Javascript’这个’

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);
  };
}

原文地址：https://www.jb51.cc/js/149910.html