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I am working with the following string below:

abcdefg^delim^location^delim^123456

And currently have the following regex:

^(.*?)(?:^delim^|$)

The delimiter between values is "^delim^". I am running in to issues however when I try to execute the below code:

var str = 'abcdefg^delim^location^delim^123456'
var regex = /^(.*?)(?:^delim^|$)/
var matches = str.match(regex)

The issue is that my "matches" variable is returning an array of 2 values: "abcdefg^delim^" AND "abcdefg". How can I edit my regex to not capture the first value there? I only want what comes before the first occurrence of "^delim^", which in this case is "abcdefg"

2

Answers


  1. var str = 'abcdefg^delim^location^delim^123456'
    var regex = /^(.*?)(?=^delim^|$)/gi
    var matches = str.match(regex)
    
    console.log(matches)

    (?:^delim^|$): This is a non-capturing group that matches either ‘^delim^’ or the end of the string (‘$’). This part allows the match to continue beyond the first ‘^delim^’.

    Instead you want to use:

    (?=^delim^|$): Positive lookahead assertion that ensures the match is followed by ‘^delim^’ or the end of the string (‘$’). However, it doesn’t consume any characters in the match.

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  2. Regular Expression: ^(.*?)(?:^delim^|$)

    • ^: Asserts the start of the string.
      (.*?): A non-greedy capturing group that matches any characters (as few as possible) until the next part of the pattern.
    • (?:^delim^|$): A non-capturing group that matches either the delimiter ^delim^ or the end of the string $.
    • String to Match: 'abcdefg^delim^location^delim^123456'

    Matching Process:

    • The regular expression is applied to the string from the beginning (^).
    • The non-greedy capturing group (.*?) captures the characters 'abcdefg'.
    • The non-capturing group (?:^delim^|$) is not yet matched.

    Resulting Array (matches):

    • The entire matched string: 'abcdefg^delim^location^delim^'
    • The content of the capturing group (.*?): 'abcdefg'
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