使用AJAX将变量发送到php

Question

如何在此代码中将rowNumber变量发送到dataSourcephp文件？

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'];
?>

Answer 1

在JavaScript中:

XMLHttpRequestObject.open("GET", dataSource + '?rowNumber=' + rowNumber);

在PHP中:

$result = mysql_query( 'CALL view_polls(`' . $_GET['rowNumber'] . '`);' );