Javascript

Question

我需要一个函数，它的参数是一个对象，如果我把它留空它会加载默认值。

类似于：

function loadMap(args) { //args is an object and its optional

   //this is what I want to do, test args and load defauts if necesary
   /*
   pseudocode:
   if args.latitude is not set then
       give it a default value

   if args.longitude is not set then
       give it a default value

    end pseudocode */

   alert("Latitude is "+args.latitude );
   alert("Longitude is "+args.longitude );
}

//outputs the default values
loadMap();

//outputs custom values
loadMap({latitude: "x.xxxx", longitude: "-x.xxxx"});

//and should work too and output one default and one custom value
loadMap({latitude: "x.xxxx"});

我在这个问题 (Is there a better way to do optional function parameters in Javascript?) 中找到了这个解决方案，但它需要 jQuery： http://jsfiddle.net/xy84kwdv/

这几乎是我想要的，但我不想依赖 jquery 函数。

Answer 1

您是否在寻找类似的东西：

function loadMap(args) { 
    args = args || {}; 
    args.latitude = args.latitude || "X.XXX"; 
    args.longitude = args.longitude || "Y.YYY"; 
    alert(args.latitude);
    alert(args.longitude);
}; 

loadMap({latitude:"custom_latitude"});

Answer 2

How to overload functions in javascript? 查看OP的第二个答案。它提供了使用 jQ 的替代方法。

注意：如果 args 通过但计算结果为真，Kim 的代码将导致逻辑错误。更好的解决方法： optionalArg = (typeof optionalArg === "undefined") ? "defaultValue"：可选参数；

Answer 3

这使得参数对象成为可选的，并且使各个属性成为可选的：

var defaults = { latitude: x, longitude: y };
function loadMap(args) {
    args = args || {};
    for (var key in defaults) {
        if (!(key in args)) {
            args[key] = default[key];
        }
    }
    ...
}

Answer 4

矫枉过正，引入全面的对象克隆 (ECMAScript 5.1)

function cloneOwnProperties(A, B, overwrite) { // this is shallow
    var i, p;
    if (!(A instanceof Object)) {
        if (undefined === A) A = {};
        else throw new TypeError('cloneOwnProperties called on non-object');
    }
    if (!B) B = {};
    p = Object.getOwnPropertyNames(A);
    for (i = 0; i < p.length; ++i)
        if (overwrite || !(p[i] in B))
            Object.defineProperty(B, p[i], Object.getOwnPropertyDescriptor(A, p[i]));
    return B;
}

// and for completeness
function cloneObject(obj_in, clone_seal, clone_frozen) {
    var obj_out;
    if (!(obj_in instanceof Object))
        throw new TypeError('cloneObject called on non-object');
    obj_out = Object.create(Object.getPrototypeOf(obj_in));
    cloneOwnProperties(obj_in, obj_out, true);
    if (clone_seal && Object.isSealed(obj_in)) Object.seal(obj_out);
    if (clone_frozen && Object.isFrozen(obj_in)) Object.freeze(obj_out);
    return obj_out;
}
// back on topic

现在您可以使用它来设置一些默认值

function foo(bar) {
    bar = cloneOwnProperties(bar, {
        fizz: true,
        buzz: false
    }, true);
    return bar;
}

foo({buzz: 'user_value'}); // {fizz: true, buzz: "user_value"}

Answer 5

这个问题肯定已经超过 3 年了（2018-03-02）。我搜索了这个主题，但我得到的大部分问题和排名靠前的答案似乎已经过时，例如 this, this, this, or .

目前，lots of browsers support ES6/ECMAScript 2015(the 6th Edition of ECMAScript). For compatibility, you can use WebpackJs(or another packer) and BabelJs(or another compiler) to compile and pack your ES6 codes to ES5. And the ES6-version answer for this question may be Object Destructuring。

function drawES2015Chart({size = 'big', cords = {x: 0, y: 0}, radius = 25} = {}) {
    console.log(size, cords, radius);
    // do some chart drawing
}

drawES2015Chart({cords: {x: 18, y: 30}, radius: 30});

function fn(requiredArg, {optionalArg = 'Default Value', a = 1, b ={some: 'object'}} = {}){
    // Do with destructured values: requiredArg, optionalArg, a, and b.
    console.log(requiredArg, optionalArg, a, b);
}

所以对于你的问题，它将是：

function loadMap({latitude = "x.xxxx", longitude = "-x.xxxx"} = {}) {
    console.log('latitude is:', latitude, 'and longitude is:', longitude);
    console.log(`latitude is: ${latitude}, and longitude is: ${longitude}`);
}

// Call the function.
loadMap({latitude: "1.2345", longitude: "-6.7890"});

Answer 6

Fisher 的答案应该是公认的答案。另外，如果你想重命名任何解构的变量，同时仍然保持默认值，你可以这样做：

function findByCoordinates({coordinates: cords = {x: 0, y: 0}} = {}) {
  console.log(cords);
  // Some calculation...
}

Answer 7

使用“??”而不是“||”或者你可能在未定义或错误的变量中有错误

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Nullish_coalescing_operator

function loadMap(args) { 
    args = args ?? {}; 
    args.latitude = args.latitude ?? "X.XXX"; 
    args.longitude = args.longitude ?? "Y.YYY"; 
    alert(args.latitude);
    alert(args.longitude);
}; 

loadMap({latitude:"custom_latitude"});

Javascript - 以可选参数作为对象的函数？

Javascript - function with optional parameters as object?

object

parameters

optional