Match multiple words in any order via one regex

Question

as stated in heading I want regex which will give me results in order based on my \'query\'.

line=\'VERSION=\"OTHER\" POWER=\"LOW\" FREQ=\"OFF\" MAXTUN=\"BLE         


        

Answer 1

You may leverage a non-capturing alternation group to match either VERSION or FREQ (optionally preceded with a word boundary, just check if it meets your requirements):

\b(?:VERSION|FREQ)="(.*?)"

See the regex demo

Details

• \b - a leading word boundary
• (?:VERSION|FREQ) - either VERSION or FREQ
• =" - a =" substring
• (.*?) - Group 1 (the actual output of findall): any 0+ chars other than line break chars, as few as possible
• " - a double quote.

See the Python demo:

import re
line='VERSION="OTHER" POWER="LOW" FREQ="OFF" MAXTUN="BLER"'
print(re.findall(r'\b(?:VERSION|FREQ)="(.*?)"', line))
# => ['OTHER', 'OFF']

A better idea, perhaps, is to capture key-value pairs and map them to a dictionary:

import re
line = 'VERSION="OTHER" POWER="LOW" FREQ="OFF" MAXTUN="BLER"'
results = re.findall(r'(VERSION|FREQ)="(.*?)"', line)
print(dict(results))
# => {'FREQ': 'OFF', 'VERSION': 'OTHER'}

See the Python demo.

Answer 2

I'm afraid there is no way to match in the order you want using regex: you could execute the part before | first, and then the part after |. Or order the result afterwards.