I need some help with a grep command (in the Bash).
In my source files, I want to list all unique parameters of a function. Background: I want to search through all files, to see, which permissions ([perm("abc")] are used.
Example.txt:
if (x) perm("this"); else perm("that");
perm("what");
I'd like to have my grep output:
this
that
what
If I do my grep with this search expression
perm\(\"(.*?)\"\)
I'll get perm("this), perm("that"), etc. but I'd like to have just the permissions: this and that and what.
How can I do that?
Use a look-behind:
$ grep -Po '(?<=perm\(")[^"]*' file
this
that
what
This looks for all the text occurring after perm(" and until another " is found.
Note -P is used to allow this behaviour (it is a Perl regex) and -o to just print the matched item, instead of the whole line.
Here is a gnu awk version (due to multiple characters in RS)
awk -v RS='perm\\("' -F\" 'NR>1 {print $1}' file
this
that
what
Related
I am trying to use grep with the pwd command.
So, if i enter pwd, it shows me something like:
/home/hrq/my-project/
But, for purposes of a script i am making, i need to use it with grep, so it only prints what is after hrq/, so i need to hide my home folder always (the /home/hrq/) excerpt, and show only what is onwards (like, in this case, only my-project).
Is it possible?
I tried something like
pwd | grep -ov 'home', since i saw that the "-v" flag would be equivalent to the NOT operator, and combine it with the "-o" only matching flag. But it didn't work.
Given:
$ pwd
/home/foo/tmp
$ echo "$PWD"
/home/foo/tmp
Depending on what it is you really want to do, either of these is probably what you really should be using rather than trying to use grep:
$ basename "$PWD"
tmp
$ echo "${PWD#/home/foo/}"
tmp
Use grep -Po 'hrq/\K.*', for example:
grep -Po 'hrq/\K.*' <<< '/home/hrq/my-project/'
my-project/
Here, grep uses the following options:
-P : Use Perl regexes.
-o : Print the matches only (1 match per line), not the entire lines.
\K : Cause the regex engine to "keep" everything it had matched prior to the \K and not include it in the match. Specifically, ignore the preceding part of the regex when printing the match.
SEE ALSO:
grep manual
perlre - Perl regular expressions
I want to grep "[calleruid]=aab01b055-89e3-49f3-839e-507bb128d07e&smscresponse"
in Below file
2014-10-15 18:38:32,831 plivo-rest[2781]: INFO: Fetching GET http://*******/outbound_callback.aspx with smscresponse[to]=8912722fsf9&smscresponse[ALegUUID]=5bb516fsd64-546c-11e4-879f-551816a551303677&smscresponse[calluid]=aab01b055-89e3-49f3-839e-507bb128d07e&smscresponse[direction]=outbosund&smscresfdsponse[endreason]=UNALLOCATED_NUMBER&smscresponse[from]=83339995896999&smscresponse[starttime]=0&smscresponse[ALegRequestUUID]=5bb4bafc-546c-11e4-891d-000c29ec6e41&smscresponse[RequestUUID]=5bb4bafc-546c-11e4-891d-000c29ec6e41&smscresponse[callstatus]=completed&smscresponse[endtime]=1413378509&smscresponse[ScheduledHangupId]=5bb4c15a-546c-11e4-891d-000c29ec6e41&smscresponse[event]=missed_call_hangup
I used this command
$ grep -oP '(calluid).*$'
this greps upto end of file
I used this command
$ grep -oP '(calluid).{40}'
it fetches 40 characters but i have 1000's of calleruid's so each have different no.s of characters
So please guide me to grep exact callerid data
Use a lookahead to force the regex engine to do the match upto a specific character or a boundary.
$ grep -oP '\[calluid\][^\]\[]*(?=\[|$)' file
[calluid]=aab01b055-89e3-49f3-839e-507bb128d07e&smscresponse
Here is an gnu awk (due to multiple characters in RS) version:
awk -v RS="[[]calluid[]]=" -F[ 'NR==2 {print $1}' file
aab01b055-89e3-49f3-839e-507bb128d07e&smscresponse
You can also set RS like this: RS="\\\[calluid]="
I am trying to parse items out of a file I have. I cant figure out how to do this with grep
here is the syntax
<FQDN>Compname.dom.domain.com</FQDN>
<FQDN>Compname1.dom.domain.com</FQDN>
<FQDN>Compname2.dom.domain.com</FQDN>
I want to spit out just the bits between the > and the <
can anyone assist?
