Match the same start and end character of a string with Regex

Question

I'm trying to match the start and end character of a string to be the same vowel. My regex is working in most scenarios, but failing in others:

var re = /([aeiou]).*\1/; re.test(str); 

Sample input:

• abcde, output - false (Valid)
• abcda, output - true (Valid)
• aabcdaa, output - true (Valid)
• aeqwae, output - true (Not valid)
• ouqweru, output - true (Not valid)

Answer 1

You need to add anchors to your string.

When you have, for example:

aeqwae 

You say the output is true, but it's not valid because a is not the same as e. Well, regex simply matches the previous character (before e), which is a. Thus, the match is valid. So, you get this:

[aeqwa]e 

The string enclosed in the brackets is the actual match and why it returns true.

If you change your regex to this:

/^([aeiou]).*\1$/ 

By adding ^, you tell it that the start of the match must be the start of the string and by adding $ you tell it that the end of the match must be the end of the string. This way, if there's a match, the whole string must be matched, meaning that aeqwae will no longer get matched.

A great tool for testing regex is Regex101. Give it a try!

Note: Depending on your input, you might need to set the global (g) or multi-line (m) flag. The global flag prevents regex from returning after the first match. The multi-line flag makes ^ and $ match the start and end of the line (not the string). I used both of them when testing with your input.

Answer 2

Just a different version of @Hristiyan Dodov answer that I have written for fun.

regex = /^(a|e|i|o|u).*\1$/    const strings = ['abcde', 'abcda', 'aabcdaa', 'aeqwae', 'ouqweru']    strings.forEach((e)=>{    const result = regex.test(e)    console.log(e, result)  })