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.)
Related
Team,
I want to grep for a substring container - and then only output that string and not whole line. how can i? I know i can awk on space and pull using $ but want to know how to do in grep?
echo $test_pods_info | grep -F 'test-'
output
test-78ac951e-89a6-4199-87a4-db8a1b8b054f export-9b55f0d5-071d-431-1d2ux0-avexport-xavierisp-sjc4--a4dd85-102 1/1 Running 0 19h
expected output
test-78ac951e-89a6-4199-87a4-db8a1b8b054f
awk is more suitable for this as you want to get first field in a matching line:
awk '/test-/{print $1}' <<< "$taxIncluded"
test-78ac951e-89a6-4199-87a4-db8a1b8b054f
If you really want to use grep then this might be what you're looking for:
grep -o 'test-\S*' <<< "$taxIncluded"
or:
grep -o 'test-[^[:space:]]*' <<< "$taxIncluded"
Try
echo $test_pods_info | grep -o 'test-'
the -o option is:
show[ing] only the part of a line matching PATTERN
according to grep --help. Of course, this will only print test-, so you'll need to rework your regex:
grep -oE '(test).*[[:space:]]\b'
Figured it out..
echo $test_pods_info | grep -o "\test-\w*-\w*\-\w*\-\w*\-\w*"
outoput
test-78ac951e-89a6-4199-87a4-db8a1b8b054f
but i wish there is simple way. like \test-*\
I want the output of the sed file edit to go into my log file name d_selinuxlog.txt. Currently, grep outputs the specified string as well as 3 other strings above and below in the edited file.
#!/bin/bash
{ getenforce;
sed -i s/SELINUX=enforcing/SELINUX=disabled /etc/selinux/config;
grep "SELINUX=*" /etc/selinux/config > /home/neb/scropts/logs/d_selinuxlog.txt;
setenforce 0;
getenforce; }
I want to be seeing just SELINUX=disabled in the log file
All the lines with the lines SELINUX are going to match, even the commented ones, so, you need to omit that ones, and the * from the match.
grep "SELINUX=" /etc/selinux/config | grep -v "#"
This is my output
17:52:07 alvaro#lykan /home/alvaro
$ grep "SELINUX=" /etc/selinux/config | grep -v "#"
SELINUX=disabled
17:52:22 alvaro#lykan /home/alvaro
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).
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
Is it possible to determine the number of times a particular word appears using grep
I tried the "-c" option but this returns the number of matching lines the particular word appears in
For example if I have a file with
some words and matchingWord and matchingWord
and then another matchingWord
running grep on this file for "matchingWord" with the "-c" option will only return 2 ...
note: this is the grep command line utility on a standard unix os
grep -o string file will return all matching occurrences of string. You can then do grep -o string file | wc -l to get the count you're looking for.
I think that using grep -i -o string file | wc -l should give you the correct output, what happens when you do grep -i -o string file on the file?
You can simply count words (-w) with wc program:
> echo "foo foo" | grep -o "foo" | wc -w
> 2