Thanks
grep can do some text extraction. however not sure if this is what you want:
grep -Po "(?<=>)[^<]*"
test
kent$ echo "<FQDN>Compname.dom.domain.com</FQDN>
dquote>
dquote> <FQDN>Compname1.dom.domain.com</FQDN>
dquote>
dquote> <FQDN>Compname2.dom.domain.com</FQDN>"|grep -Po "(?<=>)[^<]*"
Compname.dom.domain.com
Compname1.dom.domain.com
Compname2.dom.domain.com
Grep isn't what you are looking for.
Try sed with a regular expression : http://unixhelp.ed.ac.uk/CGI/man-cgi?sed
You can do it like you want with grep :
grep -oP '<FQDN>\K[^<]+' FILE
Output:
Compname.dom.domain.com
Compname1.dom.domain.com
Compname2.dom.domain.com
As others have said, grep is not the ideal tool for this. However:
$ echo '<FQDN>Compname.dom.domain.com</FQDN>' | egrep -io '[a-z]+\.[^<]+'
Compname.dom.domain.com
Remember that grep's purpose is to MATCH things. The -o option shows you what it matched. In order to make regex conditions that are not part of the expression that is returned, you'd need to use lookahead or lookbehind, which most command-line grep does not support because it's part of PCRE rather than ERE.
$ echo '<FQDN>Compname.dom.domain.com</FQDN>' | grep -Po '(?<=>)[^<]+'
Compname.dom.domain.com
The -P option will work in most Linux environments, but not in *BSD or OSX or Solaris, etc.
I have a file that possibly contains bad formatting (in this case, the occurrence of the pattern \\backslash). I would like to use grep to return only the line numbers where this occurs (as in, the match was here, go to line # x and fix it).
However, there doesn't seem to be a way to print the line number (grep -n) and not the match or line itself.
I can use another regex to extract the line numbers, but I want to make sure grep cannot do it by itself. grep -no comes closest, I think, but still displays the match.
try:
grep -n "text to find" file.ext | cut -f1 -d:
If you're open to using AWK:
awk '/textstring/ {print FNR}' textfile
In this case, FNR is the line number. AWK is a great tool when you're looking at grep|cut, or any time you're looking to take grep output and manipulate it.
All of these answers require grep to generate the entire matching lines, then pipe it to another program. If your lines are very long, it might be more efficient to use just sed to output the line numbers:
sed -n '/pattern/=' filename
Bash version
lineno=$(grep -n "pattern" filename)
lineno=${lineno%%:*}
I recommend the answers with sed and awk for just getting the line number, rather than using grep to get the entire matching line and then removing that from the output with cut or another tool. For completeness, you can also use Perl:
perl -nE 'say $. if /pattern/' filename
or Ruby:
ruby -ne 'puts $. if /pattern/' filename
using only grep:
grep -n "text to find" file.ext | grep -Po '^[^:]+'
You're going to want the second field after the colon, not the first.
grep -n "text to find" file.txt | cut -f2 -d:
To count the number of lines matched the pattern:
grep -n "Pattern" in_file.ext | wc -l
To extract matched pattern
sed -n '/pattern/p' file.est
To display line numbers on which pattern was matched
grep -n "pattern" file.ext | cut -f1 -d:
Is there a way to make grep output "words" from files that match the search expression?
If I want to find all the instances of, say, "th" in a number of files, I can do:
grep "th" *
but the output will be something like (bold is by me);
some-text-file : the cat sat on the mat
some-other-text-file : the quick brown fox
yet-another-text-file : i hope this explains it thoroughly
What I want it to output, using the same search, is:
the
the
the
this
thoroughly
Is this possible using grep? Or using another combination of tools?
Try grep -o:
grep -oh "\w*th\w*" *
Edit: matching from Phil's comment.
From the docs:
-h, --no-filename
Suppress the prefixing of file names on output. This is the default
when there is only one file (or only standard input) to search.
-o, --only-matching
Print only the matched (non-empty) parts of a matching line,
with each such part on a separate output line.
Cross distribution safe answer (including windows minGW?)
grep -h "[[:alpha:]]*th[[:alpha:]]*" 'filename' | tr ' ' '\n' | grep -h "[[:alpha:]]*th[[:alpha:]]*"
If you're using older versions of grep (like 2.4.2) which do not include the -o option, then use the above. Else use the simpler to maintain version below.
