string.search（）忽略重音字符？ [英] string.search() that ignores accented characters?

问题描述

我需要将重音字符视为与非重音字符相同。这是我的代码：

I need to treat accented characters as if they were the same as their non accented counterparts. This is my code:

var re = new RegExp(string, 'i');
if(target.search(re) == 0) { }

目前忽略在角色的情况下，我如何忽略角色的重音？

It currently ignores the character's case, how do I also ignore if the character is accented or not?

推荐答案

我认为你必须删除重音符号首先做你的RegExp。

您可以使用此功能，我发现此处：

I think you have to remove the accents first then do your RegExp.
You can use this function taht I found here :

function stripVowelAccent(str)
{
 var rExps=[
 {re:/[\xC0-\xC6]/g, ch:'A'},
 {re:/[\xE0-\xE6]/g, ch:'a'},
 {re:/[\xC8-\xCB]/g, ch:'E'},
 {re:/[\xE8-\xEB]/g, ch:'e'},
 {re:/[\xCC-\xCF]/g, ch:'I'},
 {re:/[\xEC-\xEF]/g, ch:'i'},
 {re:/[\xD2-\xD6]/g, ch:'O'},
 {re:/[\xF2-\xF6]/g, ch:'o'},
 {re:/[\xD9-\xDC]/g, ch:'U'},
 {re:/[\xF9-\xFC]/g, ch:'u'},
 {re:/[\xD1]/g, ch:'N'},
 {re:/[\xF1]/g, ch:'n'} ];

 for(var i=0, len=rExps.length; i<len; i++)
  str=str.replace(rExps[i].re, rExps[i].ch);

 return str;
}

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