CODEWITHSUNDEEP
phpjavascriptxmlajax
Will Gill
Will Gill

Reputation: 587

Using PHP to export XML, pulling it into the DOM via AJAX, but the xml is not visible

I have a PHP script that appears to be producing valid XML output, and I am attempting to pull this into the browser with an ajax XMLHttpRequest call.

After the ajax call is made, I can see in firebug that the request was succesful and the xml appears valid, but when I try to put the response into a variable I get an error saying the response is null.

Here is the PHP code:

<?php
$q=$_GET["q"];

$con = mysql_connect('address.com', 'dbnme', 'password');
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("sql01_5789willgil", $con);

$sql="SELECT * FROM place";

$result = mysql_query($sql);

$xmlResponse = "<?xml version='1.0' encoding='ISO-8859-1'?>\n";
$xmlResponse .= "<placeList>\n";

while($row = mysql_fetch_array($result))
  {
  $xmlResponse .= "<place>\n";
  $xmlResponse .= "<title>". $row['title'] . "</title>\n";
  $xmlResponse .= "<description>" . $row['description'] ."</description>\n";
  $xmlResponse .= "<latitude>" . $row['latitude'] ."</latitude>\n";
  $xmlResponse .= "<longitude>" . $row['longitude'] ."</longitude>\n";
  $xmlResponse .= "<image>" . $row['image'] ."</image>\n";
  $xmlResponse .= "</place>\n";
  }

echo $xmlResponse;

mysql_close($con);
?>

And the offending JavaScript: 

//Pull in the xml data for the bubbles
        if (window.XMLHttpRequest)
          {// code for IE7+, Firefox, Chrome, Opera, Safari
          xmlhttp=new XMLHttpRequest();
          }
        else
          {// code for IE6, IE5
          xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
          }

        xmlhttp.onreadystatechange=function()
          {
          if (xmlhttp.readyState==4 && xmlhttp.status==200)
            {
            xmlDoc=xmlhttp.responseXML;  //pull the xml into the DOM
            var x = xmlDoc.getElementsByTagName("place"); //return the contents of the xml file (places) to an array called x
            setupMap();
            }
          }

        xmlhttp.open("GET","getPlaces.php",true);
        xmlhttp.send();

Thankyou very much for any help!

Upvotes: 0

Views: 1770

Answers (2)

ajreal
ajreal

Reputation: 47331

It seems you forget to close your root node

// ensure not output before this point
header("Content-type: text/xml"); // good practice
while($row = mysql_fetch_array($result))
{
  ...;
}

$xmlResponse .= "</placeList>\n";
echo $xmlResponse;

To ensure parse DOM successfully, an additional header to indicate response is a XML might necessary

Upvotes: 2

GolezTrol
GolezTrol

Reputation: 116170

Maybe the xml isn't valid after all. Try reading responseText to see if you get any result at all. Also, PHP 's got a DOM object itself for generating xml. Using that will make it easier to create valid XML.

[edit]

As far as I can tell, you did at least forget </placelist> at the end of the output.

And it might be a good idea to escape your values. A < in the title will break your XML.

Upvotes: 1

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