I'm trying to get my filename without its extension using a regex I found on Stack Overflow. The regex is:
(.+?)(\.[^.]*$|$)
I try this on the command line
echo TestFileName.1.0.0.2.zip | grep "(.+?)(\.[^.]*$|$)"
And I get nothing in the command line. If I try it with this regex:
echo TestFileName.1.0.0.2.zip | grep "Test"
I do see the TestFileName.1.0.0.2.zip gets printed to the console with Test highlighted in red. When I tried my data in this website: http://rubular.com/r/LNrI4inMU1
It does seem to work. Am I applying the regex wrong in Bash?
You're using an extended regular expression; the standard regex language which grep uses doesn't support what you're trying to do. Change grep to be grep -E and the match will work. This specifies that your regex is an extended one.
$ echo TestFileName.1.0.0.2.zip | grep -E "(.+?)(\.[^.]*$|$)"
TestFileName.1.0.0.2.zip
See this link for more information on the distinction between regular and extended regex.
Using BASH regex:
s='TestFileName.1.0.0.2.zip'
[[ "$s" =~ ^(.*)\.[^.]+$ ]] && echo "${BASH_REMATCH[1]}"
TestFileName.1.0.0.2
Add -P (Perl-regexp) parameter to your grep along with -o (only-matching).
$ echo TestFileName.1.0.0.2.zip | grep -oP "(.+)(?=\.)"
TestFileName.1.0.0.2
Related
I have a Build called 700-I20190808-0201. I need to convert it to 7.0.0-I20190808-0201. I can do that with regular expression:
sed 's/\([0-9]\)\([0-9]\)\([0-9]\)\(.\)/\1.\2.\3\4/' abc.txt
But the solution does not work when the build ID is 7001-I20190809-0201. Can we make the regular expression dynamic so that it works for both (700 and 7001)?
Could you please try following.
awk 'BEGIN{FS=OFS="-"}{gsub(/[0-9]/,"&.",$1);sub(/\.$/,"",$1)} 1' Input_file
If you have Perl available, lookahead regular expressions make this straightforward:
$ cat foo.txt
700-I20190808-0201
7001-I20190809-0201
$ perl -ple 's/(\d)(?=\d+\-I)/\1./g' foo.txt
7.0.0-I20190808-0201
7.0.0.1-I20190809-0201
You can implement a simple loop using labels and branching using sed:
$ echo '7001-I20190809-0201' | sed ':1; s/^\([0-9]\{1,\}\)\([0-9][-.]\)/\1.\2/; t1'
7.0.0.1-I20190809-0201
$ echo '700-I20190809-0201' | sed ':1; s/^\([0-9]\{1,\}\)\([0-9][-.]\)/\1.\2/; t1'
7.0.0-I20190809-0201
If your sed support -E flag:
sed -E ':1; s/^([0-9]+)([0-9][-.])/\1.\2/; t1'
sed -e 's/\([0-9]\)\([0-9]\)\([0-9]\)\(.\)/\1.\2.\3.\4/' -e 's/\.\-/\-/' abc.txt
This worked for me, very simple one. Just needed to extract it in my ant script using replaceregex pattern
I have a bash variable, some file path (with spaces) and filename, e.g:
$ echo $tmp
/home/xyz/some/path/with spaces/AlbumArt_{random-number-sequence}_Large.jpg
When I attempt to identify the filename part with grep, e.g:
$ echo "$tmp" | egrep 'AlbumArt.*Large.jpe?g$'
/home/xyz/some/path/with spaces/**AlbumArt_{random-number-sequence}_Large.jpg**
The filename part appears to be identified correctly, but when I attempt to convert this to a sed substitution expression, e.g:
$ echo "$tmp" | sed 's#AlbumArt.*Large.jpe?g$#NewString#'
/home/xyz/some/path/with spaces/AlbumArt_{random-number-sequence}_Large.jpg
The expected substitution isn't happening. Thanks in advance for any help.
