code:
listdb ID923 -l |gawk '{if (substr($0,37,1)==1 && NR == 3)print "YES"; else if (substr ($0,37,1)==0 && NR == 3)  print "NO"}'

Did you miss semicolon ?
 

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Perl::Critic::Policy::BuiltinFunctions::ProhibitLvalueSuUser(Contributed Perl DocPerl::Critic::Policy::BuiltinFunctions::ProhibitLvalueSubstr(3pm)

NAME
Perl::Critic::Policy::BuiltinFunctions::ProhibitLvalueSubstr - Use 4-argument "substr" instead of writing "substr($foo, 2, 6) = $bar". AFFILIATION
This Policy is part of the core Perl::Critic distribution. DESCRIPTION
Conway discourages the use of "substr()" as an lvalue, instead recommending that the 4-argument version of "substr()" be used instead. substr($something, 1, 2) = $newvalue; # not ok substr($something, 1, 2, $newvalue); # ok The four-argument form of "substr()" was introduced in Perl 5.005. This policy does not report violations on code which explicitly specifies an earlier version of Perl (e.g. "use 5.004;"). CONFIGURATION
This Policy is not configurable except for the standard options. SEE ALSO
"substr" in perlfunc (or "perldoc -f substr"). "4th argument to substr" in perl5005delta AUTHOR
Graham TerMarsch <graham@howlingfrog.com> COPYRIGHT
Copyright (c) 2005-2011 Graham TerMarsch. All rights reserved. This program is free software; you can redistribute it and/or modify it under the same terms as Perl itself. perl v5.14.2 2012-06-07 Perl::Critic::Policy::BuiltinFunctions::ProhibitLvalueSubstr(3pm)
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