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javascript/jquery integer becomes object

Answer 1

A:

When the 2nd alert() executes, the name 'sched' points to the current element in the $.each() iteration, since your callback (sent to $.each) declared the sched parameter.

** EDITED **

This simplified example does what I think you're looking for...

<html>
<head>
    <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js"&gt;&lt;/script&gt;
</head>
<body>

<ul id="monkey">
  <li><span>xxx</span></li>
  <li><span>yyy</span></li>
  <li><span>zzz</span></li>
</ul>

<script type="text/javascript" charset="utf-8">

for(thisIsANumber=1; thisIsANumber < 4; thisIsANumber++){
    alert(thisIsANumber);
    $("span", $("#monkey li:nth-child(" + thisIsANumber + ")")).each(function(thisIsAnHtmlElement) {
        alert(thisIsANumber);
    });
}
</script>
</body>

</html>

I think the crux of the issue here is this: you're not in charge of the parameter that gets passed to the callback function. You can't choose that to pass -- that's done by jQuery, and it passed the DOM element at the current position of the $.each() loop.

You can, however, reference the original 'sched' variable that's defined outside your $.each() callback and managed by the for-loop.

Drew Wills 2010-01-08 21:27:20

Thanks drew, I thought it was something like that. but how do you get around this? I've tried adding a var sched2=sched, and passing that in the .each(function(sched2), but that didn't work.

pedalpete 2010-01-08 21:31:10

To me it doesn't look as though you're using the callback parameter at all. In other words, every time you reference 'sched' you simply want the numeric index -- is this accurate? If so, just change to .each(function(someNameYouDontUse) {...}), or (probably works) just drop the parameter entirely.

Drew Wills 2010-01-08 21:35:41

I am using the parameter, it defines which column gets updated and which columns the events belong to. I've tried using a different name (which is what I was alluding to with sched2), but that still returns an object.

pedalpete 2010-01-08 21:49:07

Not sure why you lost a point when I tried to add another. Now I understand though. I do find it strange that I don't pass the parameter in the function, but that jQuery can use it anyway. That's I guess what was tripping me up.

pedalpete 2010-01-08 23:47:06

Answer 2

A:

I suggest you take a look at the jquery documentation for the $.each function. http://docs.jquery.com/Utilities/jQuery.each . $.each() supports the ability to pass in a function with one parameter for just a value itself or two parameters, one for index and one for value. Remember all parameters defined in the function passed into $.each() are used by jquery exclusively if you want to use a variable defined outside the scope of the function passed into $.each() then you need to make sure the name of the variable doesn't collide with any named parameter or variable inside the function passed into $.each().

avelis 2010-01-08 21:45:25

Maybe I'm not understanding correctly. I've looked through the documentation, and I'm not seeing anything that relates to this kind of issue. As you can see from my code above, I am passing a single parameter of sched. As far as I can tell, that is being passed into the function and the name definately does not collide with anything else passed into each. I've also tried re-naming sched to sched2, and passing sched2 into each, but I get the same result.

pedalpete 2010-01-08 21:52:44

You _can't_ pass a parameter, you can only accept the parameter(s) that jQuery is willing to send you. See my edited answer.

Drew Wills 2010-01-08 22:36:34

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