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Grepping exact pattern and deleting the rows

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# 8  
Old 05-02-2016
Quote:
Originally Posted by am24
.
.
. the solution for teh second question i asked will be
grep -v '^..513... ' sample > op6
Yes

Quote:
Thanks. I have tried the code. But getting an error message as below.
.
.
.
Quote:
Originally Posted by Don Cragun
If you are using a Solaris/SunOS system, use /usr/xpg4/bin/awk or nawk instead of awk.
This User Gave Thanks to RudiC For This Post:
am24
# 9  
Old 05-02-2016
Hello am24,

Seems to be you are using Sun/Solaris box, usually nawk should work on it not sure why you are getting an error. Please on a Solaris/SunOS system, change awkto /usr/xpg4/bin/awk , /usr/xpg6/bin/awk. Kindly do let me know how it goes then.

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
am24
# 10  
Old 05-02-2016
Yes, as regular expression ^ means the start and $ the end of the line.
If it really matters for position better go with Rudi's example as it is more straight and easy.
This User Gave Thanks to zaxxon For This Post:
am24
# 11  
Old 05-02-2016
Thanks everyone.

Yes i am working on Solaris.I have tried using nawk, /usr/xpg6/bin/awk and /usr/xpg6/bin/awk. But same error again.

Code:
/usr/xpg6/bin/awk '(substr($1,3,3) != 51[35])'  sample > op9
/usr/xpg6/bin/awk: Command not found

Code:
/usr/xpg4/bin/awk '(substr($1,3,3) != 51[35])' sample > op10
/usr/xpg4/bin/awk: syntax error  Context is:
>>>     (substr($1,3,3) != 51[  <<<

I have tried below by keeping only '513' . Now the below both worked.

/usr/xpg4/bin/awk '(substr($1,3,3) != 513)' sample > op11

nawk '(substr($1,3,3) != 513)' sample > op12


May be some wrong in mentioning two patterns in the code. I am not sure.

Regards,
am24
# 12  
Old 05-02-2016
Hello am24,

Could you please use following and let me know if this helps.
Code:
/usr/xpg4/bin/awk '(substr($1,3,3) != 513 && substr($1,3,3) != 515)' sample > op10

Thanks,
R. Singh
# 13  
Old 05-02-2016
You need to double quote the bracket expression "51[35]" - it is a regex, not a number, and a string comparison, not an integer comparison.
# 14  
Old 05-02-2016
Hi Ravinder,

The below code worked.
/usr/xpg4/bin/awk '(substr($1,3,3) != 513 && substr($1,3,3) != 515)' sample > op10

Hi Rudi,

The below did not work. The output 'op16' same as the input 'sample'

nawk '(substr($1,3,3) != "51[35]")' sample > op16

Still trying on it.

Regards,
am24
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