Linux cross distribution safe answer
grep -oh "[[:alpha:]]*th[[:alpha:]]*" 'filename'
To summarize: -oh outputs the regular expression matches to the file content (and not its filename), just like how you would expect a regular expression to work in vim/etc... What word or regular expression you would be searching for then, is up to you! As long as you remain with POSIX and not perl syntax (refer below)
More from the manual for grep
-o Print each match, but only the match, not the entire line.
-h Never print filename headers (i.e. filenames) with output lines.
-w The expression is searched for as a word (as if surrounded by
`[[:<:]]' and `[[:>:]]';
The reason why the original answer does not work for everyone
The usage of \w varies from platform to platform, as it's an extended "perl" syntax. As such, those grep installations that are limited to work with POSIX character classes use [[:alpha:]] and not its perl equivalent of \w. See the Wikipedia page on regular expression for more
Ultimately, the POSIX answer above will be a lot more reliable regardless of platform (being the original) for grep
As for support of grep without -o option, the first grep outputs the relevant lines, the tr splits the spaces to new lines, the final grep filters only for the respective lines.
(PS: I know most platforms by now would have been patched for \w.... but there are always those that lag behind)
Credit for the "-o" workaround from #AdamRosenfield answer
It's more simple than you think. Try this:
egrep -wo 'th.[a-z]*' filename.txt #### (Case Sensitive)
egrep -iwo 'th.[a-z]*' filename.txt ### (Case Insensitive)
Where,
egrep: Grep will work with extended regular expression.
w : Matches only word/words instead of substring.
o : Display only matched pattern instead of whole line.
i : If u want to ignore case sensitivity.
You could translate spaces to newlines and then grep, e.g.:
cat * | tr ' ' '\n' | grep th
Just awk, no need combination of tools.
# awk '{for(i=1;i<=NF;i++){if($i~/^th/){print $i}}}' file
the
the
the
this
thoroughly
grep command for only matching and perl
grep -o -P 'th.*? ' filename
I was unsatisfied with awk's hard to remember syntax but I liked the idea of using one utility to do this.
It seems like ack (or ack-grep if you use Ubuntu) can do this easily:
# ack-grep -ho "\bth.*?\b" *
the
the
the
this
thoroughly
If you omit the -h flag you get:
# ack-grep -o "\bth.*?\b" *
some-other-text-file
1:the
some-text-file
1:the
the
yet-another-text-file
1:this
thoroughly
As a bonus, you can use the --output flag to do this for more complex searches with just about the easiest syntax I've found:
# echo "bug: 1, id: 5, time: 12/27/2010" > test-file
# ack-grep -ho "bug: (\d*), id: (\d*), time: (.*)" --output '$1, $2, $3' test-file
1, 5, 12/27/2010
cat *-text-file | grep -Eio "th[a-z]+"
You can also try pcregrep. There is also a -w option in grep, but in some cases it doesn't work as expected.
From Wikipedia:
cat fruitlist.txt
apple
apples
pineapple
apple-
apple-fruit
fruit-apple
grep -w apple fruitlist.txt
apple
apple-
apple-fruit
fruit-apple
I had a similar problem, looking for grep/pattern regex and the "matched pattern found" as output.
At the end I used egrep (same regex on grep -e or -G didn't give me the same result of egrep) with the option -o
so, I think that could be something similar to (I'm NOT a regex Master) :
egrep -o "the*|this{1}|thoroughly{1}" filename
To search all the words with start with "icon-" the following command works perfect. I am using Ack here which is similar to grep but with better options and nice formatting.
ack -oh --type=html "\w*icon-\w*" | sort | uniq
You could pipe your grep output into Perl like this:
grep "th" * | perl -n -e'while(/(\w*th\w*)/g) {print "$1\n"}'
grep --color -o -E "Begin.{0,}?End" file.txt
? - Match as few as possible until the End
Tested on macos terminal
$ grep -w
Excerpt from grep man page:
-w: Select only those lines containing matches that form whole words. The test is that the matching substring must either be at the beginning of the line, or preceded by a non-word constituent character.
ripgrep
Here are the example using ripgrep:
rg -o "(\w+)?th(\w+)?"
It'll match all words matching th.