到PHP我怎么可以在这个代码发送rowNumber
变量dataSource
PHP文件?发送变量与AJAX
function getData(dataSource, divID,rowNumber)
{
if(XMLHttpRequestObject)
{
var obj = document.getElementById(divID);
XMLHttpRequestObject.open("GET", dataSource);
XMLHttpRequestObject.onreadystatechange = function()
{
if (XMLHttpRequestObject.readyState == 4 &&
XMLHttpRequestObject.status == 200)
{
obj.value = XMLHttpRequestObject.responseText;
}
}
XMLHttpRequestObject.send(null);
}
}
PHP文件(数据源):
<?php
//mysql connection
$result = mysql_query('CALL view_polls(`rowNumber`);');
$row=mysql_fetch_array($result);
echo $row['title'];
?>
采取jQuery框架来看看......它使AJAX确实容易:) –