sed regex to match ['', 'WR' or 'RN'] + 2-4 digits

Posted by Karl on Stack Overflow See other posts from Stack Overflow or by Karl
Published on 2011-01-14T00:35:23Z Indexed on 2011/01/14 0:53 UTC
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regex

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unix

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Hi

I'm trying to do some conditional text processing on Unix and struggling with the syntax. I want to acheive

Find the first 2, 3 or 4 digits in the string
if 2 characters before the found digits are 'WR' (could also be lower case)
    Variable = the string we've found (e.g. WR1234)
    Type = "work request"
else
    if 2 characters before the found digits are 'RN' (could also be lower case)
      Variable = the string we've found (e.g. RN1234)
      Type = "release note"
    else
      Variable = "WR" + the string we've found (Prepend 'WR' to the digits)
      Type = "Work request"
    fi
fi

I'm doing this in a Bash shell on Red Hat Enterprise Linux Server release 5.5 (Tikanga)

Thanks in advance, Karl

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