In fact egrep is a variant of grep -E, allowing to 'activate' extended regular expression (you can see: https://en.wikipedia.org/wiki/Regular_expression#Standards).
Thus, you just need to use the same option with sed:
echo "$tmp" | sed -E 's#AlbumArt.*Large.jpe?g$#NewString#'
Given the following shell script, would someone be so kind as to explain the grep -Po regex please?
#!/bin/bash
# Issue the request for a bearer token, json is returned
raw_json=`curl -s -X POST -d "username=name&password=secret&client_id=security-admin-console" http://localhost:8081/auth/realms/master/tokens/grants/access`
# Strip away all but the "access_token" field's value using a Python regular expression
bearerToken=`echo $raw_json | grep -Po '"'"access_token"'"\s*:\s*"\K([^"]*)'`
echo "The bearer token is:"
echo $bearerToken
So specifically, I'm interested in understanding the parts of the regex
grep -Po '"'"access_token"'"\s*:\s*"\K([^"]*)'`
and how it works. Why so many quotes? What is the "K" for? I've some experience with grep regex but this confuses me.
This is the actual output of the curl command and the shell script (grep) works as desired returning just the contents of the "access_token" value.
{"access_token":"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.fQmQKn-xatvflHPAaxCfrrVow3ynpw0sREho7__jZo2d0g1SwZV7Lf4C26CcweNLlb3wmKHHo63HRz35qRxJ7BXyiZwHgXokvDJj13yuOb6Sirg9z02n6fwGy8Iog30pUvffnDaVnUWHfVL-h_R4-OZNf-_YUK5RcL2DHt0zUXI","expires_in":60,"refresh_expires_in":1800,"refresh_token":"eyJhbGciOiJSUzI1NiJ9.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.WeiJOC1jQ52aKgnW8UN2Lv9rJ_yKZiOhijOYKLN2EEOkYF8rvRZsSKbTPFKTIUvjnwy2A7V_N-GhhJH4C-T7F5__QPNofSXbCNyvATj52jGLxk9V0Afvk-Z5QAWi55PJRTC0qteeMRcO2Frw-0KtKYe9o3UcGICJubxhZHsXBLA","token_type":"bearer","id_token":"eyJhbGciOiJSUzI1NiJ9.eyJuYW1lIjoiIiwianRpIjoiMGIyMGI0ODctOTI4OS00YTFhLTgyNmMtM2NiOTg0MDJkMzVkIiwiZXhwIjoxNDQ2ODI4MDU5LCJuYmYiOjAsImlhdCI6MTQ0NjgyNzk5OIwouldhaveToBeNutsUiLCJwcmVmZXJyZWRfdXNlcm5hbWUiOiJhZG1pbiIsImVtYWlsX3ZlcmlmaWVkIjpmYWxzZX0.DmG8Lm4niL1djzNrLsZ2CrsB1ZzUPnR2Nm7IZnrwrmkXsrPxjl6pyXKCWSj6pbk2sgVI8NNFqrGIJmEJ7gkTZWm328VGGpJsmMuJBki0KbqBRKORGQSgkas_34rwzhcTE3Iki8h_YVs2vvNIx_eZSOvIzyEcP3IGHuBoxcR6W3E","not-before-policy":0,"session-state":"62efc05c-1bf5-4f55-b749-5e0eff94155b"}
In case anyone finds this post, this is what I ended up using:
if hash jq 2>/dev/null; then
# Use the jq command to safely parse json
bearerToken=$(echo $raw_json | jq -r '.access_token')
else
# Strip away all but the "access_token" field's value using a perl regular expression
bearerToken=$(echo $raw_json | grep -Po '"'"access_token"'"\s*:\s*"\K([^"]*)')
fi
Since not all regex flavors support lookbehind, Perl introduced the \K. In general when you have:
a\Kb
When “b” is matched, \K tells the engine to pretend that the match attempt started at this position.
In your example, you want to pretend that the match attempt started at what appears after the "access_token":" text.
