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  #1  
Old December 26th, 2004, 01:21 AM
Akdor 1154 Akdor 1154 is offline
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PHP Newbie - hover button help

Hi, just learning PHP and I'm creating a menu, I kind of cheated and got Frontpage to make the basic code for a hover button as I don't know jscript - but it works, at least when it was in HTML. Now the menu displays fine, and hovering and click the button changes the image right, but the OnMouseOut attrib seems to be dead. Here're the bits of my code:

this is in the head:
Code:
<script language="JavaScript">
<!--
function newImage(arg) {
	if (document.images) {
		rslt = new Image();
		rslt.src = arg;
		return rslt;
	}
}

function changeImages() {
	if (document.images && (preloadFlag == true)) {
		for (var i=0; i<changeImages.arguments.length; i+=2) {
			document[changeImages.arguments[i]].src = changeImages.arguments[i+1];
		}
	}
}

var preloadFlag = false;
function preloadImages() {
	if (document.images) {
		akdor_Button_over = newImage("images/menu/akdor-Button_over.png");
		akdor_Button_down = newImage("images/menu/akdor-Button_down.png");
		downloads_Button_down
newImage(images/menu/downloads-Button_over.png");
		downloads_Button_over = 
newImage(images/menu/downloads-Button_down.png");
		preloadFlag = true;
	}
}
//-->
</script>

and later on (after <body onload="preloadImages();">)
PHP Code:
<?php
    $section 
"akdor";
    function 
makemenuitem ($itemname) {
    global 
$section;
    
$menuitemcurrent '<img name="$preitemname_Button" src="images/menu/$preitemname-Button_selected.png" width="100" height="30" border="0"></a><br />' "\n";
    
$menuitembutton 
'<a href="$preitemname.php"' "\n" 'onMouseOver="changeImages(\'$preitemname_Button\', \'images/menu/$preitemname-Button_over.png\'); return true;"' "\n" .
'onMouseOut="changeImage(\'$preitemname_Button\', \'images/menu/$preitemname-Button.png\'); return true;"' "\n" .
'onMouseDown="changeImages(\'$preitemname_Button\', \'images/menu/$preitemname-Button_down.png\'); return true;"' "\n" .
'onMouseUp="changeImages(\'$preitemname_Button\', \'images/menu/$preitemname-Button_over.png\'); return true;">' "\n" .
'<img name="$preitemname_Button" src="images/menu/$preitemname-Button.png" width="100" height="30" border="0"></a><br />' "\n";
    if (
$section == $itemname) {
        
$menuitemcurrent str_replace ('$preitemname'$itemname$menuitemcurrent);;
        print (
$menuitemcurrent);
    } else {
        
$menuitembutton str_replace ('$preitemname'$itemname$menuitembutton);
        print (
$menuitembutton);
    }
    }
    
makemenuitem ('akdor');
    
makemenuitem ('downloads');
?>

For the sake of legibillity, I've just made line breaks when defining $menuitembutton, in my code it's just one line.
It's meant to create a hover button, for all menu items except the one that links to the current page; in which case it just displays an orange menu item pic (instead of blue ).

Any help would be appreciated.

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  #2  
Old December 27th, 2004, 05:15 PM
Madpawn Madpawn is offline
My beat is correct.
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Your onMouseOut is calling changeImage() instead of changeImages().

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  #3  
Old December 27th, 2004, 06:01 PM
Akdor 1154 Akdor 1154 is offline
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Ah, so it is.

Thanks

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  #4  
Old December 28th, 2004, 10:38 AM
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MadCowDzz MadCowDzz is offline
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Madpawn, good eye!

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