Regex: get number formated n.n.n from string using javascript?

Question

I am trying to get number i.e. 1.0.3 from string . I only want numbers that are formatted with two dots and have ver# before them. Is my regex implementation correct. It is work but will it fail in any condition?

var str = "https://example.x.y.com/z/ver#1.5.0"; 
var res = str.match(/ver#.\.(?:(\d+)\.)?(?:(\d+)\.)?(\*|\d+)/g);
return res;

https://jsfiddle.net/tthfkzjt/

I guess even a simpler regex like this #\d+\.\d+.\d+ will be suffice. Demo — user2705585
– user2705585, Commented Mar 11, 2016 at 14:26
As @noob posted - regex101.com/r/vJ4qU1/2. That one will not match ver#a.0.5 whereas your original one does match ver#a.0.5. You are probably better off using theirs. — JosephGarrone
– JosephGarrone, Commented Mar 11, 2016 at 14:28

score 1 · Accepted Answer · 2016-03-14 16:54:49Z

1

I am trying to get number i.e. 1.0.3 from string . I only want numbers that are formatted with two dots and have ver# before them

This could be done by simple regex: /ver#(\d+\.\d+\.\d+)/

Capture the first group using \1 or $1.

Regex101 Demo

JS Fiddle

var str = "https://example.x.y.com/z/ver#1.5.0"; 
var res = str.match(/ver#(\d+\.\d+\.\d+)/);
document.getElementById("res").innerHTML = res ? res[1] : "";

<div id="res"/>

edited Mar 14, 2016 at 16:54

answered Mar 11, 2016 at 14:47

user2705585

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4 Comments

bitecoder Over a year ago

Second dot should be escaped so it matches dot literally instead of matching any character. /ver#(\d+\.\d+\.\d+)/. @noob can you edit it. Thanks!

user2705585 Over a year ago

@Wiktor Stribiżew: How did we missed this obvious thing ? Was there any reason not to escape second dot ?

Wiktor Stribiżew Over a year ago

If there is more than one issue in a solution, the possibility of overlooking one of them is very high. In this case, I was not focused on the expression itself, just on the JS code: I just blindly copied the pattern to jsfiddle.net from regex101.com and pasted here.

user2705585 Over a year ago

@bitecoder: Thanks for pointing out the mistake. Appreciate it !

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