Passing a function in javascript

Question

I am trying to pass a function to another function to use in its call back, I call the function using...

call("login", {email:user,password:pw}, alert(getResponse()));

call is defined as:

function call(url, p, f){
    $.getJSON(baseUrl + url, p, f(data));
};

and getResponse is defined as:

function getResponse(d){
    return d.result.success;
};

data is returned by getJSON (see here: http://api.jquery.com/jQuery.getJSON/)

How am I meant to pass data to that function?

The function is called, because it hits a break point I set.

When getResponse is called though I get Uncaught ReferenceError: data is not defined

FabianCook · Accepted Answer · 2012-12-18 20:42:55Z

3

Use

call("login", {email:user,password:pw}, function(data){alert(getResponse(data)}));

The expression alert(getResponse()) is the call of the alert function (with getResponse called without any parameter, and thus this would throw an error as d.result would mean taking result from undefined), but function(data){alert(getResponse(data)}) is the definition of a new function taking data as parameter.

Also on a note to that you do not need to have f(data) in $.getJSON(baseUrl + url, p, f(data)); you only require f, so you would call $.getJSON(baseUrl + url, p, f);

edited Dec 18, 2012 at 20:42

FabianCook

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answered Dec 18, 2012 at 20:36

Denys Séguret

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1 Comment

FabianCook Over a year ago

Sweet as, I knew it would be something simple like that. Just haven't done enough js in my time

Samuel Rossille · Accepted Answer · 2012-12-18 20:38:21Z

1

The expression alert(getResponse()) pops an alert and doesn't evaluate to a function but to undefined.

You have to pass a function which implementation is alert(getResponse(d)) where d is it's parameter.

Such as the expression function(d){alert(getResponse(d);} for example

answered Dec 18, 2012 at 20:38

Samuel Rossille

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