This example will better demonstrate the \K usage:
~$ echo 'hello world' | grep -oP 'hello \K(world)'
world
~$ echo 'hello world' | grep -oP 'hello (world)'
hello world
In addition, \K allows a variable-length look-behind:
$ echo foooooo bar | grep -oP "(?<=foo+) \Kbar"
grep: lookbehind assertion is not fixed length
$ echo foooooo bar | grep -oP "foo+ \Kbar"
bar
My solution was: sed -n 's/cut off this part \(display this part only\) cut off this part/\1/gp'
References:
https://www.cyberciti.biz/faq/unix-linux-sed-print-only-matching-lines-command/
info sed (texinfo package)
man 1 sed
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.
I just want to match some text in a Bash script. I've tried using sed but I can't seem to make it just output the match instead of replacing it with something.
echo -E "TestT100String" | sed 's/[0-9]+/dontReplace/g'
Which will output TestTdontReplaceString.
Which isn't what I want, I want it to output 100.
Ideally, it would put all the matches in an array.
edit:
Text input is coming in as a string:
newName()
{
#Get input from function
newNameTXT="$1"
if [[ $newNameTXT ]]; then
#Use code that im working on now, using the $newNameTXT string.
fi
}
You could do this purely in bash using the double square bracket [[ ]] test operator, which stores results in an array called BASH_REMATCH:
[[ "TestT100String" =~ ([0-9]+) ]] && echo "${BASH_REMATCH[1]}"
echo "TestT100String" | sed 's/[^0-9]*\([0-9]\+\).*/\1/'
echo "TestT100String" | grep -o '[0-9]\+'
The method you use to put the results in an array depends somewhat on how the actual data is being retrieved. There's not enough information in your question to be able to guide you well. However, here is one method:
index=0
while read -r line
do
array[index++]=$(echo "$line" | grep -o '[0-9]\+')
done < filename
Here's another way:
array=($(grep -o '[0-9]\+' filename))
Pure Bash. Use parameter substitution (no external processes and pipes):
string="TestT100String"
echo ${string//[^[:digit:]]/}
Removes all non-digits.
I Know this is an old topic but I came her along same searches and found another great possibility apply a regex on a String/Variable using grep:
# Simple
$(echo "TestT100String" | grep -Po "[0-9]{3}")
# More complex using lookaround
$(echo "TestT100String" | grep -Po "(?i)TestT\K[0-9]{3}(?=String)")
With using lookaround capabilities search expressions can be extended for better matching. Where (?i) indicates the Pattern before the searched Pattern (lookahead),
\K indicates the actual search pattern and (?=) contains the pattern after the search (lookbehind).
https://www.regular-expressions.info/lookaround.html
The given example matches the same as the PCRE regex TestT([0-9]{3})String
Use grep. Sed is an editor. If you only want to match a regexp, grep is more than sufficient.
using awk
linux$ echo -E "TestT100String" | awk '{gsub(/[^0-9]/,"")}1'
100
I don't know why nobody ever uses expr: it's portable and easy.
newName()
{
#Get input from function
newNameTXT="$1"
if num=`expr "$newNameTXT" : '[^0-9]*\([0-9]\+\)'`; then
echo "contains $num"
fi
}
Well , the Sed with the s/"pattern1"/"pattern2"/g just replaces globally all the pattern1s to pattern 2.
Besides that, sed while by default print the entire line by default .
I suggest piping the instruction to a cut command and trying to extract the numbers u want :
If u are lookin only to use sed then use TRE:
sed -n 's/.*\(0-9\)\(0-9\)\(0-9\).*/\1,\2,\3/g'.
I dint try and execute the above command so just make sure the syntax is right.
Hope this helped.
using just the bash shell
declare -a array
i=0
while read -r line
do
case "$line" in
*TestT*String* )
while true
do
line=${line#*TestT}
array[$i]=${line%%String*}
line=${line#*String*}
i=$((i+1))
case "$line" in
*TestT*String* ) continue;;
*) break;;
esac
done
esac
done <"file"
echo ${array[#]}