你能解释一下为什么fn的第二次调用会出错吗？代码如下.
function Test(n) {
this.test = n;
var bob = function (n) {
this.test = n;
};
this.fn = function (n) {
bob(n);
console.log(this.test);
};
}
var test = new Test(5);
test.fn(1); // returns 5
test.fn(2); // returns TypeError: 'undefined' is not a function
这是一个重现错误http://jsfiddle.net/KjkQ2/的JSfiddle
解决方法
您的bob函数是从全局范围调用的.因此,this.test指向一个名为test的全局变量,它覆盖您创建的变量.如果你运行console.log(window.test),你会发生什么.
为了使代码按预期运行,您需要以下其中一项
function Test(n) {
this.test = n;
// If a function needs 'this' it should be attached to 'this'
this.bob = function (n) {
this.test = n;
};
this.fn = function (n) {
// and called with this.functionName
this.bob(n);
console.log(this.test);
};
}
要么
function Test(n) {
this.test = n;
var bob = function (n) {
this.test = n;
};
this.fn = function (n) {
// Make sure you call bob with the right 'this'
bob.call(this,n);
console.log(this.test);
};
}
OR基于闭包的对象
// Just use closures instead of relying on this
function Test(n) {
var test = n;
var bob = function (n) {
test = n;
};
this.fn = function (n) {
bob(n);
console.log(test);
};
}
