格式化字符串并用正则表达式替换

Question

I have a number that's at least 7 digits long. Typical examples: 0000123 , 00001234 , 000012345

I want to transform them so that they become respectively: 01:23 , 12:34 , 23:45

Which mean replacing the whole string by the last 4 characters and putting a colon in the middle.

I can get the last 4 digits with (\\d{4})$ And I can get 2 groups with this: (\\d{2})(\\d{2})$

With the last option, on a string 0000123 $1:$2 match gives me 00001:23 where I want 01:23

I replace the string like so:

newVal = val.replace(/regex/, '$1:$2');

Answer 1

You need to match the beginning digits with \\d* (or with just .* if there can be anything):

 var val = "0001235"; var newVal = val.replace(/^\\d*(\\d{2})(\\d{2})$/, '$1:$2'); console.log(newVal); 

Pattern details :

• ^ - start of string
• \\d* - 0+ digits (or .* will match any 0+ chars other than line break chars)
• (\\d{2}) - Group 1 capturing 2 digits
• (\\d{2}) - Group 2 capturing 2 digits
• $ - end of string.

Answer 2

As Alex K. said , no need for a regular expression, just extract the parts you need with substr :

val = val.substr(-4, 2) + ":" + val.substr(-2);

Note that when the starting index is negative, it's from the end of the string.

Example:

 function update(val) { return val.substr(-4, 2) + ":" + val.substr(-2); } function test(val) { console.log(val + " => " + update(val)); } test("0000123"); test("0001234"); test("000012345"); 

Answer 3

Use this regex: ^(\\d+?)(\\d{2})(\\d{2})$ :

 var newVal = "0000123".replace(/^(\\d+?)(\\d{2})(\\d{2})$/, '$2:$3'); console.log(newVal); 

Answer 4

您可以丢掉第一个字符，而仅替换最后一个匹配的部分。

 console.log('00000001234'.replace(/^(.*)(\\d{2})(\\d{2})$/, '$2:$3'));