I have a log-file. Most of the lines in it ends with some sub-string (for example it 0.02). I can filter that lines with pattern 0\.02$. But how I can filter rest lines, other than ending with 0.02?
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Loom
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do you use some tool? or some programming language? there could be easier way. – Kent Jul 24 '13 at 14:13
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1Wait, do you mean get lines that DON'T end with `0.02` – MDEV Jul 24 '13 at 14:14
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3`grep -v '0\.02$' my.log` – RichieHindle Jul 24 '13 at 14:15
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Why not just invert the search? Match against the pattern but reject those lines instead of allowing them? (See, for example, `grep -v`) – Platinum Azure Jul 24 '13 at 14:17
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@SmokeyPHP Yes, you are right. Looks like my poor English obfuscated my question :) I'l try fix it. – Loom Jul 24 '13 at 14:28
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2sed has the `d` command which removes matching lines. So something like this `sed '/0\.02$/d' logfile` might work for you. – doubleDown Jul 24 '13 at 14:30
3 Answers
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Find end-of-line, then look back to check if your string is not there:
.(?<!0\.02)$
Jongware
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I'm not sure whether this would be a problem (or a benefit) in the asker's scenario, but having the leading dot means this pattern would not match an empty line, which technically meets the requirement of not ending with "0.02". – Wiseguy Jul 24 '13 at 14:31
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Some GREP dialects don't like to look for just-a-location, hence the period. – Jongware Jul 24 '13 at 15:23
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@Jongware, your statement about some grep dialects not liking to look for just-a-location is true about the regex searching in NetBeans 7.4. Your solution worked there. – idclaar Mar 18 '14 at 23:03
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@idclaar: I know this to be true for InDesign and for TextPad (Windows). TextWrangler (Mac), on the other hand, has no problem using only `^` or `$` in find-and-replaces. – Jongware Mar 18 '14 at 23:11
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If you're using grep then grep has -v switch just for that:
egrep -v "(^|[^0-9])0\.02$" my.log
anubhava
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