javascript正则表达式排除字符

Question

I have text like this

ab:_c

a:_c

The regex that I am using is /:(.*)/g , but it also captures ':_'

I want to achieve--> c

What I've got--> :_c

How do I exclude them?

PS: replaced space with underscore , for easy understanding.

EDIT: I want to capture everything behind ': '

EDIT2: here is regexr with text, as u can see, it also captures ': '

Answer 1

You can use: /:\\s(.*)/g .

That will capture a single space after a : before starting to capture:

a b: c => "c"
a: c   => "c"

 var re = /:\\s(.*)/; var tests = [ 'a: c', 'a: cat', 'ba: test', 'a: c', 'Test: this is the rest of the sentence', ]; for (var i = 0; i < tests.length; i++) { console.log('TEST: ', '"' + tests[i] + '"'); console.log('RESULT: "' + tests[i].match(re)[1] + '"'); } 

Note: You must get the first capture group from the match. If you do not, you will not get the desired result. See How do you access the matched groups in a JavaScript regular expression? for information on how to do this.

Answer 2

If you know that underscore will always prefix the c you could do this: /:_?(.*)/g

The _? matches 0 or 1 underscores, so you will only get c as a result.

 console.log('ab:_c'.match(/:_?(.*)/)[1]) 

Answer 3

Do you want to get the string "c" or the last character?

var a = "a:_c a b:_c";

/c/g.exec(a)

/.$/g.exec(a)

Edit to match comment:

Use the following to remove everything before and including ":_"

a.replace(/.*:_*/, '')

For space:

a.replace(/.*:\s*/, '')