Most of the existing answers predate the introduction of lookbehind assertions in JavaScript in 2018. You didn't specify how you wanted the delimiters to be included in the result. One typical use case would be sentences delimited by punctuation ([.?!]), where one would want the delimiters to be included at the ends of the resulting strings. This corresponds to the fourth case in the accepted answer, but as noted there, that solution only works for single characters. Arbitrary strings with the delimiters appended at the end can be formed with a lookbehind assertion:

'It is. Is it? It is!'.split(/(?<=[.?!])/)
/* [ 'It is.', ' Is it?', ' It is!' ] */

我还想到了这个解。不需要正则表达式，非常易读。

Const STR = "你好，世界，今天多么美好的一天balbla" const separatorIndex = str.indexOf("great") const parsedString = str.slice(separatorIndex) console.log (parsedString)

使用(正)前向，这样正则表达式就断言特殊字符存在，但实际上并不匹配它:

string.split(/<br \/>(?=&#?[a-zA-Z0-9]+;)/g);

看看它的实际应用:

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _bbbb < br / > &Dagger;亿”; 游戏机log(管柱。斯普利特(/ < br \/>(?=&#?[ a-zA-Z0-9] +) / (g));

如果将分隔符括在parantheses中，它将是返回数组的一部分。

string.split(/(<br \/>&#?[a-zA-Z0-9]+);/g);
// returns ["aaaaaa", "<br />&dagger;", "bbbb", "<br />&Dagger;", "cccc"]

根据你想保留的部分来改变你匹配的子组

string.split(/(<br \/>)&#?[a-zA-Z0-9]+;/g);
// returns ["aaaaaa", "<br />", "bbbb", "<br />", "cccc"]

您可以通过忽略字母的大小写来改进表达式 string.split (/ () & # ? [a-z0-9] +, / gi);

你可以这样匹配预定义的组:\d =[0-9]和\w = [a-zA-Z0-9_]。这意味着表达式可以是这样的。

string.split(/<br \/>(&#?[a-z\d]+;)/gi);

JavaScriptKit上有一个很好的正则表达式参考。

我对jichi的回答做了修改，并把它放在一个支持多字母的函数中。

String.prototype.splitAndKeep = function(separator, method='seperate'){
    var str = this;
    if(method == 'seperate'){
        str = str.split(new RegExp(`(${separator})`, 'g'));
    }else if(method == 'infront'){
        str = str.split(new RegExp(`(?=${separator})`, 'g'));
    }else if(method == 'behind'){
        str = str.split(new RegExp(`(.*?${separator})`, 'g'));
        str = str.filter(function(el){return el !== "";});
    }
    return str;
};

jichi的答案第三种方法在这个函数中不起作用，所以我采用了第四种方法，并删除了空白区域以获得相同的结果。

编辑: 第二种方法，除了一个数组来拆分char1或char2

String.prototype.splitAndKeep = function(separator, method='seperate'){
    var str = this;
    function splitAndKeep(str, separator, method='seperate'){
        if(method == 'seperate'){
            str = str.split(new RegExp(`(${separator})`, 'g'));
        }else if(method == 'infront'){
            str = str.split(new RegExp(`(?=${separator})`, 'g'));
        }else if(method == 'behind'){
            str = str.split(new RegExp(`(.*?${separator})`, 'g'));
            str = str.filter(function(el){return el !== "";});
        }
        return str;
    }
    if(Array.isArray(separator)){
        var parts = splitAndKeep(str, separator[0], method);
        for(var i = 1; i < separator.length; i++){
            var partsTemp = parts;
            parts = [];
            for(var p = 0; p < partsTemp.length; p++){
                parts = parts.concat(splitAndKeep(partsTemp[p], separator[i], method));
            }
        }
        return parts;
    }else{
        return splitAndKeep(str, separator, method);
    }
};

用法:

str = "first1-second2-third3-last";

str.splitAndKeep(["1", "2", "3"]) == ["first", "1", "-second", "2", "-third", "3", "-last"];

str.splitAndKeep("-") == ["first1", "-", "second2", "-", "third3", "-", "last"];

扩展函数将字符串与子字符串或RegEx分开，分隔符根据第二个参数放在前面或后面。

    String.prototype.splitKeep = function (splitter, ahead) {
        var self = this;
        var result = [];
        if (splitter != '') {
            var matches = [];
            // Getting mached value and its index
            var replaceName = splitter instanceof RegExp ? "replace" : "replaceAll";
            var r = self[replaceName](splitter, function (m, i, e) {
                matches.push({ value: m, index: i });
                return getSubst(m);
            });
            // Finds split substrings
            var lastIndex = 0;
            for (var i = 0; i < matches.length; i++) {
                var m = matches[i];
                var nextIndex = ahead == true ? m.index : m.index + m.value.length;
                if (nextIndex != lastIndex) {
                    var part = self.substring(lastIndex, nextIndex);
                    result.push(part);
                    lastIndex = nextIndex;
                }
            };
            if (lastIndex < self.length) {
                var part = self.substring(lastIndex, self.length);
                result.push(part);
            };
            // Substitution of matched string
            function getSubst(value) {
                var substChar = value[0] == '0' ? '1' : '0';
                var subst = '';
                for (var i = 0; i < value.length; i++) {
                    subst += substChar;
                }
                return subst;
            };
        }
        else {
            result.add(self);
        };
        return result;
    };

测试:

    test('splitKeep', function () {
        // String
        deepEqual("1231451".splitKeep('1'), ["1", "231", "451"]);
        deepEqual("123145".splitKeep('1', true), ["123", "145"]);
        deepEqual("1231451".splitKeep('1', true), ["123", "145", "1"]);
        deepEqual("hello man how are you!".splitKeep(' '), ["hello ", "man ", "how ", "are ", "you!"]);
        deepEqual("hello man how are you!".splitKeep(' ', true), ["hello", " man", " how", " are", " you!"]);
        // Regex
        deepEqual("mhellommhellommmhello".splitKeep(/m+/g), ["m", "hellomm", "hellommm", "hello"]);
        deepEqual("mhellommhellommmhello".splitKeep(/m+/g, true), ["mhello", "mmhello", "mmmhello"]);
    });