My hostname details are as below after showing command of hostname in linux
my-host-test-db-10001.dns.biz.xyz.com
my-host-test2-db-10002.dns.biz.xyz.com
my-host-test3-db-10003.dns.biz.xyz.com
I want to fetch the 3rd string from these above (test/test2/test3). how can I achieve it?
In addition to the simpler solution using cut, you can use more flexible grep:
hostname | grep -Po '^[^-]+-[^-]+-\K[^-]+'
For example:
grep -Po '^[^-]+-[^-]+-\K[^-]+' <<< 'my-host-test2-db-10002.dns.biz.xyz.com'
Output:
test2
Here, GNU 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:
perlre - Perl regular expressions
I am trying to extract the version from a colon delimited list. The value I want is for foo, however there is another value in the list called foo-bar causing both values to return. This is what I am doing:
LIST="foo:1.0.0
foo-bar:1.0.1"
VERSION=$(echo "${LIST}" | grep "\bfoo\b" | cut -s -d':' -f2)
echo -e "VERSION: ${VERSION}"
Output:
VERSION: 1.0.0
1.0.1
NOTE: Sometimes LIST will look like the following, which should result in version being empty (this is expected).
LIST="foo
foo-bar:1.0.1"
You may use a PCRE regex enabled with -P option and use a (?!-) negative lookahead that will fail the match in case there is a - after a whole word foo:
grep -P "\bfoo\b(?!-)"
See online demo
This regex should extract any number and optional dots at the end of each line. If the line ends with a colon, then it won't match.
grep -oE '(([[:digit:]]+[.]*)+)$
(other quote/grep questions are about bash interpretation, this is not)
Apparently grep handles escaped single quotes differently than other escaped regex characters, but I don't understand why.
$ grep --version
grep (GNU grep) 2.25
$ cat data
a']
b']
c']
d]
e\']
$ cat patterns
a']
b\'\]
c'\]
d\']
e\']
$ grep -Ef patterns data
a']
c']
Because c is matched but b isn't, apparently grep does not interpret an escaped single quote \'as a single quote. But as what then?
d isn't matched, so it is not ignored.
e isn't matched, so it is not taken literally
TIA for solving this x-mas mystery! PS. Yes in this case I can use -F for literal matching, but my application requires regex.
\' in GNU tools means "end of string". See http://www.regular-expressions.info/gnu.html:
Additional GNU Extensions
....
The anchor \` (backtick) matches at the very start of the subject string,
while \' (single quote) matches at the very end.
Don't ask me why they introduced that as it seems to be exactly the same as $.
I'll use an example to illustrate my problem. Suppose we have the file name 'file.txt' that contains the following string:
AooYoZooYZoAoooooYZ
I'd like to use grep to find all substrings that begin with 'A' and end with 'YZ' but do not contain 'YZ' in between the 'A' and 'YZ'. The desired output would be:
AooYoZooYZ
AoooooYZ
My best guess is to do the following:
$grep -E -o 'A[^(YZ)]*YZ' file.txt
But the output is only:
AoooooYZ
I'd like the parentheses to hold their meaning for the YZ but I read in the GNU grep manual (http://www.gnu.org/software/grep/manual/grep.html) that:
"Most meta-characters lose their special meaning inside bracket expressions." I've also tried:
$grep -E -o 'A.*YZ file.txt
But this outputs the entire line:
AooYoZooYZoAoooooYZ
Is there a way to override this or another way of solving my problem?
Maybe you can use non-greedy match which can be used in Perl regexp
echo 'AooYoZooYZoAoooooYZ' | grep -P -o 'A.*?YZ'
However, note that the manual for GNU grep says that -P option is highly experimental.
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.