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  1. Prepare
  2. Tutorials
  3. 10 Days of Javascript
  4. Day 7: Regular Expressions I
  5. Discussions

Day 7: Regular Expressions I

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  • Andrewdwatters1
     + 23 comments
    function regexVar(str) {
//  ^ => first item matches:
// () => stores matching value captured within
// [aeiou] => matches any of the characters in the brackets
// . => matches any character:
// + => for 1 or more occurrances (this ensures str length > 3)
// \1 => matches to previously stored match. 
    // \2 looks for matched item stored 2 instances ago 
    // \3 looks for matched item stored 3 ago, etc

//  $ ensures that matched item is at end of the sequence

    let re = /^([aeiou]).+\1$/;
    return re;
}

  • yahyaessam
     + 4 comments
    let re =  /^([aeiou]).*\1$/;

    1- ^ asserts position at start of the string
    2- 1st Capturing Group ([aeiou])
    3- .* matches any character (except newline) - between zero and unlimited times, as many times as possible, giving back as needed (greedy)
    4- \1 matches the same text as most recently matched by the 1st capturing group
    5- $ asserts position at the end of the string

  • angel_shtarbev
     + 10 comments
    let re = /^(a|e|i|o|u).*\1$/;

return re;

  • justnut007
     + 1 comment

    not a elegant but a simple solution

    let re = /^a.*a$|^e.*e$|^u.*u$|^i.*i$|^o.*o$/

  • hoWan
     + 3 comments

    matches only strings with small case at least 3 chars long, same vowel start and end.

    const re = /^([aeiou])[a-z]+\1$/;
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