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
In the tail output i have following string...Using grep command how can i search for the string "contentState\":\"STOPPED\". I have to search for the whole string "contentState\":\"STOPPED\" rather
than searching for STOPPED or contentState only.
I tried following command: But it is not working.
grep -e ""contentState\":\"STOPPED\" /opt/logs/out.log | tail -1
{\"eventType\":\"appAction\",\"action\":\"CONTENT_STATE_CHANGE\",\"evt\":{\"contentState\":\"STOPPED\"}}}
To search for "contentState\":\"STOPPED\", you need to (a) put the whole string in single-quotes to protect it, and (b) escape (double) the backslashes. Thus:
grep -e '"contentState\\":\\"STOPPED\\"' /opt/logs/out.log
Without the protection provided the outer single-quotes, the unescaped double-quotes would be subject to the shell's quote removal and grep would never see them.
Example
Consider this test file:
$ cat log
good "contentState\":\"STOPPED\"
bad contentState\":\"STOPPED\"
bad "contentState\":\"STOPPED"
Let's run our command:
$ grep -e '"contentState\\":\\"STOPPED\\"' log
good "contentState\":\"STOPPED\"
As we can see, the good line is returned and only the good line.
I want to match tags in files (with optional brackets) ... easy one would think ... the regex is something like ^\[?MyTag\]?. But ... Grep doesn't like it. None of the lines that would be valid matches are actually matched.
The interesting part is: if I replace the ? with a * (so zero to infinite matches, not zero or one) it matches everything like it should, but really that would mean the feature is broken and I don't believe that.
Any input?
Using grep (GNU grep) 2.22 on Windows.
PS: so grep is like this ...
grep -e "^\[?MyTag\]?" file.txt
and my test file is like this
[MyTag] hello
NotMyTag ugly
[NotMyTag] dumb
MyTag world
which obviously should result in 1st and 4th line showing but shows nothing.
First off, ? is not supported in vanilla grep, so you need to use the -E flag to enable extended regex. You can easily verify this by running grep '?' <<< 'a' and grep -E '?' <<< 'a'. Only the latter will match. -e just explicitly indicates what your regex is. It is not the same as -E.
Your initial command works fine if you change the -e to upper case:
grep -E '^\[?MyTag\]?'
Example:
$ grep -E '^\[?MyTag\]?' <<< '[MyTag] hello
> NotMyTag ugly
> [NotMyTag] dumb
> MyTag world'
Output:
[MyTag] hello
MyTag world
Credit goes to the answers of this question on SuperUser.
? is not part of the basic regular expressions, which grep supports. GNU grep supports them as an extension, but you have to escape them:
$ grep '^\[\?MyTag\]\?' file.txt
[MyTag] hello
MyTag world
Or, as pointed out, use grep -E to enable extended regular expressions.
For GNU grep, the only difference between grep and grep -E, i.e., using basic and extended regular expressions, is what you have to escape and what not.
Basic regular expressions
Capture groups and quantifying have to be escaped: \( \) and \{ \}
Zero or one (?), one or more (+) and alternation (|) are not part of BRE, but supported by GNU grep as an extension (but need to be escaped: \? \+ \|)
Extended regular expressions
Capture groups and quantifying don't have to be escaped: ( ) and { }
?, + and | are supported and don't need be be escaped
I have a very large file 100Mb+ where all the content is on one line.
I wish to find a pattern in that file and a number of characters around that pattern.
For example I would like to call a command like the one below but where -A and -B are number of bytes not lines:
cat very_large_file | grep -A 100 -B 100 somepattern
So for a file containing content like this:
1234567890abcdefghijklmnopqrstuvwxyz
With a pattern of
890abc
and a before size of -B 3
and an after size of -A 3
I want it to return:
567890abcdef
Any tips would be great.
Many thanks.
You could try the -o option:
-o, --only-matching
Show only the part of a matching line that matches PATTERN.
and use a regular expression to match your pattern and the 3 preceding/following characters i.e.
grep -o -P ".{3}pattern.{3}" very_large_file
In the example you gave, it would be
echo "1234567890abcdefghijklmnopqrstuvwxyz" > tmp.txt
grep -o -P ".{3}890abc.{3}" tmp.txt
Another one with sed (you may need it on systems where GNU grep is not available):
sed -n '
s/.*\(...890abc...\).*/\1/p
' infile
Best way I can think of doing this is with a tiny Perl script.
#!/usr/bin/perl
$pattern = $ARGV[0];
$before = $ARGV[1];
$after = $ARGV[2];
while(<>) {
print $& if( /.{$before}$pattern.{$after}/ );
}
You would then execute it thusly:
cat very_large_file | ./myPerlScript.pl 890abc 3 3
EDIT: Dang, Paolo's solution is much easier. Oh well, viva la Perl!
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.