尽管这不是最简单的方法，但你可以这样做:

var addressString = "~john smith~123 Street~Apt 4~New York~NY~12345~",
    keys = "name address1 address2 city state zipcode".split(" "),
    address = {};

// clean up the string with the first replace
// "abuse" the second replace to map the keys to the matches
addressString.replace(/^~|~$/g).replace(/[^~]+/g, function(match){
    address[ keys.unshift() ] = match;
});

// address will contain the mapped result
address = {
    address1: "123 Street"
    address2: "Apt 4"
    city: "New York"
    name: "john smith"
    state: "NY"
    zipcode: "12345"
}

更新ES2015，使用解构

const [address1, address2, city, name, state, zipcode] = addressString.match(/[^~]+/g);

// The variables defined above now contain the appropriate information:

console.log(address1, address2, city, name, state, zipcode);
// -> john smith 123 Street Apt 4 New York NY 12345

因为分隔逗号的问题被复制到这个问题上，所以在这里添加这个。

如果你想对一个字符进行拆分，并处理该字符后面可能出现的额外空格，通常使用逗号，你可以使用替换然后拆分，如下所示:

var items = string.replace(/,\s+/, ",").split(',')

嗯，最简单的方法是:

var address = theEncodedString.split(/~/)
var name = address[0], street = address[1]

根据ECMAScript6 ES6，干净的方法是解构数组:

const input = 'john smith~123 Street~Apt 4~New York~NY~12345'; Const[名称，街道，单位，城市，州，zip] = input.split('~'); console.log(名称);// John Smith console.log(街);// 123街 console.log(单位);// 4号房间 console.log(城市);//纽约 console.log(状态);/ /纽约 console.log (zip);/ / 12345

输入字符串中可能有额外的项。在这种情况下，你可以使用rest操作符获取一个数组，或者直接忽略它们:

const input = 'john smith~123 Street~Apt 4~New York~NY~12345'; Const[名字，街道，…Others] = input.split('~'); console.log(名称);// John Smith console.log(街);// 123街 console.log(他人);// ["Apt 4"， "New York"， "NY"， "12345"]

我假设值是只读引用，并使用了const声明。

党委ES6 !

扎克是对的。使用他的方法，你也可以做出一个看似“多维”的数组。我在JSFiddle http://jsfiddle.net/LcnvJ/2/上创建了一个快速示例

// array[0][0] will produce brian
// array[0][1] will produce james

// array[1][0] will produce kevin
// array[1][1] will produce haley

var array = [];
    array[0] = "brian,james,doug".split(",");
    array[1] = "kevin,haley,steph".split(",");

这个答案不如破坏性的答案好，但鉴于这个问题是12年前提出的，我决定给出一个12年前也适用的答案。

function Record(s) {
    var keys = ["name", "address", "address2", "city", "state", "zip"], values = s.split("~"), i
    for (i = 0; i<keys.length; i++) {
        this[keys[i]] = values[i]
    }
}

var record = new Record('john smith~123 Street~Apt 4~New York~NY~12345')

record.name // contains john smith
record.address // contains 123 Street
record.address2 // contains Apt 4
record.city // contains New York
record.state // contains NY
record.zip // contains zip