获取字符串中的重复字符 [英] Get duplicate characters in string

问题描述

我尝试匹配/获取字符串中的所有重复。这是我到目前为止所做的：

  var str ='abcabc123123'; 
 var REPEATED_CHARS_REGEX = /(.).*\1/gi; 
 
 console.log（str.match（REPEATED_CHARS_REGEX））; // => ['abca'，'1231'] 
  

正如你可以看到的匹配结果是 ['abca'，'1231'] ，但我可以得到 ['abc'，'123'] 。

，是可以更改字符串中需要多长时间获得匹配的持续时间 ...

例如如果字符串 abcabcabc 和重复时间设置为 2 在 ['abcabc'] 中。如果设置为 3 ，则应为 ['abc'] 。

$ b b

更新

非 - RegExp / p>

解决方案

好吧，我想 falsetru 有一个好主意，零宽度前瞻。

 'abcabc123123'.match（/(.+ ）（？= \1）/ g）
 // [abc，123] 
  

这允许它只匹​​配初始子字符串，同时确保至少重复1次。

对于 M42 的后续示例中，可以使用。*？修改以允许重复之间的间隙。 / p>

 'abc123ab12'.match（/(.+)(?=.*？\1）/ g）
 // [ab，12] 
  

然后，可以为捕获组添加量词（ {n} ）：

 'abcabc1234abc'.match（/（。+）{2}（？=。*？\1）/ g）
 // [abcabc] 
  

或者，为了仅匹配初始值与后面的重复次数，请在预测内添加量词。



 'abc123ab12ab'.match（/(.+)(?=(.*？\1）{2}）/ g）
 // [ab] 
  

它也可以匹配最小重复次数不带max的范围量词符 - {2，}

 'abcd1234ab12cd34bcd234 '.match（/(.+)(?=(.*？\1）{2，}）/ g）
 // [b，cd，2，34 ] 
  




I try to match/get all repetitions in a string. This is what I've done so far:

var str = 'abcabc123123';
var REPEATED_CHARS_REGEX = /(.).*\1/gi;

console.log( str.match(REPEATED_CHARS_REGEX) ); // => ['abca', '1231']

As you can see the matching result is ['abca', '1231'], but I excpect to get ['abc', '123']. Any ideas to accomplish that?

2nd question:

Another thing I excpect, is to make it possible to change the duration how often a char needs to be in the string to get matched...

For example if the string is abcabcabc and the repetation-time is set to 2 it should result in ['abcabc']. If set to 3 it should be ['abc'].

Update

A non-RegExp solution is perfectly alright!

解决方案

Well, I think falsetru had a good idea with a zero-width look-ahead.

'abcabc123123'.match(/(.+)(?=\1)/g)
// ["abc", "123"]

This allows it to match just the initial substring while ensuring at least 1 repetition follows.

For M42's follow-up example, it could be modified with a .*? to allow for gaps between repetitions.

'abc123ab12'.match(/(.+)(?=.*?\1)/g)
// ["ab", "12"]

Then, to find where the repetition starts with multiple uses together, a quantifier ({n}) can be added for the capture group:

'abcabc1234abc'.match(/(.+){2}(?=.*?\1)/g)
// ["abcabc"]

Or, to match just the initial with a number of repetitions following, add the quantifier within the look-ahead.

'abc123ab12ab'.match(/(.+)(?=(.*?\1){2})/g)
// ["ab"]

It can also match a minimum number of repetitions with a range quantifier without a max -- {2,}

'abcd1234ab12cd34bcd234'.match(/(.+)(?=(.*?\1){2,})/g)
// ["b", "cd", "2", "34"]

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