2014-01-20 21 views
0

我有以下代码,如何识别scheduleVariableTest参数?当我console.log this.scheduleSettings它显示我的对象,我似乎无法找到正确的选择器来显示测试。显示内部数组的对象参数javascript

我已经试过this.scheduleSettings.scheduleVariableTest,this.scheduleSettings [0] .scheduleVariableTest,它们都返回undefined。 =/

var ReservationSchedulePicker = function(reservationType){ 
    this.scheduleSettings = []; // Main settings for Schedule Picker 
    this.scheduleSettings[''] = { // Default Main Settings for Schedule Picker 
     scheduleReservationType : reservationType, 
     scheduleVariableTest : "Testing", 
     schedulePickerID : $("#schedulePicker"), 
     scheduleMarkup : "", 
     schedulePickerTotalCostID : $("#schedulePickerTotal"), 
     schedulePickerTotalCost : "", 
     schedulePickerErrors : [], 
     schedulePickerErrorsID : $("#schedulePickerErrors") 
    }; 

感谢您提前提供任何帮助!

+0

如果你在一个函数里面定义了一个'var scheduleSettings'变量,它只能在这个函数中使用。如果你在函数之外定义它,或者没有'var',它就是无处不在的。看看我的答案 - 我在外面定义它,因为我不知道你想在哪里使用这个对象。 – Jurik

回答

0
scheduleSettings = {}; 

var ReservationSchedulePicker = function(reservationType){ 
    scheduleSettings = { // Default Main Settings for Schedule Picker 
     scheduleReservationType : reservationType, 
     scheduleVariableTest : "Testing", 
     schedulePickerID : $("#schedulePicker"), 
     scheduleMarkup : "", 
     schedulePickerTotalCostID : $("#schedulePickerTotal"), 
     schedulePickerTotalCost : "", 
     schedulePickerErrors : [], 
     schedulePickerErrorsID : $("#schedulePickerErrors") 
    }; 
}; 

ReservationSchedulePicker('wohoo'); 
console.log(scheduleSettings.scheduleReservationType) // output: 'wohoo'