Question

在Javascript中如何在回调上传递变量大小的参数是否有灵活而通用的方法？

var myCallback = function (param1, param2){
    ...
};
var myOtherCallback = function (param1, param2, param3){
    ...
};

var func = function(callback, callbackParams){
    // how can I put the values on callbackParams as the actual callback parameters down here?
    callback( /* callbackParams */ );
};

func(myCallback, [myparam1, myparam2]);
func(myOtherCallback, [newparam1, newparam2, newparam3]);

Answer 1

您可以在函数中使用spread operator。话虽这么说，我的例子中没有看到任何回调用途。我假设你故意简化你的问题，向我们解释。

var myCallback = function (param1, param2){
    ...
};
var myOtherCallback = function (param1, param2, param3){
    ...
};

var func = function(callback, callbackParams){
    // how can I put the values on callbackParams as the actual callback parameters down here?
    callback(...callbackParams);
};

func(myCallback, [myparam1, myparam2]);
func(myOtherCallback, [newparam1, newparam2, newparam3]);

Answer 2

我找到了我要找的东西......

var myCallback = function (param1, param2){
    console.log("myCallback", param1, param2)
};
var myOtherCallback = function (param1, param2, param3){
    console.log("myCallback", param1, param2, param3)
};

var func = function(callback){
    var additionalParams = Array.prototype.slice.call(arguments, 2);
    callback.apply(this, additionalParams);
};

func(myCallback, "myparam1", "myparam2");
func(myOtherCallback, "newparam1", "newparam2", "newparam3");

虽然对于更现代的方法，Kuu Aku的答案可能会更好

如何在Javascript中的回调上传递变量大小的参数

2 个答案: