与如果在JavaScript

Question

声明，我决定努力增加以下if语句的JavaScript

if (form.address == "Vancouver, BC") { 
     user.set("address", "#complete address"); 
} else if (form.address == "Edmonton, AB") { 
     user.set("address", "complete address"); 
} else if (form.address == "Vancouver + United States") { 
     user.set("address", "#complete address And 1300 Boblett Street, US"); 
} else if (form.address == "Edmonton + United States") { 
     user.set("address", "complete address"); 
} 

，并接收在控制台上出现以下错误：

TypeError: object is not a function 

下面是整个代码

$scope.signUp = function(form) { 
    var user = new Parse.User(); 
    user.set("password", form.password); 
     user.set("username", form.username); 
       user.set("fullName", form.fname); 

     user.set("companyName", form.cname); 
     if (form.address == "Vancouver, BC") { 
     user.set("address", "#1001 - 610 Granville Street Vancouver, BC V6C 3J3"); 
} else if (form.address == "Edmonton, AB") { 
     user.set("address", "10915 65th Ave NW Edmonton, AB T6H 1W1"); 
} else if (form.address == "Vancouver + United States") { 
     user.set("address", "#1001 - 610 Granville Street Vancouver, BC V6C 3J3 And 1300 Boblett Street, Unit-B1569 Blaine, WA 98230"); 
} else if (form.address == "Edmonton + United States") { 
     user.set("address", "10915 65th Ave NW Edmonton, AB T6H 1W1 And 1300 Boblett Street, Unit-B1569 Blaine, WA 98230"); 
} 

      user.set("phoneNumber", form.pnumber); 
      user.set("mailingAddress", form.maddress); 
      user.set("plan", form.plan); 


    user.signUp(null, { 
     success: function(user) { 
    window.location = 'payment.php' 
     }, 
     error: function(user, error) { 
     var div = document.getElementById("signUpError"); 
    div.textContent = "Unable to sign up: " + error.message; 
    scroll(0,0); 
     } 
    });  
    }; 

本质上有一种形式，并在其中有一个选项输入来选择一个地址，当某些例如选择纽约，它会存储分析该位置的完整地址（用户只能看到纽约，但后端存储完整的地址）。

Answer 1

当你看到这种类型的错误时，你将某个函数作为一个未被定义为函数的函数来调用。 例如：

var user = { 
set: { 
     phone: '', 
     address:'' 
    } 
}; 
user.set("phoneNumber", "form.pnumber");