Javascript not php - please help

aewarnick

Why doesn't this work?

function over(typeName, n)
{
  if(document.images)
    document.images.typeName.src = ImArr[n].src;
}
function out(typeName, n)
{
  if(document.images)
    document.images.typeName.src = ImArr[n].src;
}

<img src='Blend1.jpg' name='Blend' onMouseOver=over('Blend', 1) onMouseOut=out('Blend', 0)>

It works if I do not use the variable typeName but why?

MarkR

Your HTML is incorrect. HTML attributes must be enclosed in quotes (either single or double)

Mark

aewarnick

That was one thing...

Why did I have to do this:
document.images[typeName].src = ImArr[n].src;

And use name instead of 'Blend':
<img src='Blend1.jpg' name='Blend' onMouseOver='over(name, 1)' ...

I'm annoyed and confused...C++ is not annoying like this.

Can someone help me understand?

LoganK

JavaScript has some of the loosest variables I've ever seen in a programming language...since it doesn't prefix variables with anything, you can't call a function which expects a variable "typeName" and do this: document.images.typeName, because JavaScript will think you're looking for something on the page called typeName. It will not recognize a variable. Use getElementByID() to ensure the script works.

aewarnick

Thanks. I have made the decision to always use [] instead of . so I never have the problem again.

I'm still really confused about this though:

I HAD to use:
<img src='Blend1.jpg' name='Blend' onMouseOver='over(name, 1)' ...
instead of:
<img src='Blend1.jpg' name='Blend' onMouseOver='over('Blend', 1)' ...

LoganK

I don't think it's just "name"...I think it's "this.name".

aewarnick

What are you talking about? name works, 'Blend' does not.

LoganK

Well, the way you coded it, the apostrophe ' wasn't escaped properly...maybe this would work:

<img src='Blend1.jpg' name='Blend' onMouseOver="over('Blend', 1)"

aewarnick

I can't do that because I already used "" around most of my code because I'm sending it as a php variable.

Drakla

Then put a backslash before inner double quotes to escape them.

aewarnick

The backslash does work but I'll have to try it on the code posted above later because I already used a workaround to fix the problem.

LoganK

Allright then..it'd be something like this:

echo "<img src='Blend1.jpg' name='Blend' onMouseOver=\"over('Blend', 1)\">";

aewarnick

So if I don't get an error because of that line, the code will work?

LoganK

If you're talking about the line I posted, you should get no errors, and it should work just fine. You can even check it out for yourself - once the PHP script is done parsing it, view the HTML source and it'll look the code I posted in post #8.