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
Related
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 '$3 == "chr1"' file
There is a possibility to search using grep in TextWrangler
I want to find and replace the following word: bauvol, but not bauvolumen.
I tried typing ^bauvol$ 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.
I am trying to grep the output of a command that outputs unknown text and a directory per line. Below is an example of what I mean:
.MHuj.5.. /var/log/messages
The text and directory may be different from time to time or system to system. All I want to do though is be able to grep the directory out and send it to a variable.
I have looked around but cannot figure out how to grep to the end of a word. I know I can start the search phrase looking for a "/", but I don't know how to tell grep to stop at the end of the word, or if it will consider the next "/" a new word or not. The directories listed could change, so I can't assume the same amount of directories will be listed each time. In some cases, there will be multiple lines listed and each will have a directory list in it's output. Thanks for any help you can provide!
If your directory paths does not have spaces then you can do:
$ echo '.MHuj.5.. /var/log/messages' | awk '{print $NF}'
/var/log/messages
It's not clear from a single example whether we can generalize that e.g. the first occurrence of a slash marks the beginning of the data you want to extract. If that holds, try
grep -o '/.*' file
To fetch everything after the last space, try
grep -o '[^ ]*$' file
For more advanced pattern matching and extraction, maybe look at sed, or Awk or Perl or Python.
Your line can be described as:
^\S+\s+(\S+)$
That's assuming whitespace is your delimiter between the random text and the directory. It simply separates the whitespace from the non-whitespace and captures the second part.
Or you might want to look into the word boundary character class: \b.
I know you said to use grep, but I can't help to mention that this is trivially done using awk:
awk '{ print $NF }' input.txt
This is assuming that a whitespace is the delimiter and that the path does not contain any whitespaces.
Is there any way to do the opposite of showing only the matching part of strings in grep (the -o flag), that is, show everything except the part that matches the regex?
That is, the -v flag is not the answer, since that would not show files containing the match at all, but I want to show these lines, but not the part of the line that matches.
EDIT: I wanted to use grep over sed, since it can do "only-matching" matches on multi-line, with:
cat file.xml|grep -Pzo "<starttag>.*?(\n.*?)+.*?</starttag>"
This is a rather unusual requirement, I don't think grep would alternate the strings like that. You can achieve this with sed, though:
sed -n 's/$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: $&"
I am trying the following query, but without success
grep -nr "[[:alnum:]]+\.[[:alnum:]]+\(\)" .
So, according to my logic, a method call would be one or more alphanumeric characters
[[:alnum:]]+
followed by a dot
\.
followed by one or more alphanumeric characters
[[:alnum:]]+
followed by paranthesis (for void return type only)
\(\)
But this query isn't working. How to write such a query?
grep provides several types of regex syntax.
Your pattern is written is the extended syntax and works with -E
extended-regexp has an easier/better syntax, and perl-regexp is, well, quite powerful.
-E, --extended-regexp
-F, --fixed-strings
-G, --basic-regexp (the default)
-P, --perl-regexp
grep -nrE "[[:alnum:]]+\.[[:alnum:]]+\(\)" .
You need to use "\+" instead of "+" otherwise it'll directly match the character "+".