Javascript undefined is not a function

Question

anytime I'm trying to reference my global variable i get an Uncaught TypeError: undefined is not a function error and nothing happens. I'm probably missing something obvious since I'm new to javascript. I've simplified the code below so that you can give it a look. All help is much appreciated.

var popup;

function first(){
    popup = window.open(...); //opens a popup
    popup.moveTo(10,10) //moves the window
    second();
}

function second(){
    .
    .
    .
    //does an XMLHttpRequest and when it's done (simplified) it tries to close the window
    xmlHttp.onreadystatechange = function(){
        popup.close();
    }
}

Whenever popup.close() is fired I get the error above. Am I missing something really obvious here?

For future problems as this, here is a solution:

instead of popup.close();

var win = window.open("","popup");//open a new window with the same name as before
win.close();

By doing so it will reference the same window object and will be able to close it.

I suspect the same as @BenjaminGruenbaum. But you could easily avoid such mistakes by defining popup inside first and passing it to second as argument. — jgillich
– jgillich, Commented Jul 31, 2014 at 8:19
@user2071225 I see what you're thinking but first(); has to run when I want it to and not on startup.. Shouldn't it be a global reference? — user2970038
– user2970038, Commented Jul 31, 2014 at 8:19
@user2970038 You need to post how you call these functions if you want to get this solved. Can you create a JSFiddle that demonstrates the problem? — jgillich
– jgillich, Commented Jul 31, 2014 at 8:24

Fizer Khan · Accepted Answer · 2014-07-31 08:19:27Z

2

From your error, there is two possibility

pop is not accessible
close does not exist in pop object

I tried both in console

When object pop does not exist

  pop.close()

Error you will get

  ReferenceError: pop is not defined

When no function close in the object

 var pop = {}
 pop.close()

Error you will get

 TypeError: undefined is not a function

From your error, there is no function name close in the pop object.

answered Jul 31, 2014 at 8:19

Fizer Khan

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Sushmit Patil · Accepted Answer · 2014-07-31 08:22:36Z

0

Try this one..

var popup;

function first(){
    popup = window.open(...); //opens a popup
    popup.moveTo(10,10) //moves the window
    second(popup);
}

function second(popup_){
    .
    .
    .
    //does an XMLHttpRequest and when it's done (simplified) it tries to close the window
    xmlHttp.onreadystatechange = function(){
        popup_.close();
    }
}

answered Jul 31, 2014 at 8:22

Sushmit Patil

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1 Comment

user2970038 Over a year ago

It's giving me the same error which is weird since popup.moveTo works :/

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