Angular regular expression that identifies two words enclosed by ->

Does anyone have a regular expression that can match various types of text strings, including empty chains and word combinations?

"" (empty chain)

word1->word2

word1 -> word2

succesives \r\s\t

\n

etc. The words can contain digits, # symbols, or any symbol except \n.

I've tried using the following regex:

const expreg = new RegExp('([^"->"\n]->[^"->"\n])|(([^"->"\n]->[^"->"\n])\n)*');

But it's not capturing all the variations I need, such as

word1->

and it doesn't work for hello1->hello2 either.

((^(\s\r\t|\n|->)^(->|\n)->^(\s|\r|\t|\n|->)^(->|\n))|(^(\s|\r|\t|\n|->)^(->|\n)->^(\s|\r|\t|\n|->)^(->|\n)\n)|(\n)|([\s\r\t]+)|([\s\r\t]+\n))

angularregextypescript

Answer №1

To match one or more word characters, followed by a repeating pattern of "->" and one or more word characters again.

The entire expression can also match an empty string if needed.

^(?:[^->]+[^\S\r\n]*->[^\S\r\n]*[^->]+)?$

Check out the regex demo

const regex = /^(?:[^->]+[^\S\r\n]*->[^\S\r\n]*[^->]+)?$/m;
[
  "word1->word2",
  "",
  "word1  ->   word2",
  "my django db->my django db",
  "tds->dfds\nmy->computer",
  "word1->",
  "word1  ->   word2 ->word3"
].forEach(str => console.log(`${regex.test(str)} "${str}"`));

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