未定義存取jQuery選擇

我使用對象屬性作爲jQuery選擇有一個問題未定義存取jQuery選擇

define(["models/security/user", 'text!templates/security/registration.html'], function(SecurityUserModel, Template){ 

    var SecurityRegistrationView; 

    SecurityRegistrationView = Backbone.View.extend({ 
     initialize: function(){ 
      // setting model 
      this.model = new SecurityUserModel(); 

      // binding validation 
      Backbone.Validation.bind(this); 
      this.model.bind("validated:valid", this.valid); 
      this.model.bind("validated:invalid", this.invalid); 

      this.render(); 
     }, 
     form: { 
      "username": "#_user_username" 
      , "email": "#_user_email" 
      , "password": "#_user_password" 
     }, 
     render: function(){ 
      $(this.el).append(Template); 
     }, 
     events: { 
      "submit form": "submit" 
     }, 
     submit: function(e){ 
      e.preventDefault(); 

      this.model.set("username", $(this.form.username).val()); 
      this.model.set("password", $(this.form.email).val()); 
      this.model.set("email", $(this.form.password).val()); 
      this.model.validate(); 

      if (this.model.isValid) { 
       this.model.save(); 
      } 
     }, 
     valid: function(model, attrs){ 
      console.log(attrs[0]); 
      $(this.form[attrs[0]]).parent("div.control-group").addClass("success"); 
     }, 
     invalid: function(model, attrs){ 
      console.log(attrs[0]); 
      $(this.form[attrs[0]]).parent("div.control-group").addClass("error"); 
     } 
    }); 

    return SecurityRegistrationView; 
});

在鉻

控制檯輸出：

username registration.js:45 
Uncaught TypeError: Cannot read property 'username' of undefined registration.js:46 
password registration.js:45 
Uncaught TypeError: Cannot read property 'password' of undefined registration.js:46 
email registration.js:45 
Uncaught TypeError: Cannot read property 'email' of undefined registration.js:46 
username registration.js:45 
Uncaught TypeError: Cannot read property 'username' of undefined registration.js:46 



$(this.form.username).val(); // works 
$(this.form[someVar]); // does not work

來源

2012-07-10 dev.pus

id正確嗎？發佈HTML代碼與表格 – Naigel 2012-07-10 11:29:55

this在valid和invalid回調可能並不是指你的看法。將您的initialize方法中的綁定修改爲適當的範圍：

this.model.bind("validated:valid", this.valid, this); 
this.model.bind("validated:invalid", this.invalid, this);

來源

2012-07-10 12:00:53 nikoshr

這適用，至少爲console.log;）。父級的jQuery命令文件（「div.control.group」）語句。 .parent（「div」）。parent（「div」）解決了這個問題:) – 2012-07-10 12:05:45

如果你想避免.parent（）。parent（）我發現.closest（「div.control-group）只是旁邊:) – 2012-07-10 12:07:50

嘗試更換：

$(this.form.username)

With：

$(this.form['username'])

或者說，沒有「這個」：

$(form['username'])

同爲「密碼」和「電子郵件」。

來源

2012-07-10 11:58:32 CoursesWeb

未定義存取jQuery選擇

回答

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