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10-04-2007
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awk + last occurrence
Hi,
I'm attempting to search, using awk, a pattern range in a file. Something like: >awk '/first bit of text.../,/...last bit of text/' file Is it possible to print only the last (or first) occurrence of the pattern range this way? Thanks for any suggestions. Al |
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10-04-2007
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from the top of my head - not easly and/or awk-natively, but:
Code:
/123/ {a=""}
# last
#/123/,/456/ {a=(a=="") ? $0 : a RS $0}
# first
/123/,/456/ {a=(a=="") ? $0 : a RS $0}
/456/ {exit}
END {
print a
}
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10-06-2007
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Works great, thanks!
Al |
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