I have a massive amount of text I just need the contents of whats inside the single quotes (excluding the single quotes).

for example, here's a cutdown version of what I am searching.

output line from channel: [2021-11-14 15:59:20] config='954'!
output line from channel: [2021-11-14 15:59:21] DEBUG: job_name='test' disabled=true
output line from channel: [2021-11-14 15:59:25] DEBUG: job_id='a185' configsized

and I would like to return

a185

The regular expression I have so far is this, but it returns the jobid='' - as well as the data i required. I tried to use a capture group and I thought you could delete it?

My regex skills are old and out of touch lol :-)

(job_id=)'[^']*'

Note that the line has to have DEBUG on it somewhere to match everything.

CodePudding user response：

You can use

DEBUG.*job_id='([^']*)'

and get the Group 1 value. See the regex demo. Details:

• DEBUG - a DEBUG string
• .* - any zero or more chars other than line break chars, as many as possible
• job_id=' - a job_id=' string
• ([^']*) - Capturing group 1: any zero or more chars other than '
• ' - a ' char.

See the Go demo online:

package main

import (
    "fmt"
    "regexp"
)

func main() {
    markdownRegex := regexp.MustCompile(`DEBUG.*job_id='([^']*)'`)
    results := markdownRegex.FindStringSubmatch(`output line from channel: [2021-11-14 15:59:25] DEBUG: job_id='a185' configsized`)
    fmt.Printf("%q", results[1])
}
// => "a185"

CodePudding user response：

If you just want to grab a185 then you can use regex as:

(?<=job_id=')[^'] 

and if you want to get the regex where you need to have DEBUG also then you can do as:

(?<=DEBUG: job_id=')[^']