How to initialize all methods inside a namespace?namespaces javascript INITIALIZING ALL methods

Question

Maybe this is a simple question but I could not find anything on google or this site. So if I have

var myNamespace = {

    foo: function() {
    },

    bar: function() {
    }
};

how would i load both foo() and bar() without having to do it one by one, e.g.

<script type="text/javascript">
    $(document).ready(function(){
        myNamespace.foo();
        myNamespace.bar();
        .
        .
        .
    });
</script>

bjornd · Accepted Answer · 2011-09-09 15:50:41Z

2

for (var k in namespace) {
    if (typeof namespace[k] === 'function') {
        namespace[k]();    
    }
}

answered Sep 9, 2011 at 15:50

bjornd

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1 Comment

Joel A. Villarreal Bertoldi Over a year ago

I think this will fail if something in the namespace isn't a function.

user278064 · Accepted Answer · 2011-09-09 16:07:48Z

2

var func;

for (func in namespace) {
   if (namespace.hasOwnProperty(func) {
       if (typeof namespace == 'Function') {
           namespace[func]();
       }
   }
}

edited Sep 9, 2011 at 16:07

answered Sep 9, 2011 at 15:53

user278064

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1 Comment

bjornd Over a year ago

I think it's a bad idea to use reserved word function as a variable name.

Joel A. Villarreal Bertoldi · Accepted Answer · 2011-09-09 15:58:37Z

1

I've not tested this, but something like this should prove useful. Every function in JavaScript has a call() method for invoking it. Then:


$(document).ready(function() {
  $.each(myNamespace, function(index, obj) {
    if (typeof(obj) == 'function')
    {
       obj.call(myNamespace);
    }
  });
});

If you need to replace the this scope reference, you could patch this up with jQuery.proxy.

edited Sep 9, 2011 at 15:58

answered Sep 9, 2011 at 15:49

Joel A. Villarreal Bertoldi

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1 Comment

bjornd Over a year ago

1. 'Function' should be 'function'. 2. You need to provide the context namespace to the call method.

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