How to query database from JavaScript using Ajax and PHP correctly?

Question

I'm trying to do a simple exercise querying a database from JS using XMLHTTPrequest Object and POST method. Basically I pass a string to PHP server, it recieves it, queries DB and returns an XML from where I read the information with JS, but something is not working: the callback function that should be executed after request is ready.

Here's the JS code:

function leerDNI(dni){

var params="dni="+dni;

downloadUrl(params,"genxml.php", function(data) {
    var xml = parseXml(data);//THIS IS NOT BEING EXECUTED
    var dnis = xml.documentElement.getElementsByTagName("dni");//THIS IS NOT BEING EXECUTED
    for (var i = 0; i < dnis.length; i++) {//THIS IS NOT BEING EXECUTED
    var name = dnis[i].getAttribute("name");
    alert(name); //THIS IS NOT BEING EXECUTED
    document.getElementById("name").value=name; //THIS IS NOT BEING EXECUTED
 }
 });

 }

function downloadUrl(params,url, callback) {
 var request = window.ActiveXObject ?
 new ActiveXObject('Microsoft.XMLHTTP') :
 new XMLHttpRequest; 

request.onreadystatechange = function() {
 if (request.readyState == 4) {
 request.onreadystatechange = doNothing;
 callback(request.responseText, request.status);

 }
 };
 request.open("POST", url, true);
 request.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");//creación de headers
 request.send(params);//enviamos la petición POST
}

function parseXml(str) {
 if (window.ActiveXObject) {
 var doc = new ActiveXObject('Microsoft.XMLDOM');
 doc.loadXML(str);
 return doc;
 } else if (window.DOMParser) {
 return (new DOMParser).parseFromString(str, 'text/xml');
 }
}

function doNothing() {}

The XML is being generated correctly. Anyone can light me up with what's wrong in my code? Thank you very much!

grifos · Accepted Answer · 2012-03-26 22:43:32Z

0

May not be the reason but in downloadUrl(param, url, callback) function, the callback call is:

callback(request.responseText, request.status);

So 2 parameters.

However when you call the function:

downloadUrl(params,"genxml.php", function(data){...} )

The callback you gave has only 1 parameter. Check that.

answered Mar 26, 2012 at 22:43

grifos

3,4011 gold badge18 silver badges14 bronze badges

Yes, I've checked that deleting request.status parameter before, didn't make a change. Thanks anyway!

danielrozo
– danielrozo

03/27/2012 09:16:09
Commented Mar 27, 2012 at 9:16
Did you check with an addon like firebug if the ajax request is executed succefully and that your browser received the data?

grifos
– grifos

03/27/2012 14:41:25
Commented Mar 27, 2012 at 14:41

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Melsi · Accepted Answer · 2012-03-26 23:29:45Z

This one uses post and is ready to go and change to your needs!

html file (file name does not matter)

<html>
<head>
<script type="text/javascript">

var request = false;
try { 
  request = new XMLHttpRequest(); 
} catch (trymicrosoft) {                         
  try { 
    request = new ActiveXObject("Msxml2.XMLHTTP"); 
  } catch (othermicrosoft) {
    try {
      request = new ActiveXObject("Microsoft.XMLHTTP");
    } catch (failed) {                  
      request = false;       
    }
  }
}

if (!request) 
  alert("Error initializing XMLHttpRequest!"); 
</script>

<script type="text/javascript">

 var fileOption;


   function runPhp(Args) 
   {  
        var url = "script.php"; 
        fileOption = Args;
        var params = "Args=" +Args+"";
        request.open("POST", url, true);  

        //Some http headers must be set along with any POST request.
        request.setRequestHeader("Content-type", "application/x-www-form-urlencoded;charset=utf-8");
        request.setRequestHeader("Content-length", params.length);
        request.setRequestHeader("Connection", "close");

        request.onreadystatechange = updatePage;
        request.send(params);

   }////////////////////

   function getXml( ) 
   { 
        var url = 'temp.xml'; 
        var params = null; 
        request.open("POST", url, true);     
        request.setRequestHeader("Connection", "close");    
        request.onreadystatechange = displayFile;
        request.send(params); 
   }////////////////////

   //You're looking for a status code of 200 which simply means okay.
   function updatePage() {
     if (request.readyState == 4) {
       if (request.status == 200) 
       {   

            if(fileOption==1)  
                {fileName=request.responseText;  return;}
            document.getElementById('divResults').innerHTML=request.responseText;
            document.getElementById('textareaResults').innerHTML=request.responseText;   
       } 
       else{
         //alert("status is " + request.status);
         }
     }
   }

      function displayFile() {
     if (request.readyState == 4) {
       if (request.status == 200) 
       {   
            document.getElementById('textareaResults').innerHTML=request.responseText;
            document.getElementById('divResults').innerHTML='File loaded in text area above.';
       } 
       else{
         //alert("status is " + request.status);
         }
     }
   }

</script>
</head>
<body >


<span style="background-color:blue;color:yellow;"  
onClick="runPhp('GetFromMysql')"/>
Click this to get data from MySql.<br> 
</span><br><br>

<span style="background-color:blue;color:yellow;"  
onClick="runPhp('MysqlToFile')"/>
Click this to get data from MySql and store them in a file instead of displaying.<br> <br> <br> 
</span>

<span style="background-color:blue;color:yellow;"  
onClick="getXml()"/>
Click to read the xml file.<br> 
</span>

<textarea rows="10" cols="88"  id="textareaResults">
</textarea>
 <br><br>
<pre><div    id="divResults"></div></pre>
<br><br>




</body>
</html>

and this is script.php

<?PHP

    $xml_output = routine();
    if($_POST['Args']==='GetFromMysql')
        echo $xml_output .= "</records>"; 

    elseif($_POST['Args']==='MysqlToFile')
    {
        $fileName =  'temp.xml';
        $fp = fopen($fileName, 'w');
        fwrite($fp,  $xml_output); 
        fclose($fp); 
        echo "temp.xml";//return the file name
    }

    function routine()
    {
        mysql_connect('localhost', 'root',''); 
        mysql_select_db("mysql");
        $query="select * from help_category;"; 
        $resultID = mysql_query($query ) or die("Data not found."); 
        $xml_output = "<?xml version=\"1.0\"?>\n"; 
        $xml_output .= "<records>\n"; 

        for($x = 0 ; $x < mysql_num_rows($resultID) ; $x++)
        { 
            $row = mysql_fetch_assoc($resultID); 
            $xml_output .= "\t<record>\n"; 
            $xml_output .= "\t\t<help_category_id>" . $row['help_category_id'] . "</help_category_id>\n";  
            $xml_output .= "\t\t<name>" . $row['name'] . "</name>\n";  
            $xml_output .= "\t\t<parent_category_id>" . $row['parent_category_id'] . "</parent_category_id>\n"; 
            $xml_output .= "\t</record>\n"; 
        }
        return $xml_output;
    }

?>

danielrozo · Accepted Answer · 2012-03-27 17:21:13Z

0

I've solved the issue, it was this line:

var dnis = xml.documentElement.getElementsByTagName("dni");

That was supossed to be substituted for:

var dnis = xml.documentElement.getElementsByTagName("persona");

Because I was generating persona nodes.

Thanks for your answers!

answered Mar 27, 2012 at 17:21

danielrozo

1,4421 gold badge10 silver badges22 bronze badges

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