使用AJAX时出现Javascript错误

Question

I am trying to do a connection with the database using AJAX. I am totally new to AJAX...just saw a couple of tutorials and took a code snippet and tried to adapt it to my code.. but when ever I press the button I get the following error Error: Unable to get property 'documentElement' of undefined or null reference

php file

<?php
header('Content-Type: text/xml');
echo '<?xml version="1.0" encoding="UTF-8" standalone="yes" ?>';
echo '<response>';
$title = $_GET['title'];
$conn=("localhost","root","","askthedoctor");
$sql="select patient_text, doctor_text where title='".$title."')";
$result=mysqli_query($conn,$sql);
$row=mysqli_fetch_array($result);
echo $row[0];
echo $row[1];
echo '</response>';
?>

javasscript

var xmlHttp = createXmlHttpRequestObject();

function createXmlHttpRequestObject(){
var xmlHttp;

if(window.ActiveXObject){ 
try{
    xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
}catch(e){
    xmlHttp = false;
}
}else{ 
try{
    xmlHttp = new XMLHttpRequest();
}catch(e){
    xmlHttp = false;
}
}

if(!xmlHttp)
alert("Cant create that object !")
else
return xmlHttp;
}

function process(){
if(xmlHttp.readyState==0 || xmlHttp.readyState==4){
title= encodeURIComponent(document.getElementById("title").value);
xmlHttp.open("GET", "displaypatientmessage.php?title="+title,true);
xmlHttp.onreadystatechange = handleServerResponse;
xmlHttp.send(null);
}else{
setTimeout('process()',1000);
}
}

function handleServerResponse(){
if(xmlHttp.readyState==4){ 
if(xmlHttp.status==200){
    xmlResponse = xmlHttp.responseXML; 
   ////////////////////////////////error////////////////////
    xmlDocumentElement = xmlResponse.documentElement;
   /////////////////////////////error/////////////////////////
    message1 = xmlDocumentElement.firstChild.data;
    document.getElementById("question").innerHTML = message1;
    message2 = xmlDocumentElement.firstChild.data;
    document.getElementById("answer").innerHTML = message2;
    setTimeout('process()', 1000);
}else{
    alert('Someting went wrong !');
}
}
}

the form containing the button which calls the javascript

<table id="table" class="table">

<tr>
<th>Messages</th>   
<th>Problem Description</th>
<th>Doctor's Answer</th>
<th></th>
</tr>
<tr>
<th><select id="title"> 
<?php   
$sql="select title from messages where paitient_id=(select id from login where username='".$username."');";
$result=mysqli_query($conn,$sql);
while($row=mysqli_fetch_array($result))
{
?>  
<?php echo "<option value=\"mesazhi1\">".$row[0]."</option>";}?>
</select>
</th>
<td><textarea rows="17.95" col="100" id ="question" > </textarea></td>
<td><textarea rows="17.95" col="100" id ="answer" readonly> </textarea></td>
</tr>
<tr>
<td></td>
</tr>

<tr>
<td><input type="button" name="openmessage" value="Display Selected Message" onClick="process()"></td>
</tr>

</table>
</form>

Answer 1

1] One more thing, see the

<?php echo "<option value=\"mesazhi1\">".$row[0]."</option>";}?>

as value of each <option> is same. You will get same value of title at

title= encodeURIComponent(document.getElementById("title").value);   

so no matter what option you select, you will always get same value ie \\mesazhi\\ .

2] Also, take a look,

$sql="select patient_text, doctor_text where title='".$title."')";

you have not mention the table name.

3] Also why is ) at the end of query?

4] And finally

$conn=("localhost","root","","askthedoctor");

should be

$conn=mysqli_connect("localhost","root","","askthedoctor");