调用obj函数不是正确的方法吗？

Question

I have a function which is return a object. in the object i have two function to show the popup and close it. it works within the parent function, but it's not from out side.. is it not a right way to call that.. else how can i call the obj function from out side?

my function :

var popupHandler = function(handler,msg,popUp){

    msg =  msg == "help" ? "help" : "results"
    $(handler).click(function(){
        popObj.showPop(); //works
    })
    $('.cls-how2play').click(function(){
        if(msg == 'help') popObj.closePop(); //works
    });

    var popObj = {
        showPop  : function(){
                    if(!(popUp).is(':visible')) $(popUp).fadeIn().children().find('.'+msg).show().siblings().hide();
                },
        closePop : function(){
                    $(popUp).fadeOut()
                }
    }
    return popObj;
}

from calling ouside like this :

 $('.ui-footer').click( function(){ 
  var closeIt = popupHandler(); 
  closeIt.popObj.closePop() }) //not works.. why?
}

any one can help me the right way to call the obj functions from outside of the returning function?

thanks.

Answer 1

Rather than

closeIt.popObj.closePop()

You want

closeIt.closePop()

Your popupHandler function returns the popObj object, which has the showPop and closePop functions on it. So closeIt is a reference to that same object.

Answer 2

As you are returning the popObj , your closeId will get only the two functions, not wrapped in the popObj object. Therefor you will call the function like so, without popObj :

closeIt.closePop();

Answer 3

您应该致电：

closeIt.closePop();

Answer 4

There is no need to wrap this in an object since you immediately return it.

you can write

return{
showPop  : function(){
                    if(!(popUp).is(':visible')) $(popUp).fadeIn().children().find('.'+msg).show().siblings().hide();
                },
        closePop : function(){
                    $(popUp).fadeOut()
                }
}

Now closeIt.closePop(); should work very well.

Answer 5

As i can c

popupHandler is a function, and popObj is a return result of the function "popupHandler"

when the program run to

var closeIt = popupHandler();

it means that the "closeIt" assigned by the result of the function "popupHandler", a obj as the same as the "popObj".

you can consider that "closeIt" is a copy of "popObj".

and "popObj" is not a property of "closeIt" , them are the same.

so you should code closeIt.closePop(), but not closeIt.popObj.closePop(),

and not popObj.closePop() as well.

because popObj was "var"ed in the declare of the popupHandler, it belonged that scope.