I have versions like:
v1.0.3-preview2
v1.0.3-sometext
v1.0.3
v1.0.2
v1.0.1
I am trying to get the latest version that is not preview (doesn't have text after version number) , so result should be:
v1.0.3
I used this grep: grep -m1 "[v\d+\.\d+.\d+$]"
but it still outputs: v1.0.3-preview2
what I could be missing here?
To return first match for pattern v<num>.<num>.<num>, use:
grep -m1 -E '^v[0-9]+(\.[0-9]+){2}$' file
v1.0.3
If you input file is unsorted then use grep | sort -V | head as:
grep -E '^v[0-9]+(\.[0-9]+){2}$' f | sort -rV | head -1
When you use ^ or $ inside [...] they are treated a literal character not the anchors.
RegEx Details:
^: Start
v: Match v
[0-9]+: Match 1+ digits
(\.[0-9]+){2}: Match a dot followed by 1+ dots. Repeat this group 2 times
$: End
To match the digits with grep, you can use
grep -m1 "v[[:digit:]]\+\.[[:digit:]]\+\.[[:digit:]]\+$" file
Note that you don't need the [ and ] in your pattern, and to escape the dot to match it literally.
With awk you could try following awk code.
awk 'match($0,/^v[0-9]+(\.[0-9]+){2}$/){print;exit}' Input_file
Explanation of awk code: Simple explanation of awk program would be, using match function of awk to match regex to match version, once match is found print the matched value and exit from program.
Regular expressions match substrings, not whole strings. You need to explicitly match the start (^) and end ($) of the pattern.
Keep in mind that $ has special meaning in double quoted strings in shell scripts and needs to be escaped.
The boundary characters need to be outside of any group ([]).
Related
I have a log file contains some information like below
"variable1=XXX, emotionType=sad, sentimentType=negative..."
What I want is to grep only the matched string, the string starts with emotionType and ends with the first occurrence of comma.
E.g.
emotionType=sad
emotionType=joy
...
What I have tried is
grep -e "/^emotionType.*,/" file.log -o
but I got nothing. Anyone can tell me what should I do?
You need to use
grep -o "emotionType[^,]*" file.log
Note:
Remove ^ or replace with \<, starting word boundary construct if your matches are not located at the beginning of each line
Remove the / chars on both ends of the regex since grep does not use regex delimiters (like sed)
[^,] is a negated bracket expression that matches any char other than a comma
* is a POSIX BRE quantifier that matches zero or more occurrences.
See an online demo:
#!/bin/bash
s="variable1=XXX, emotionType=sad, sentimentType=negative, emotionType=happy"
grep -o "emotionType=[^,]*" <<< "$s"
Output:
emotionType=sad
emotionType=happy
1st solution: With awk you could try following program. Simple explanation would be using awk's match function capability and using regex to match string emotionType till next occurrence of , and printing all the matches in awk program.
var="variable1=XXX, emotionType=sad, sentimentType=negative, emotionType=happy"
Where var is a shell variable.
echo "$var" |
awk '{while(match($0,/emotionType=[^,]*/)){print substr($0,RSTART,RLENGTH);$0=substr($0,RSTART+RLENGTH)}}'
2nd solution: Or in GNU awk using RS variable try following awk program.
echo "$var" | awk -v RS='emotionType=[^,]*' 'RT{sub(/\n+$/,"",RT);print RT}'
Can you please tell me how to get the token value correctly? At the moment I am getting: "1jdq_dnkjKJNdo829n4-xnkwe",258],["FbtResult
echo '{"facebookdotcom":true,"messengerdotcom":false,"workplacedotcom":false},827],["DTSGInitialData",[],{"token":"aaaaaaa"},258],["FbtResult' | sed -n 's/.*"token":\([^}]*\)\}/\1/p'
You need to match the full string, and to get rid of double quotes, you need to match a " before the token and use a negated bracket expression [^"] instead of [^}]:
sed -n 's/.*"token":"\([^"]*\).*/\1/p'
Details:
.* - any zero or more chars
"token":" - a literal "token":" string
\([^"]*\) - Group 1 (\1 refers to this value): any zero or more chars other than "
.* - any zero or more chars.
This replacement works:
echo '{"facebookdotcom":true,"messengerdotcom":false,"workplacedotcom":false},827],["DTSGInitialData",[],{"token":"aaaaaaa"},258],["FbtResult'
| sed -n 's/.*"token":"\([a-z]*\)"\}.*/\1/p'
Key capture after "token" found between quotes via \([a-z]*\), followed by a closing brace \} and remaining characters after that as .* (you were missing this part before, which caused the replacement to include the text after keyword as well).
Output:
aaaaaaa
A grep solution:
echo '{"facebookdotcom":true,"messengerdotcom":false,"workplacedotcom":false},827],["DTSGInitialData",[],{"token":"aaaaaaa"},258],["FbtResult' | grep -Po '(?<="token":")[^"]+'
yields
aaaaaaa
The -P option to grep enables the Perl-compatible regex (PCRE).
