正则表达式捕获，带有可选的下划线和数字

Question

I'm trying to find a regular expression that will match the base string without the optional trailing number ( _123 ). eg:

lorem_ipsum_test1_123 -> capture lorem_ipsum_test1

lorem_ipsum_test2 -> capture lorem_ipsum_test2

I tried using the following expression, but it would only work when there is a trailing _number. /(.+)(?>_[0-9]+)/ /(.+)(?>_[0-9]+)?/ Similarly, adding the ? (zero or more) quantifier only worked when there is no trailing _number, otherwise, the trailing _number would just be part of the first capture.

Any suggestions?

Answer 1

You may use the following expression:

^(?:[^_]+_)+(?!\d+$)[^_]+

• ^ Anchor beginning of string.
• (?:[^_]+_)+ Repeated non capturing group. Negated character set for anything other than a _ , followed by a _ .
• (?!\\d+$) Negative lookahead for digits at the end of the string.
• [^_]+ Negated character set for anything other than a _ .

Regex demo here .

Please note that the \\n in the character sets in the Regex demo are only for demonstration purposes, and should by all means be removed when using as a pattern in Javascript.

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Javascript demo:

 var myString = "lorem_ipsum_test1_123"; var myRegexp = /^(?:[^_]+_)+(?!\\d+$)[^_]+/g; var match = myRegexp.exec(myString); console.log(match[0]); var myString = "lorem_ipsum_test2" var myRegexp = /^(?:[^_]+_)+(?!\\d+$)[^_]+/g; var match = myRegexp.exec(myString); console.log(match[0]); 

Answer 2

You might match any character and use a negative lookahead that asserts that what follows is not an underscore, one or more digits and the end of the string:

^(?:(?!_\\d+$).)*

Explanation

• ^ Assert start of the string
• (?: Non capturing group
  • (?! Negative lookahead to assert what is on the right side is not
    • _\\d+$ Match an underscore, one or more digits and assert end of the string
  • .) Match any character and close negative lookahead
• )* Close non capturing group and repeat zero or more times

Regex demo

 const strings = [ "lorem_ipsum_test1_123", "lorem_ipsum_test2" ]; let pattern = /^(?:(?!_\\d+$).)*/; strings.forEach((s) => { console.log(s + " ==> " + s.match(pattern)[0]); }); 

Answer 3

You are asking for

/^(.*?)(?:_\d+)?$/

See the regex demo . The point here is that the first dot pattern must be non-greedy and the _\\d+ should be wrapped with an optional non-capturing group and the whole pattern (especially the end) must be enclosed with anchors.

Details

• ^ - start of string
• (.*?) - Capturing group 1: any zero or more chars other than line break chars, as few as possible due to the non-greedy ( "lazy" ) quantifier *?
• (?:_\\d+)? - an optional non-capturing group matching 1 or 0 occurrences of _ and then 1+ digits
• $ - end of string.

However, it seems easier to use a mere replacing approach,

s = s.replace(/_\d+$/, '')

If the string ends with _ and 1+ digits, the substring will get removed, else, the string will not change.

See this regex demo .

Answer 4

Try to check if the string contains the trailing number. If it does you get only the other part. Otherwise you get the whole string.

var str = "lorem_ipsum_test1_123"

if(/_[0-9]+$/.test(str)) {
   console.log(str.match(/(.+)(?=_[0-9]+)/g))
} else {
   console.log(str)
}

Or, a lot more concise:

str = str.replace(/_[0-9]+$/g, "")