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I am trying to use egrep on a file like the following:
ABCD(something): Some very good code;
ABCD(somethingElse): Some other very good code;
ABCD(somethingElseElse): Some other very good code;
I want to produce output like the following:
Some very good code;
Some other very good code;
Some other very good code;
I am using the following grep command:
egrep -RIn --color "ABCD(.+):" grep_log_test.txt
Which works but is also outputting the matched pattern. How can I exclude the pattern from the output? I see the -o option but that just prints the patten match only...
As I've mentioned in the comments above, sed is probably the best and easiest tool to use here, as in:
sed 's/ABCD(.\+): *//'
It uses the s command to match ABCD(.\+): * between first set of slashes and replace it with (empty string) between second set of slashes.
Related
I have found on this answer the regex to find a string between two characters. In my case I want to find every pattern between ‘ and ’. Here's the regex :
(?<=‘)(.*?)(?=’)
Indeed, it works when I try it on https://regex101.com/.
The thing is I want to use it with grep but it doesn't work :
grep -E '(?<=‘)(.*?)(?=’)' file
Is there anything missing ?
Those are positive look-ahead and look behind assertions. You need to enable it using PCRE(Perl Compatible Regex) and perhaps its better to get only matching part using -o option in GNU grep:
grep -oP '(?<=‘)(.*?)(?=’)' file
This is a common problem I encounter when using grep. Say the pattern is 'chr1' in a third column of a file, when I do the following:
grep 'chr1' file
How can I avoid getting the results including chr10, chr11, chr13 etc as well?
Thanks!
It seems this works:
grep -w 'chr1' file
Since you're interested in values in specific columns, you're much better off using awk:
awk '$3 == "chr1"' file
Any help would be greatly appreciated. I can read code and figure it out, but I have trouble writing from scratch.
I need help starting a ksh script that would search a file for multiple strings and write each line containing one of those strings to an output file.
If I use the following command:
$ grep "search pattern" file >> output file
...that does what I want it to. But I need to search multiple strings, and write the output in the order listed in the file.
Again... any help would be great! Thank you in advance!
Have a look at the regular expression manuals. You can specify multiple strings in the search expression such as grep "John|Bill"
Man grep will teach you a lot about regular expressions, but there are several online sites where you try them out, such as regex101 and (more colorful) regexr.
Sometimes you need egrep.
egrep "first substring|second substring" file
When you have a lot substrings you can put them in a variable first
findalot="first substring|second substring"
findalot="${findalot}|third substring"
findalot="${findalot}|find me too"
skipsome="notme"
skipsome="${skipsome}|dirty words"
egrep "${findalot}" file | egrep -v "${skipsome}"
Use "-f" in grep .
Write all the strings you want to match in a file ( lets say pattern_file , the list of strings should be one per line)
and use grep like below
grep -f pattern_file file > output_file
There is a possibility to search using grep in TextWrangler
I want to find and replace the following word: bauvol, but not bauvolumen.
I tried typing ^bauvol$ into the search field but that didn't do the trick, it didn't find anything, although the word is clearly there.
I think it's because, in grep, the ^and $signify start and end of line, not a word?!
You want to use \b as word boundaries, as #gromi08 said:
\bbauvol\b
If you want to copy any portion of this word (so you can replace it, modify it, change the case, etc.) it is usually best to wrap it in ( and ) braces so you can reference them in the Replace box:
Find:
(\bbauvol\b)
Replace:
<some_tag>\1</some_tag>
Did you have anything specific you were trying to do with the result once you found it (cut it, duplicate it, etc.)?
Use the -w option of grep (see grep man-page.
This option searches for the expression as a word.
Therefore the command will be:
cat file.txt | grep -w bauvol
And yes, ^ and $ are for start and end of line.
Is there any way to do the opposite of showing only the matching part of strings in grep (the -o flag), that is, show everything except the part that matches the regex?
That is, the -v flag is not the answer, since that would not show files containing the match at all, but I want to show these lines, but not the part of the line that matches.
EDIT: I wanted to use grep over sed, since it can do "only-matching" matches on multi-line, with:
cat file.xml|grep -Pzo "<starttag>.*?(\n.*?)+.*?</starttag>"
This is a rather unusual requirement, I don't think grep would alternate the strings like that. You can achieve this with sed, though:
sed -n 's/$PATTERN//gp' file
EDIT in response to OP's edit:
You can do multiline matching with sed, too, if the file is small enough to load it all into memory:
sed -rn ':r;$!{N;br};s/<starttag>.*?(\n.*?)+.*?<\/starttag>//gp' file.xml
You can do that with a little help from sed:
grep "pattern" input_file | sed 's/pattern//g'
I don't think there is a way in grep.
If you use ack, you could output Perl's special variables $` and $' variables to show everything before and after the match, respectively:
ack string --output="\$`\$'"
Similarly if you wanted to output what did match along with other text, you could use $& which contains the matched string;
ack string --output="Matched: $&"