如何从字符串中删除重音字符? 特别是在IE6中,我有这样的东西:

accentsTidy = function(s){
    var r=s.toLowerCase();
    r = r.replace(new RegExp(/\s/g),"");
    r = r.replace(new RegExp(/[àáâãäå]/g),"a");
    r = r.replace(new RegExp(/æ/g),"ae");
    r = r.replace(new RegExp(/ç/g),"c");
    r = r.replace(new RegExp(/[èéêë]/g),"e");
    r = r.replace(new RegExp(/[ìíîï]/g),"i");
    r = r.replace(new RegExp(/ñ/g),"n");                
    r = r.replace(new RegExp(/[òóôõö]/g),"o");
    r = r.replace(new RegExp(/œ/g),"oe");
    r = r.replace(new RegExp(/[ùúûü]/g),"u");
    r = r.replace(new RegExp(/[ýÿ]/g),"y");
    r = r.replace(new RegExp(/\W/g),"");
    return r;
};

但是IE6让我很烦,它好像不喜欢我的正则表达式。


当前回答

$scope.legal_name = $sanitize($scope.legal_name);
    console.log("Name before function...",$scope.legal_name);

    var str = "";
    for(var i=0; i < $scope.legal_name.length; i++) {
        var charName = $scope.legal_name.charAt(i);
        if((charName == '&') && ($scope.legal_name.charAt(i + 1) == '#')){
            var count = 0;
            var subString = "";
            while(true) {
                if(($scope.legal_name.charCodeAt(i + 2 + count) > 47) && ($scope.legal_name.charCodeAt(i + 1 + count) < 58 )) {
                    subString = subString + $scope.legal_name.charAt(i + 2 + count);
                    count++;
                } else {
                    if(subString.length > 0) {
                        var value = parseInt(subString);
                        str = str + String.fromCharCode(value);
                        i = i + 1 + count; 
                        break;
                    }
                }
            }               
        } else {
            str = str + charName;
        }
    }
    $scope.legal_name = str;
    console.log("Name After function...",str);

其他回答


function removeAccents(strAccents){
    strAccents = strAccents.split('');
    strAccentsOut = new Array();
    strAccentsLen = strAccents.length;
    var accents = 'ÀÁÂÃÄÅàáâãäåÒÓÔÕÕÖØòóôõöøÈÉÊËèéêëðÇçÐÌÍÎÏìíîïÙÚÛÜùúûüÑñŠšŸÿýŽž';
    var accentsOut = ['A','A','A','A','A','A','a','a','a','a','a','a','O','O','O','O','O','O','O','o','o','o','o','o','o','E','E','E','E','e','e','e','e','e','C','c','D','I','I','I','I','i','i','i','i','U','U','U','U','u','u','u','u','N','n','S','s','Y','y','y','Z','z'];
    for (var y = 0; y < strAccentsLen; y++) {
        if (accents.indexOf(strAccents[y]) != -1) {
            strAccentsOut[y] = accentsOut[accents.indexOf(strAccents[y])];
        }
        else
            strAccentsOut[y] = strAccents[y];
    }
    strAccentsOut = strAccentsOut.join('');
    return strAccentsOut;
}

假设你知道你在做什么,我怀疑IE6没有正确地解释文件的编码,因此不能识别文件中的非ascii字符:

确保文件保存为UTF-8格式(例如) 使用Fiddler或其他工具检查web服务器是否发送正确的内容编码HTTP报头。

(虽然它“闻起来”不对,但我会考虑做排序,比如在服务器上使用一些locale感知的东西……但无论如何…)

通过给定的测试,一个解决方案似乎要快得多:http://jsperf.com/diacritics/9

function removeDiacritics(str) {
   return str.replace(/[^A-Za-z0-9\s]+/g, function(a){
      return diacriticsMap[a] || a; 
   });
}
removeDiacritics(teste);

工作示例:http://jsbin.com/sovorute/1/edit

推理:这样做速度更快的一个原因是,我们只遍历由负正则表达式模式选择的特殊字符。最快的测试(不带in的字符串迭代)在给定文本上迭代1001,这意味着每个字符。这个函数只迭代了35次,输出了相同的结果。请记住,这将只替换地图中所指示的内容。

关于这个主题的经典文章:http://alistapart.com/article/accent-folding-for-auto-complete

来源:http://semplicewebsites.com/removing-accents-javascript,也提供了一个不错的人物地图。

我发现所有这些都有点笨拙,而且我不太擅长正则表达式,所以这里有一个更简单的版本。将它翻译成你最喜欢的服务器端语言是很容易的,假设字符串已经在Unicode中:

// String containing replacement characters for stripping accents 
var stripstring = 
    'AAAAAAACEEEEIIII'+
    'DNOOOOO.OUUUUY..'+
    'aaaaaaaceeeeiiii'+
    'dnooooo.ouuuuy.y'+
    'AaAaAaCcCcCcCcDd'+
    'DdEeEeEeEeEeGgGg'+
    'GgGgHhHhIiIiIiIi'+
    'IiIiJjKkkLlLlLlL'+
    'lJlNnNnNnnNnOoOo'+
    'OoOoRrRrRrSsSsSs'+
    'SsTtTtTtUuUuUuUu'+
    'UuUuWwYyYZzZzZz.';

function stripaccents(str){
    var answer='';
    for(var i=0;i<str.length;i++){
        var ch=str[i];
        var chindex=ch.charCodeAt(0)-192;   // Index of character code in the strip string
        if(chindex>=0 && chindex<stripstring.length){
            // Character is within our table, so we can strip the accent...
            var outch=stripstring.charAt(chindex);
            // ...unless it was shown as a '.'
            if(outch!='.')ch=outch;
        }
        answer+=ch;
    }
    return answer;
}

基于Ian Elliott的优秀解决方案的缩短代码:

accentsTidy = function(s){
    var r = s.toLowerCase();
    non_asciis = {'a': '[àáâãäå]', 'ae': 'æ', 'c': 'ç', 'e': '[èéêë]', 'i': '[ìíîï]', 'n': 'ñ', 'o': '[òóôõö]', 'oe': 'œ', 'u': '[ùúûűü]', 'y': '[ýÿ]'};
    for (i in non_asciis) { r = r.replace(new RegExp(non_asciis[i], 'g'), i); }
    return r;
};

编辑:修正的无效代码