AJAX returning null XML - AJAX returning null XML

我第一次尝试使用AJAX，很难掌握它的窍门

function requestData(j) {
    var xmlhttp;
        if (window.XMLHttpRequest) {
            xmlhttp=new XMLHttpRequest();
        }
        xmlhttp.onreadystatechange = function() {
                if (xmlhttp.readyState==4 && xmlhttp.status==200) {
                    xml = xmlhttp.responseXML;
                    post = "";
                    title = "";
                    postdata = xml.getElementsByTagName("post");
                    titledata = xml.getElementsByTagName("title");
                    datedata = xml.getElementsByTagName("date");
                    timedata = xml.getElementsByTagName("time");
                    document.getElementById("post").value = postdata[0].childNodes[0].nodeValue;
                    document.getElementById("heading").value = titledata[0].childNodes[0].nodeValue;
                    document.getElementById("date").value = datedata[0].childNodes[0].nodeValue;
                    document.getElementById("time").value = timedata[0].childNodes[0].nodeValue;
                    document.getElementById("id").value = j;
                    document.getElementById("update").value = "true";
                }
            };
        xmlhttp.open("POST","../script/getnewsdata.php",true);
        xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
        xmlhttp.send("ID=" + j);
        return false;
    }

Firebug在postdata=xml.getElementsByTagName（"post"）行告诉我"xml为null"；这意味着xmlhttp.responseXML为null。

这是服务器端脚本：

<?php
$db = mysql_connect("wadafw","awfawf","awfsgv");
if(!$db)
{
  die("Could not connect: " . mysql_error());
}
mysql_select_db("afggbare", $db);
$updata = mysql_query('SELECT * FROM News WHERE NewsID='.$_POST['ID']);
$blog = mysql_fetch_array($updata);
$post = $blog['Content'];
$regex = Array('/<br />/', '/<('/?)(b|i|u)>/', '/<a href="(http://[www.]?'w+)">('w+)<'/a>/', '/<div class="media"><iframe title="YouTube video player" width="425" height="265" src="http://www.youtube.com/embed/('w+)hd=1" frameborder="0" allowfullscreen></iframe></div>/', '/<div class="media"><img width="425" src="(http://[www.]?['w+])" /></div>/');
$regReplace = Array(''r'n', '[$1$2]', '[link=$1]$2[/link]', '[youtube]http://www.youtube.com/watch?v=$1[/youtube]',
                '[img]$1[/img]');
$post = preg_replace($regex, $regReplace, $post);
echo    '<newsItem>
        <title>'.$blog['Heading'].'</title>
        <post>'.$post.'</post>
        <date>'.$blog['time'].'</date>
        <time>'.$blog['time'].'</time>
    </newsItem>';
?>

正则表达式可能不好。。。但现在这并不重要。

已更改为。。。现在得到"函数未定义"错误：

function requestData(j) {
        $.ajax("../script/getnewsdata.php", {
            data: {ID: j},
            type: "POST",
            dataType: "xml",
            success: function(data, status, jqXHR){
                var xml = jqXHR.responseXML;
                postdata = xml.getElementsByTagName("post");
                titledata = xml.getElementsByTagName("title");
                datedata = xml.getElementsByTagName("date");
                timedata = xml.getElementsByTagName("time");
                document.getElementById("post").value = postdata[0].childNodes[0].nodeValue;
                document.getElementById("heading").value = titledata[0].childNodes[0].nodeValue;
                document.getElementById("date").value = datedata[0].childNodes[0].nodeValue;
                document.getElementById("time").value = timedata[0].childNodes[0].nodeValue;
                document.getElementById("id").value = j;
                document.getElementById("update").value = "true";
            }

        }
    }

好的，找到那个问题了。这只是一个拼写错误。。。没有关闭$.ajax（）参数括号。现在我没有犯错误。但它什么都没做。。。

与其尝试让AJAX从头开始运行，我建议使用现有的库。jQuery使这项任务变得非常容易。

http://api.jquery.com/jQuery.ajax/

function requestData(j) {
    $.ajax('../script/getnewsdata.php', {
        data: {ID: j},
        type: 'POST',
        dataType: 'xml',
        success: function(data, status, jqXHR){
            // consume data here
        }
    });
}

它需要看起来像这样：

var xmlhttp;
if (window.XMLHttpRequest)
    xmlhttp=new XMLHttpRequest();
else
    xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
xmlhttp.onreadystatechange=function()
{
    if (xmlhttp.readyState==4 && xmlhttp.status==200)
       document.getElementById("my").innerHTML=xmlhttp.responseText;
 }
 xmlhttp.open("POST","page.php,true);
 xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
 xmlhttp.send("firstname=lolo&lastname=koko");