check if function exists in javascript

Question

How can I check If a prototyped function exists?

A little more explanation:

As you can see in the sample code, I will always have a commonFunction() for both X1 and X2.

I would like to tell if X1 and X2 have their own myOwnFunction().

It's important to notice that at first hand I don't know which function I will be calling. That's why I need a dynamic way to gather that information.

CODE:

function FunctionMain (){};

FunctionMain.FunctionSub = new FunctionSub();

function FunctionX1()
{
    FunctionX1.prototype.commonFunction = function()
    {
        console.log("Hello, I'm X1");
    }

    FunctionX1.prototype.myOwnFunctionX1 = function()
    {
        console.log("This my own function");
    }    
}

function FunctionX2()
{
    FunctionX2.prototype.commonFunction = function()
    {
        console.log("Hello, I'm X2");
    }

    //I don't have myOwnFunctionX2()
}

function FunctionSub()
{
    FunctionSub.prototype.FunctionX1 = new FunctionX1();
    FunctionSub.prototype.FunctionX2 = new FunctionX2();
}

//This call works!
FunctionMain.FunctionSub.FunctionX1.commonFunction();
FunctionMain.FunctionSub.FunctionX2.commonFunction();


//what kind of test should I use?
if(typeof "FunctionMain.FunctionSub.FunctionX1.myOwnFunctionX1" == "function")
{
    console.log("It exists!");
}

if(typeof window["FunctionMain.FunctionSub.FunctionX1.myOwnFunctionX1"] == "function")
{
    console.log("It exists!");
}

FIDDLE: http://jsfiddle.net/matias/FTzjW/

Mulan · Accepted Answer · 2013-07-19 22:58:48Z

7

This is weird, don't do this

function FunctionX2()
{
    FunctionX2.prototype.commonFunction = function()
    {
        console.log("Hello, I'm X2");
    }

    //I don't have myOwnFunctionX2()
}

Do this instead

var FunctionX2 = function() {
  // constructor
};

FunctionX2.prototype.commonFunction = function() {
  console.log("Hello, I'm X2");
};

Check if it exists directly

typeof FunctionX2.prototype.commonFunction === 'function';
// => true

Or with an instance

var f2 = new FunctionX2();
typeof f2.commonFunction === 'function';
// => true

Here's a demonstration that checking for the function dynamically is possible

var functionExists = function(receiver, functionName) {
  return typeof receiver[functionName] === 'function';
};

var commonFunctionExists = function(receiver) {
  return functionExists(receiver, 'commonFunction');
};

Some tests

var f1 = new FunctionX1();
commonFunctionExists(f1);
// => true

var f2 = new FunctionX2();
commonFunctionExists(f2);
// => true

var obj = new Object();
commonFunctionExists(obj);
// => false

edited Jul 19, 2013 at 22:58

answered Jul 19, 2013 at 20:49

Mulan

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4 Comments

Matías Cánepa Over a year ago

Remember what I said: "It's important to notice that at first hand I don't know which function I will be calling. That's why I need a dynamic way to gather that information". You are explictly checking for FunctionX2

Mulan Over a year ago

It's clear to me that you're structuring your program wrong. I mean, just by the look of the way you're coding the classes in JavaScript alone. It seems as though you have a misunderstanding of the fundamentals somewhere. I'm willing to help you, but how about instead of fixing the code above, let's look at what your goal is first. Here's an example: Instead of asking, "how do I weigh an elephant with a bathroom scale?" ask "How can I be sure my pet elephant is not overweight?"

Mulan Over a year ago

The difference being: ask how you should accomplish your goal, not how to implement a solution you think will accomplish your goal. The former opens the doors for many possible solutions whereas the latter pigeonholes you into a single solution that might be flawed from the start.

Mulan Over a year ago

For what it's worth, I added a demonstration at the end of my answer

Matías Cánepa · Accepted Answer · 2013-07-20 00:47:51Z

-3

This is the solution that worked for me.

Check this jsbin (don't know why this doesn't work in jsfiddle)

FULL CODE:

function FunctionMain (){}

FunctionMain.FunctionSub = new FunctionSub();

function FunctionX1()
{
    FunctionX1.prototype.commonFunction = function()
    {
        console.log("Hello, I'm X1");
    }

    FunctionX1.prototype.myOwnFunctionX1 = function()
    {
        console.log("This my own function");
    }    
}

function FunctionX2()
{
    FunctionX2.prototype.commonFunction = function()
    {
        console.log("Hello, I'm X2");
    }

    //I don't have myOwnFunctionX2()
}

function FunctionSub()
{
    FunctionSub.prototype.FunctionX1 = new FunctionX1();
    FunctionSub.prototype.FunctionX2 = new FunctionX2();
}

//This call works!
FunctionMain.FunctionSub.FunctionX1.commonFunction();
FunctionMain.FunctionSub.FunctionX2.commonFunction();


//use this test
function testFunction(function_to_find)
{
    var context = window;
    var functions = function_to_find.split(".");
    var method = functions.pop();

    for (var i = 0; i < functions.length; i++)
    {
        context = context[functions[i]];
    }

    return typeof context[method];
}

if(testFunction("FunctionMain.FunctionSub.FunctionX1.myOwnFunctionX1") == "function") console.log("yes x1!");

if(testFunction("FunctionMain.FunctionSub.FunctionX2.myOwnFunctionX2") == "function") console.log("yes x2!");

edited Jul 20, 2013 at 0:47

answered Jul 20, 2013 at 0:30

Matías Cánepa

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4 Comments

Adam Jenkins Over a year ago

What happens when context becomes undefined before reaching the end? EDIT: I see that's exactly what happened in your fiddle.

Matías Cánepa Over a year ago

@Adam I'm sorry, I didn't understand you. Can you explain me?

HMR Over a year ago

Still setting the prototype within the constructor body, that is wrong. Do you have any idea what the prototype is for? stackoverflow.com/questions/16063394/… to check what properties an object has you can use for(something in myObject) hasOwnProperty and go up the constructor myObject=myObject.constructor untill you reach Function. But the way you inherit now you're destroying the constructor property on the object's prototype.

Mulan Over a year ago

This is so far beyond how you should be attempting this stuff. I made a ton of effort to help you via the comments and demonstration in my answer. If you want to force a broken solution instead of learn something, no one will be able to help you; not even yourself.

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