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Count number of character occurence but not from quotation marks

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# 1  
Old 02-03-2012
Count number of character occurence but not from quotation marks
I have the following string:
Code:
31-01-2012, 09:42:37;OK;94727132638;"Mozilla/5.0 (Linux; U; Android 2.2.1)";3G;WAP;

I need a script which is counting the occurrence of semicolons ( ; ) but exclude the ones from the quotation marks.
In the string given as example there are 8 semicolons but the script should return only 6, while 2 semicolons are in quotation marks (they are marked with red color).


Moderator's Comments:
Mod Comment Please use next time code tags for your code and data
# 2  
Old 02-03-2012
Hi calinlicj,

One way. Remove all characters between double-quotes and then count semicolons:
Code:
$ perl -ne 's/"[^"]*"//g; printf qq[%d\n], tr/;/;/' <<<'31-01-2012, 09:42:37;OK;94727132638;"Mozilla/5.0 (Linux; U; Android 2.2.1)";3G;WAP;'
6

Regards,
Birei
# 3  
Old 02-03-2012
Code:
perl -lne 's/\".*?\"//g;print s/;//g;' inputfile

# 4  
Old 02-03-2012
Thank you both, Birei, balajesuri.
Last edited by calinlicj; 02-03-2012 at 11:34 AM..
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