javascript variable scope in callbacks

Question

I'm curious how could this be written better:

function Klass(variable) {
    this.variable = variable;

    this.callAjax = function() {
        $.get('/url', { }, function(json) {
            console.log(variable); //! <-- shows undefined
        }, "json");
    }
}

so I create a local variable: _variable

function Klass(variable) {
    this.variable = variable;

    this.callAjax = function() {
        var _variable = this.variable;
        $.get('/url', { }, function(json) {
            console.log(_variable); //! <-- its ok
        }, "json");
    }
}

and its fine, but I really don't this solutions,

Does someone of you have a better code?

I can't see why console.log(variable); should show undefined. variables should refer to the constructor argument. Do you pass undefined to the constructor and set this.variable later? — John Dvorak
– John Dvorak, Commented Feb 28, 2013 at 7:47

Min Lin · Accepted Answer · 2013-02-28 08:15:05Z

4

That's quite the way.

function(json){console.log(_variable);}

forms a closure with "_variable". "_variable" keeps the original value forever.

If your "variable" should be updated later, and you want the updated "variable" You define

var self = this;

and call self.variable to get it.

In this way you'll get the updated "variable" each time the callback is executed.

The complete code:

function Klass(variable) {
    var self = this;
    this.variable = variable;
    this.callAjax = function() {
        $.get('/url', { }, function(json) {
            console.log(self.variable);
        }, "json");
    }
}

edited Feb 28, 2013 at 8:15

answered Feb 28, 2013 at 7:46

Min Lin

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2 Comments

John Dvorak Over a year ago

You can declare self in the constructor as well, since callAjax is a privileged method (declared in the constructor). Your approach is better since it works in prototype methods as well, though.

Min Lin Over a year ago

You're right, I wonder why the first case in the question is not working. It should be working, just the variable will not change.

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