The -o option tells grep to print only the matched substring, not the entire line.
The regex (?<="token":") is a PCRE-specific feature called a zero-width positive lookbehind assertion. The expression (?<=pattern) matches a pattern without including it in the matched result.
I have the following rule:
https://regex101.com/r/noX9lj/4
I want to make this work in a script so I'm using grep like this:
echo "\$this->table('test')" | grep -Po "qr/\$this->table\(\'(test)\'\);/"
The output should be "test"
It's not working, not sure why..
You may use
echo "\$this->table('test');" | grep -oP "\\\$this->table\\('\\K[^']+(?='\\);)"
Or, if you feed a file path to grep:
grep -oP "\\\$this->table\\('\\K[^']+(?='\\);)" file
See the online grep demo
To match $, you need to escape it with a literal backslash, and inside a double quoted string, you need to escape $ itself with one backslash char in order to stop variable expansion, and then you need to add two more backslashes to regex-escape the literal $ char, hence is the "\\\$" in the pattern.
To match any text between two single quotes, you may use [^']+ - 1 or more chars other than '.
See the regex demo
Pattern details
\$this->table\(' - $this->table(' string
\K - match reset operator that discards the text matched so far from the overall match buffer
[^']+ - one or more chars other than '
(?='\);) - a positive lookahead that requires '); string to be present immediately to the right of the current position.
There were multiple issues:
had to use "cat" instead of echo for some reason
used this rule instead:
grep -oP "this->table\('\K\w+(?='\);)"
I'm trying to come up with a regex for the following case: I need to find any matching paths using grep for the following paths:
Include all uppercase matching paths.
Example:
com/foo/Bar/1.2.3-SNAPSHOT/Bar-1.2.3-SNAPSHOT.jar
Notice the capital B in Bar.
Exclude all uppercase matching paths that only contain SNAPSHOT and have no other uppercase letters.
Example:
com/foo/bar/1.2.3-SNAPSHOT/bar-1.2.3-SNAPSHOT.jar
Is this possible with grep?
Something like this might do:
grep -vE '^([^[:upper:]]*(SNAPSHOT)?)*$'
Breakdown:
-v will reverse the match (show all non matched lines. -E enabled Extended Regular Expressions.
^ # Start of line
( )* # Capturing group repeated zero or more times
[^[:upper:]]* # Match all but uppercase zero or more times
(SNAPSHOT)? # Followed by literal SNAPSHOT zero or one time
$ # End of line
Just use awk:
$ cat file
com/foo/Bar/1.2.3-SNAPSHOT/Bar-1.2.3-SNAPSHOT.jar
com/foo/bar/1.2.3-SNAPSHOT/bar-1.2.3-SNAPSHOT.jar
With GNU awk or mawk for gensub():
$ awk 'gensub(/SNAPSHOT/,"","g")~/[[:upper:]]/' file
com/foo/Bar/1.2.3-SNAPSHOT/Bar-1.2.3-SNAPSHOT.jar
With other awks:
$ awk '{r=$0; gsub(/SNAPSHOT/,"",r)} r~/[[:upper:]]/' file
com/foo/Bar/1.2.3-SNAPSHOT/Bar-1.2.3-SNAPSHOT.jar
Well, you need find to list all paths. Then you can do it with grep with two runs. One includes all capital cases. The other one excludes that contain no capitals except SNAPSHOT:
find . | grep '[A-Z]' | grep -v '.*\/[^A-Z]*SNAPSHOT[^A-Z]*$'
I think only the last grep needs some explanation:
grep -v excludes the matching lines
.*\/ greedily matches everything up to the first slash. There'll always be a slash due to find .
[^A-Z]* finds all characters that are non-capital letters. So we apply it before and after the SNAPSHOT literal, up to the end of the string.
Here you can play with it online.
If you only want to get the matching files. I'll do it like this.
find . -type f -regex '.*[A-Z].*' | while read -r line; do echo "$line" | sed 's/SNAPSHOT//g' | grep -q '.*[A-Z].*' && echo "$line"; done
Here's a basic regex technique that I've never managed to remember. Let's say I'm using a fairly generic regex implementation (e.g., grep or grep -E). If I were to do a list of files and match any that end in either .sty or .cls, how would I do that?
ls | grep -E "\.(sty|cls)$"
\. matches literally a "." - an unescaped . matches any character
(sty|cls) - match "sty" or "cls" - the | is an or and the brackets limit the expression.
$ forces the match to be at the end of the line
Note, you want grep -E or egrep, not grep -e as that's a different option for lists of patterns.
egrep "\.sty$|\.cls$"
This regex:
\.(sty|cls)\z
will match any string ends with .sty or .cls
EDIT:
for grep \z should be replaced with $ i.e.
\.(sty|cls)$
as jelovirt suggested.