使用正则表达式匹配正则表达式，而不是 JavaScript 中的字符串顺序

Question

I'm looking to extract a specific pattern from a string, and if its not there then extract another pattern.

The string in question may have both patterns in any order. Regardless of the order, I want the first pattern to take priority.

I know I could do this with multiple lines/ match calls but I am wondering if it is possible to do this with a single match call.

var s1 = "hello A123 B456"
var s2 = "hello B123 A456"

I want to capture A### pattern first, and only if its not there , then capture the B### pattern

console.log(s1.match(/((A|B)\d{3})/)[1]); // A123
console.log(s2.match(/((A|B)\d{3})/)[1]); // B123 -- but I want it to capture the A123 first

Answer 1

I guess you can achieve that by anchoring the preferred option to the start of the subject:

 re = /^.*?(A\\d+)|(B\\d+)/ test = [ "hello A456 B123", "hello B123 A456", "hello A456 zzz", "hello B123 zzz", ]; for (t of test) { m = t.match(re) console.log(t, '=', m[1] || m[2]) }

A drawback is that you have two groups to choose from.

Answer 2

You can accomplish this with either of the following regex patterns:

^(?:.*(A\d{3})|.*(B\d{3}))        # pattern 1
^(?:.*(?=A)|.*(?=B))([AB]\d{3})   # pattern 2

---

Pattern 1

Regex

Pros/Cons: Easy, but uses two capture groups

^(?:.*(A\d{3})|.*(B\d{3}))

How this works:

• ^ anchors it to the start of the string
• (?:.*(A\\d{3})|.*(B\\d{3})) match either of the following options
  • .* matches any character (except newline characters) any number of times (it's greedy so it'll match as much as possible)
  • (A\\d{3}) matches A followed by 3 digits

The second options is the same as the first, but this works with backtracking:

^(?:.*(A\d{3})|.*(B\d{3}))

hello B123 A456
^                  # anchor to the start of the string (to the location before the h)
                   # now attempt option 1 of the alternation: .*(A\d{3})
    .*             # match any character any number of times (greedy)
hello B123 A456    # this is what we currently match
                   # now backtrack to find A\d{3}
hello B123 A       # we found A, check to see if \d{3} matches
hello B123 A456    # pattern fulfilled; result you're looking for in group 1

Code

 s = ["hello A123 B456","hello B123 A456", "hello B123"] r = /^(?:.*(A\\d{3})|.*(B\\d{3}))/ for (x of s) { m = x.match(r) if (m) console.log(m[1] || m[2]) }

---

Pattern 2

Regex

Pros/Cons: Less comprehensible, but uses only one capture group

^(?:.*(?=A)|.*(?=B))([AB]\d{3})

How this works:

• ^ anchors it to the start of the string
• (?:.*(?=A)|.*(?=B)) match either of the following options
  • .* matches any character (except newline characters) any number of times (it's greedy so it'll match as much as possible)
  • (?=A) ensures A follows the current position
• The second alternation is the same as above, but uses (?=B) instead
• ([AB]\\d{3}) match A or B followed by 3 digits

Code

 s = ["hello A123 B456","hello B123 A456", "hello B123"] r = /^(?:.*(?=A)|.*(?=B))([AB]\\d{3})/ for (x of s) { m = x.match(r) if (m) console.log(m[1]) }

Answer 3

You can also achieve the result with a negative lookahead.

 var s1 = "hello A123 B456"; var s2 = "hello B123 A456"; const regex = /A\\d{3}|B\\d{3}(?!.*A\\d{3})/ console.log(s1.match(regex)[0]); console.log(s2.match(regex)[0]);

The above regex is saying, A with 3 digits or B with 3 digits if not followed by A with 3 digits.