BASH -- grep'ing for perl-regex

Question

Beginner here and I'm trying to understand this. Can someone please break down the part in between the single quotes and describe what it does?

grep -oP '(?<=\S\/1\.\d.\s)[345]\d+'

Many thanks in advance!

Answer 1

• Positive Lookbehind (?<=\S/1.\d.\s) Assert that the Regex below matches
• \S matches any non-whitespace character (equal to [^\r\n\t\f\v ])
• \/ matches the character / literally (case sensitive)
• 1 matches the character 1 literally (case sensitive)
• \. matches the character . literally (case sensitive)
• \d matches a digit (equal to [0-9])
• . matches any character (except for line terminators)
• \s matches any whitespace character (equal to [\r\n\t\f\v ]) Match a single character present in the list below [345]
• 345 matches a single character in the list 345 (case sensitive)
• \d+ matches a digit (equal to [0-9])
• + Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)

Output simply copied from https://regex101.com/r/HfJSNm/1 : very handy to test/share/have automatic explications on regexes.