AngularJS 过滤器去除某个正则表达式

Question

I am attempting to make an angularJS filter which will remove timestamps that look like this: (##:##:##) or ##:##:##.

This is a filter to remove all letters:

.filter('noLetter', function()  {
//this filter removes all letters
        return function removeLetters(string){
        return string.replace(/[^0-9]+/g, " ");
        }
        })

This is my attempt to make a filter that removes the time stamps, however it is not working, help is much appreciated.

.filter('noStamps', function () {
  return function removeStamps(item) {
  return item.replace(/^\([0-9][0-9]:[0-9][0-9]:[0-9][0-9]\)$/i, "");
 }
})

My goal is for it to delete the timestamps it finds and leave nothing in their place.

edit based on question in comments: The time stamps are in the text so it would say "this is an example 21:20:19 of what I am 21:20:20 trying to do 21:20:22"

I would want this to be converted into "this is an example of what I am trying to do" by the filter.

Answer 1

You may use

/\s*\(?\b\d{2}:\d{2}:\d{2}\b\)?/g

See regex demo

Thre main points:

• The ^ (start of string) and $ (end of string) anchors should be removed so that the expression becomes unanchored, and can match input text partially.
• Global flag to match all occurrences
• Limiting quantifier {2} to shorten the regex (and the use of a shorthand class \\d helps shorten it, too)
• \\)? and \\(? are used with ? quantifier to match 1 or 0 occurrences of the round brackets.
• \\s* in the beginning "trims" the result (as the leading whitespace is matched).

JS snippet:

 var str = 'this is an example (21:20:19) of what I am 21:20:20 trying to do 21:20:22'; var result = str.replace(/\\s*\\(?\\b\\d{2}:\\d{2}:\\d{2}\\b\\)?/g, ''); document.getElementById("r").innerHTML = result;

 <div id="r"/>