CODEWITHSUNDEEP
javascriptregex
Alexandru Nutu
Alexandru Nutu

Reputation: 461

Javascript regex return true if custom pattern is 100% matched

In vanilla Javascript I want to check a string.

const myRegex = /^\d+|#\d+/g;
console.log(`${myRegex.test("3#123#432#555")}`); // pattern is ok -> true
console.log(`${myRegex.test("3#123#432##555")}`); // two ## -> patter wrog -> but result is true (would like this to be false)
console.log(`${myRegex.test("3#123#432#55a5")}`); // a character in the string -> pattern wrong -> but result is true (should also be false)

I played around in https://regex101.com/r/tI1sOa/1/ I get a perfect match using this regex but I want it to return false when the pattern is altered.

Pattern definition should be: number#number#number#number#number (so first we have a number followed by #number, as many times as I want)

If the pattern is number### or #number or numberLetter#number or any other combination that does not respect the pattern should return false for the test.

How can I check this with regex? Why is the one I came up with not working as I expect it?

Thank you!

Upvotes: 0

Views: 47

Answers (1)

Felix Kling
Felix Kling

Reputation: 816324

^\d+|#\d+ means

  • match one or more digits at the start of the string
  • or match a # followed by one or more digits anywhere in the string.

In other words, the ^ is part of the first alternative, it doesn't apply to second one.

A pattern that fulfills your requirements would be:

^\d+(#\d+)+$
  • match one or more digits a the start of the string followed by one or more sequences consisting of a # followed by one or more digits, till the end of the string

Upvotes: 3

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