using Javascript in a php page

Anon

I am having a terible time trying to get my mouseover event to work in a php page. I dont know how to correct the problem. Here is what I have in my header.php page that is included in a content php page.

<head>

<title>About Us</title>

<?php $page=getenv(SCRIPT_NAME); ?>

<?php $nav1off = "../images/nav1_on.gif"; ?>

</head>

function turnOn(imgName) {

if (document.images) {

	document[imgName].src = eval(imgName + "over.src");

	}

}

function turnOff(imgName) {

if (document.images) {

	document[imgName].src = eval(imgName + "off.src");

	}

}

if (document.images) {

nav1over = new Image();

nav1over.src = "../images/nav1_over.gif";

nav2over = new Image();

nav2over.src = "../images/nav2_over.gif";

nav3over = new Image();

nav3over.src = "../images/nav3_over.gif";

nav4over = new Image();

nav4over.src = "../images/nav4_over.gif";

nav1off = new Image();

nav1off.src = <?php echo $nav1off; ?>;

nav2off = new Image();

nav2off.src = "../images/nav2_off.gif";

nav3off = new Image();

nav3off.src = "../images/nav3_off.gif";

nav4off = new Image();

nav4off.src = "../images/nav4_off.gif";

}

</script>

calling the mouseover event in this code in another php page

<td width="80" height="15" align="center"><a href="index.php" onMouseOver="turnOn('nav1')" onMouseOut="turnOff('nav1')"><img src= <?php echo $nav1off; ?> name="nav1" width="64" height="15" border="0" alt="Home"></a></td>

do I have to do anything to help parse the javascript?

thx in advance,

alex

bunka

Remember that if you are in a PHP block and you are using quotation marks (besides the PHP required ones) you need to escape them with the \ character. This includes the quotation marks for JAVASCRIPT!

Example:

<head>
<title>My Link Variable</title>
</head>
<body>

$mycode = "<a href=\"http://www.phpbuilder.com\">click here</a>";

echo "$mycode";

</body>

turbo99

since I am trying to figure out php and I need some clarification then Bob.

I dont have php blocks. The javascript is in a .php page. If I am trying to understand your response, I need to escape even inside the Javascript block?

so my

if (document.images) {
nav1over = new Image();
nav1over.src = "../images/nav1_over.gif";

would be

if (document.images) {
nav1over = new Image();
nav1over.src = /"../images/nav1_over.gif/";

or <script language=/"JavaScript/">

Thanks for your help.

Alex

bunka

It would be:

if (document.images) {
nav1over = new Image();
nav1over.src = \"../images/nav1_over.gif\";

A PHP block is anything basically <? to ?>

So if you have any coding in there that is not PHP related (HTML, Javascript, etc) and has quotation marks it must be escaped.

I hope I understood you correctly.