Pb de code php dans javascript

Salut,,

écrit ton code javascripte dans un echo php:

<?php echo "
<script>
var obj = new AW.UI.Grid;
obj.setCellData([
for($i=0;$i<4;$i++){
echo'[';
for($j=0;$j<4;$j++){
print \"\"\";
echo $tab[$i][$j];
print \"\"\";
if($j!=3) echo',';
}
echo ']';
if($j!=3) echo',';
}
]); "; ?>

Si tu as vraiment besoin d exécuter du php dans du javascript recherche l'objet XHR

regardes ici

function showResultsTableRight()
{
//alert("xhr_url = "+xhr_url);
//alert("enter showResultsTableRight");
// 1. catch dynamic data from server thanks to Ajax
// 1.1 clear area before displaying result(s)
document.getElementById("resultsTableDiv").innerHTML = null;

// 1.2 collect data to do correct operation in db
var act = 'getrightlist';
var dataArray = collectSearchFormDataRight();
var data = "act="+act+"&label="+dataArray[0];

// 1.3 use xmlhttprequest object
var xhr_object = GetXmlHttpRequest();

// 1.4 open 'facade' file to access data (via an object)
xhr_object.open("POST", "/efstc/dev/php/facade/Facade_right.php", true);

// 1.5 server is ready
xhr_object.onreadystatechange = function()
{
/*if(xhr_object.readyState == 4 && xhr_object.status == 200)
{
//alert(xhr_object.responseText);
document.getElementById("resultsTableDiv").innerHTML = xhr_object.responseText;
}*/

// waiting until request is achieved
if((xhr_object.readyState == 1) ||(xhr_object.readyState == 2) || (xhr_object.readyState == 3))
document.getElementById("resultsTableDiv").innerHTML = 'LOADING...';
// now we can read server response
else if(xhr_object.readyState == 4)
{
if(xhr_object.status == 200) // continue only if HTTP status is 'OK'
document.getElementById("resultsTableDiv").innerHTML = xhr_object.responseText;
else // display error message
alert('in js_user.js file\nThere was a problem retrieving the data:\n'+xhr_object.statusText);
}
}

xhr_object.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); // mandatory if method = POST

// 2. send data to server
xhr_object.send(data); // in case of method = POST

// 3. display div which is in charge of results table
document.getElementById("resultsTableDiv").style.visibility = "visible";
}