我有下面的字符串

var a = "1,2,3,4";

当我…

var b = a.split(',');

我得到b为["1","2","3","4"]

我能做点什么让b等于[1,2,3,4]吗?


你可以使用数组。映射将每个元素转换为一个数字。

var a = "1,2,3,4";

var b = a.split(',').map(function(item) {
    return parseInt(item, 10);
});

检查文档


或者像User: thg435所指出的那样更优雅

var b = a.split(',').map(Number);

Number()将完成其余的工作:检查这里


注意:对于不支持map的旧浏览器,你可以自己添加一个实现:

Array.prototype.map = Array.prototype.map || function(_x) {
    for(var o=[], i=0; i<this.length; i++) { 
        o[i] = _x(this[i]); 
    }
    return o;
};

+string将尝试将字符串更改为数字。然后使用Array。映射函数来改变每个元素。

"1,2,3,4".split(',').map(function(el){ return +el;});

将其映射为整数:

a.split(',').map(function(i){
    return parseInt(i, 10);
})

Map查看每个数组项,将其传递给所提供的函数,并返回一个包含该函数返回值的数组。map在旧的浏览器中是不可用的,但是大多数库,如jQuery或underscore都有一个跨浏览器版本。

或者,如果你喜欢循环:

var res = a.split(",");
for (var i=0; i<res.length; i++)
{
    res[i] = parseInt(res[i], 10);
}

作为一种变体,你可以使用组合_。Map和_。来自lodash库的多种方法。整个转型将更加紧凑。以下是官方文档中的一个例子:

_.map(['6', '8', '10'], _.ary(parseInt, 1));
// → [6, 8, 10]

没有必要使用lambdas和/或给parseInt的基数参数,只需使用parseFloat或Number代替。

原因:

It's working: var src = "1,2,5,4,3"; var ids = src.split(',').map(parseFloat); // [1, 2, 5, 4, 3] var obj = {1: ..., 3: ..., 4: ..., 7: ...}; var keys= Object.keys(obj); // ["1", "3", "4", "7"] var ids = keys.map(parseFloat); // [1, 3, 4, 7] var arr = ["1", 5, "7", 11]; var ints= arr.map(parseFloat); // [1, 5, 7, 11] ints[1] === "5" // false ints[1] === 5 // true ints[2] === "7" // false ints[2] === 7 // true It's shorter. It's a tiny bit quickier and takes advantage of cache, when parseInt-approach - doesn't: // execution time measure function // keep it simple, yeah? > var f = (function (arr, c, n, m) { var i,t,m,s=n(); for(i=0;i++<c;)t=arr.map(m); return n()-s }).bind(null, "2,4,6,8,0,9,7,5,3,1".split(','), 1000000, Date.now); > f(Number) // first launch, just warming-up cache > 3971 // nice =) > f(Number) > 3964 // still the same > f(function(e){return+e}) > 5132 // yup, just little bit slower > f(function(e){return+e}) > 5112 // second run... and ok. > f(parseFloat) > 3727 // little bit quicker than .map(Number) > f(parseFloat) > 3737 // all ok > f(function(e){return parseInt(e,10)}) > 21852 // awww, how adorable... > f(function(e){return parseInt(e)}) > 22928 // maybe, without '10'?.. nope. > f(function(e){return parseInt(e)}) > 22769 // second run... and nothing changes. > f(Number) > 3873 // and again > f(parseFloat) > 3583 // and again > f(function(e){return+e}) > 4967 // and again > f(function(e){return parseInt(e,10)}) > 21649 // dammit 'parseInt'! >_<

注意:在Firefox中,parseInt的工作速度大约快4倍,但仍然比其他的慢。总共:+e < Number < parseFloat < parseInt


一个更简短的解决方案:映射并传递参数给Number:

Var a = "1,2,3,4"; Var b = a.s split(','); console.log (b); var c = b.map(Number); console.log (c);


下划线js的方式-

var a = "1,2,3,4",
  b = a.split(',');

//remove falsy/empty values from array after split
b = _.compact(b);
//then Convert array of string values into Integer
b = _.map(b, Number);

console.log('Log String to Int conversion @b =', b);

由于所有的答案都允许包含NaN,所以我想补充一句,如果您想快速地将一个混合值的数组转换为数字,您可以这样做。

var a = "1,2,3,4,foo,bar";

var b = a.split(',');

var result = b.map(_=>_|0) // Floors the number (32-bit signed integer) so this wont work if you need all 64 bits.

// or b.map(_=>_||0) if you know your array is just numbers but may include NaN.

使用arraw函数的Matt Zeunert版本(ES6)

const nums = a.split(',').map(x => parseInt(x, 10));

Array.from()详情请访问MDN

let a = "1,2,3,4";
let b = Array.from(a.split(','),Number);

or

let c = ["1", "2", "3", "4"].map(Number);

B和c是一个数字数组。

演示:

设a = "1,2,3,4"; let b = Array.from(a.s split(','),Number); let c = ["1", "2", "3", "4"].map(Number); Console.log (' b: ${b}, c: ${c} ');


这很简单。如:

["1", "2", "3", "4"].map(i=>Number(i))

您可以运行演示。 让结果=(“1”、“2”、“3”、“4”). map (i = > (i)); console.log(结果);


我对高尔夫球手的看法:

b="1,2,3,4".split`,`.map(x=>+x)

反引号是字符串字面值,所以我们可以省略圆括号(因为split函数的性质),但它相当于split(',')。字符串现在是一个数组,我们只需要用一个返回字符串整数的函数来映射每个值,因此x=>+x(它甚至比Number函数还要短(5个字符而不是6个字符))相当于:

function(x){return parseInt(x,10)}// version from techfoobar
(x)=>{return parseInt(x)}         // lambda are shorter and parseInt default is 10
(x)=>{return +x}                  // diff. with parseInt in SO but + is better in this case
x=>+x                             // no multiple args, just 1 function call

我希望它能更清楚一点。


一个衬套

Array.from(a.split(','), Number)

您可以使用JSON。解析,添加括号格式化数组

常量 a = “1,2,3,4”; const myArray = JSON.parse('[${a}]') console.log(myArray) console.info('pos 2 = ', myArray[2])


你可以在一行中将字符串数组转换为数字数组:

const arrayOfNumbers = arrayOfStrings.map(e => +e);

如果您需要将字符串数组转换为数字,这将非常有效。

const numbers = arr => arr.map(Number);
numbers(['1', '2', '3','4']);     // [1, 2, 3, 4]

let ar = [ '682',    '874',    '906',    '11168',  '73714',
  '74377',  '74034',  '138860', '138891', '139161', '139562',
  '139733', '139560', '74049',  '139759', '139934', '140104',
  '141335', '141356', '141334', '141337', '141360', '141358',
  '141365', '141419', '143333', '151477', '147342', '141355',
  '167847', '192141', '196760', '191687', '197351', '197055',
  '198852', '198731', '198816', '199034', '200053', '199226',
  '217818', '200055', '222039', '230533', '230530', '231127',
  '222042', '231100', '236171', '236913', '236980', '237015',
  '237016', '237052', '237551', '237560', '237590', '237637',
  '237733', '237731', '237655', '238890', '238910', '238837',
  '238926', '238972', '238925', '239755', '239696', '239898',
  '240037', '239909', '240036', '240082', '240097', '240526',
  '240770', '678151', '678950', '678985'];
let arry=[]
ar.map(arr=>{
arry.push(parseInt(arr))
});

console.log(arry);

使用Array.from,试试这个:

Let b = ["1", "2", "3", "4"]; b = Array.from(b,Number); console.log (b);