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Alternative command to grep -w option

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# 1  
Old 10-29-2015
Oracle Alternative command to grep -w option
Hi All,

We have few scripts where we are using grep -w option to do exact matching of the pattern. This works fine on most of our servers.
But I have encounter a very old HP-UX System(HP-UX B.11.00) where grep -w option is not available.

This is causing my scripts to fail. I need to change my code to make it work on these old servers.

Can you guys help me with an alternative for grep -w to do exact word matching.

Thanks in Advance

Regards,
Veeresham
# 2  
Old 10-29-2015
Something like this perhaps:

Code:
$ echo "ABC123" | awk '$0 ~ "(^|[ \r\n\t])"P"($|[ \r\n\t])"' P="ABC"
$ echo "123ABC123" | awk '$0 ~ "(^|[ \r\n\t])"P"($|[ \r\n\t])"' P="ABC"
$ echo "ABC 123" | awk '$0 ~ "(^|[ \r\n\t])"P"($|[ \r\n\t])"' P="ABC"
ABC 123

$ echo "123 ABC 123" | awk '$0 ~ "(^|[ \r\n\t])"P"($|[ \r\n\t])"' P="ABC"
123 ABC 123

$ echo "ABC 123" | awk '$0 ~ "(^|[ \r\n\t])"P"($|[ \r\n\t])"' P="ABC"
ABC 123

$

This User Gave Thanks to Corona688 For This Post:
veeresh_15
# 3  
Old 10-29-2015
Try perl:

Code:
[user@host ~]$ cat file
the hello world pattern
the pattern hello world
the line without that word
pattern the hello world
the patternhello world
the hellopattern world
the hellopatternworld
[user@host ~]$
[user@host ~]$
[user@host ~]$ perl -ne '/\bpattern\b/ && print' file
the hello world pattern
the pattern hello world
pattern the hello world
[user@host ~]$

This User Gave Thanks to balajesuri For This Post:
veeresh_15
# 4  
Old 10-30-2015
man grep:
Quote:
-w, --word-regexp
Select only those lines containing matches that form whole words. The test is that the matching substring must either be at the beginning of the line, or
preceded by a non-word constituent character. Similarly, it must be either at the end of the line or followed by a non-word constituent character. Word-
constituent characters are letters, digits, and the underscore.
So Corona688's proposal should be slightly modified like
Code:
echo "ABC,123" | awk '$0 ~ "(^|[^A-Za-z0-9_])"P"($|[^A-Za-z0-9_])"' P="ABC"
ABC,123

---------- Post updated at 10:13 ---------- Previous update was at 10:11 ----------

Does your grep understand extended regexes? Try
Code:
echo "ABC,123" | grep -E "(^|[^A-Za-z0-9_])ABC($|[^A-Za-z0-9_])"
ABC,123

These 3 Users Gave Thanks to RudiC For This Post:
balajesuri Corona688 veeresh_15
# 5  
Old 10-30-2015
Hi RudiC,

Your solution works perfectly for me. Can you please explain me how grep -E "(^|[^A-Za-z0-9_])ABC($|[^A-Za-z0-9_])" is acting like -w option
# 6  
Old 10-30-2015
[^A-Za-z0-9_] replicates the "... non-word constituent character. Word- constituent characters are letters, digits, and the underscore" (cf man grep). This is ORed with begin-of-line ("^") or end-of-line ("$"), respectively. The -E option is necessary to accept the EREs.

---------- Post updated at 12:41 ---------- Previous update was at 12:39 ----------

In non-C locales, that regex may not be sufficient. The [[:alnum:]] class will take care of that, but may not be available on your system.
This User Gave Thanks to RudiC For This Post:
veeresh_15
# 7  
Old 10-30-2015
Thanks a lot RudiC for the solution and explaination!

Regards,
Veeresham
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