code does not work inside $(document).ready(function() {

can some one tell me why may code does not work when I put it inside $(document).ready(function() { });
it works fine if I do not put it in there.

// JavaScript Document
<!--
$(document).ready(function() { ////makes sure document is ready before excuting any jquery  
/*--------------------------------------------------------------------------------------*/
function bizType(type)
        {
        if(type=='MEDI') {
			$("#medical_type_box").show();
			$("#profile_1_box").hide();
			$("#profile_1").val("");  
            $("#profile_2_box").hide();
			$("#profile_2").val("");
			$("#restaurant_type_box").hide();
			$("#restaurant_type").val("");
		    }
		
		if(type=='REST') {
			$("#restaurant_type_box").show();
			$("#profile_1_box").hide();
			$("#profile_1").val("");
            $("#profile_2_box").hide();
			$("#profile_2").val("");
			$("#medical_type_box").hide();
			$("#medical_type").val("");
		    }
			
		if(!(type=='MEDI' || type=='REST')) {
			$("#profile_1_box").show();
			$("#profile_2_box").show();
			$("#restaurant_type_box").hide();
			$("#restaurant_type").val("");
			$("#medical_type_box").hide();
			$("#medical_type").val("");
		    }
         }

/*--------------------------------------------------------------------------------------*/
}); //// end of document ready
///End jquery
/*--------------------------------------------------------------------------------------*/
-->

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LueyAsked:
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designatedinitializerCommented:
that's because you are calling the function bizType somewhere else, and that somewhere else gets executed before $(document).ready.

You actually don't need it inside $(document).ready
That's a function. It will only execute when you call it.
What you need inside $(document).ready is stuff that needs to initialize and act upon the DOM, which only exists on $(document).ready.

PS: You can convert those 3 'if' s to if(...){---} else if(...) {---} else {---}
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LueyAuthor Commented:
ok now i understand thanks
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