I have this text https://bitbucket.com/user/repo.git and I want to print repo, the content between / and .git, without including delimiters. I have this:
echo https://bitbucket.com/user/repo.git | grep -E -o '\/(.*?)\.git'
But it prints /repo.git. How can I print just repo?
Use the [^/]+(?=\.git$) pattern with -P option:
echo https://bitbucket.com/user/repo.git | grep -P -o '[^/]+(?=\.git$)'
See the online demo
The [^/]+(?=\.git$) pattern matches 1+ chars other than / that are followed with .git at the end of the string.
You can use sed to do that
echo https://bitbucket.com/user/repo.git | sed -e 's/^.\*\\/\\(.\*\\).git$/\1/g'
Related
How to use sed and regex to replace the text between a variable number of one token?
Example of input:
/abc/bcd/cde/
Expected output:
/../../../
Tried:
Command: echo "/abc/bcd/cde/" | sed 's/\/.*\//\/..\//g' output: /../
Using perl and look around assertions :
$ perl -pe 's|(?<=/)\w{3}(?=/)|..|g' file
/../../../
Using sed :
$ echo "/abc/bcd/cde/" | sed -E 's|[a-z]{3}|..|g'
/../../../
Replace every substring of non-slashes ([^/]\+) with two dots:
$> echo "/abc/bcd/cde/" | sed 's$[^/]\+$..$g'
# => /../../../
Base on #Gilles Quenot implementation but, capturing any alpha numeric chars between //
$ echo "/abddc/bcqsdd/cdde/" | sed -E 's|(/)?[^/]+/|\1../|g'
I have an output like Johny-Smith, Juarez-Hugo, etc. and I need instead S, H, etc. Basically, I need the last uppercase letter in a string and that's it. If this is possible in any built in Linux tools (ex awk, sed, grep, etc.) it would be greatly appreciated.
Do you need like this ?
echo "Johny-Smith" | sed 's/^.*\([A-Z]\)[^A-Z]*$/\1/g'
Test:
$ echo "Johny-Smith-Hello Johny-Smith" | sed 's/.*\([A-Z]\)[^A-Z]*/\1/g'
S
With GNU grep and if PCRE option is available
$ echo 'Johny-Smith' | grep -oP '.*\K[A-Z]'
S
$ echo 'Juarez-Hugo' | grep -oP '.*\K[A-Z]'
H
-o prints only matched portion
-P Perl regular expression
.*\K positive lookbehind, not part of output
[A-Z] any uppercase character
with perl, see perldoc for command line options explanation
$ # prints the string within captured group
$ echo 'Johny-Smith' | perl -lne 'print /.*([A-Z])/'
S
$ echo 'Juarez-Hugo' | perl -lne 'print /.*([A-Z])/'
H
In Bash:
$ var="Johny-Smith-Hello Johny-Smith"; var="${var//[^[:upper:]]/}";echo "${var: -1}"
S
${var//[^[:upper:]]/} remove all non-upper case letter chars
echo ${var: -1} output the last one
I have an input string in the following format:
bugfix/ABC-12345-1-00
I want to extract "ABC-12345". Regex for that format in C# looks like this:
.\*\\/([A-Z]+-[0-9]+).\*
How can I do that in a bash script? I've tried sed and awk but had no success because I need to extract value from the capturing group and skip the rest.
If your grep supports -P then you could use the below grep commands.
$ echo 'bugfix/ABC-12345-1-00' | grep -oP '/\K[A-Z]+-\d+'
ABC-12345
\K keeps the text matched so far out of the overall regex match.
$ echo 'bugfix/ABC-12345-1-00' | grep -oP '(?<=/)[A-Z]+-\d+'
ABC-12345
(?<=/) Positive lookbehind which asserts that the match must be preceded by a / symbol.
Through sed,
$ echo 'bugfix/ABC-12345-1-00' | sed 's~.*/\([A-Z]\+-[0-9]\+\).*~\1~'
ABC-12345
echo "bugfix/ABC-12345-1-00"| perl -ane '/.*?([A-Z]+\-[0-9]+).*/;print $1."\n"'
You could try something like:
echo "bugfix/ABC-12345-1-00" | egrep -o '[A-Z]+-[0-9]+'
OUTPUT:
ABC-12345
If you do not like to use regex, you can use this awk:
echo "bugfix/ABC-12345-1-00" | awk -F\/ '{print $NF}'
ABC-12345-1-00
Or just this:
awk -F\/ '$0=$NF'
I'm trying to get a pattern over multiple lines. I would like to ensure the line I'm looking for ends in \r\n and that there is specific text that comes after it at some point. The two problems I've had are I often get unmatched parenthesis in groupings or I get a positive match when there is none. Here are two simple examples.
echo -e -n "ab\r\ncd" | grep -U -c -z -E $'(\r\n)+.*TEST'
grep: Unmatched ( or \(
What exactly is unmatched there? I don't get it.
echo -e -n "ab\r\ncd" | grep -U -c -z -E $'\r\n.*TEST'
1
There is no TEST in the string, so why does this return a count of 1 for matches?
I'm using grep (GNU grep) 2.16 on Ubuntu 14. Thanks
Instead of -E you can use -P for PCRE support in gnu grep to use advanced regex like this:
echo -ne "ab\r\ncd" | ggrep -UczP '\r\n.*TEST'
0
echo -ne "ab\r\ncd" | ggrep -UczP '\r\n.*cd'
1
grep -E matches only in single line input.
How to replace the last matching dot?
for example, I'd like to change test.jpg to test.th.jpg
what I've tried:
echo "test.jpg" | sed 's#[^\.]*$#\.th.#g'
This should work:
$ echo "test.jpg" | sed 's/\.\([^.]*\)$/.th.\1/'
It gives:
test.th.jpg
Explanation:
\. literal dot
\( start capturing group
[^.] any character except dot
* zero or more
\) close capturing group
$ end of line
In the replacement \1 replaces the content of the first capturing group :)
kent$ sed 's/[^.]*$/th.&/' <<<"test.jpg"
test.th.jpg
or
kent$ sed 's/.*\./&th./' <<<"test.jpg"
test.th.jpg
or
kent$ awk -F. -v OFS="." '$NF="th."$NF' <<< "test.jpg"
test.th.jpg
You can also use awk, prepend "th." to the last field
$ awk 'BEGIN{FS=OFS="."}$NF="th."$NF' <<< "test.jpg"
test.th.jpg
Using pure bash
$ str="test.abc.jpg"
$ echo ${str%.*}.th.jpg
test.abc.th.jpg
if you have Ruby on your system
echo test.jpg | ruby -e 'x=gets.split(".");x.insert(-2,"th"); puts x.join(".")'