I want to grep line like this
12121 \tab something
However, grep don't recognize \t, someone in stackoverflow says we can use -P, but it's hard for me the remember, is there more obvious way?
This is how you can grep for it
grep -P '\t' *
OR
grep -P 'A\tB' "File.tsv"
If you really don't want to use -P
You can use [[:space:]] to represent a tab or whitespace in grep, hope it's clearer.
Related
I can't see what I'm missing in my grep command, can you?
http://regexr.com/5shri
echo "2021-05-09 15:38:56.888 T:1899877296 NOTICE: VideoPlayer::OpenFile:plugin://plugin.video.arteplussept/play/SHOW/069083-002-A" | grep -oE "\w+(?=\/play)/g" -
Expect: arteplussept
You need to
Use the PCRE regex engine, with -P option, not -E (which stands for POSIX ERE)
Remove /g, grep -o extracts all matches and there is no need to "embed" this modifier into the pattern
There is no need to escape /
So, you can just use
grep -oP '\w+(?=/play)'
I want to print the filename if only ALL the matches are present... on different lines
grep -l -w '10B\|01A\|gencode' */$a*filename.vcf
this prints out the filename, but not only if ALL three matches are present.
Would you consider to try awk? awk may solve it in following method,
awk '/10B/&&/01A/&&/gencode/{print FILENAME}' */$a*filename.vcf
try following, just edited your solution a bit.
grep -l '10B.*01A.*gencode' Input_file
With grep and its -P (Perl-Compatibility) option and positive lookahead regex (?=(regex)), to match patterns if in any order.
grep -lwP '(?=.*?10B)(?=.*?01A)(?=.*?gencode)' /path/to/infile
grep -l 'pattern1' files ... | xargs grep -l 'pattern2' | xargs grep -l 'pattern3'
From the grep manual:
-l, --files-with-matches
Suppress normal output; instead print the name of each input file from which output would normally have been printed. The scanning will stop on the first match. (-l is specified by POSIX.)
If I use a grep command like this ls | grep '^[-[:alnum:]\._]+$' to match filenames, it outputs no result but when the command changes to ls | grep '^[-[:alnum:]\._]*$', it works correctly. What's going on?
grep uses "basic" regexes by default, where + is a normal character. You need \+ to match 1 or more things (or use grep -E).
echo $'one\ntwo\nthree' | grep -F -v $(echo three$'\n'one)
Output should in theory be the string two
I've read that the -F command lets grep interpret each line as a list connected by 'or' qualifier.
Only mistake is some missing double-quotes:
echo $'one\ntwo\nthree' | grep -F -v "$(echo three$'\n'one)"
Also, keep in mind that this will also filter out "threesome", "someone", etc...
(#etan-reisner points out that running set -x before the original and the fixed command can be used to observe the difference the double-quotes make here, and, more generally, is a useful way to debug bash commands.)
I would like to grep digits inside a set of parentheses after a match.
Given foo.txt below,
foo: "32.1" bar: "42.0" misc: "52.3"
I want to extract the number after bar, 42.0.
The following line will match, but I'd like to extract the digit. I guess I could pipe the output back into grep looking for \d+.\d+, but is there a better way?
grep -o -P 'bar: "\d+.\d+"' foo.txt
One way is to use look ahead and look-behind assertions:
grep -o -P '(?<=bar: ")\d+.\d+(?=")'
Another is to use sed:
sed -e 's/.*bar: "\([[:digit:]]\+.[[:digit:]]\+\)".*/\1/'
You could use the below grep also,
$ echo 'foo: "32.1" bar: "42.0" misc: "52.3"' | grep -oP 'bar:\s+"\K[^"]*(?=")'
42.0