sasori Posted February 13, 2010 Share Posted February 13, 2010 hi, am trying to learn how php + ajax works, but this example from w3schools website doesn't work for me , i have re-written the code manually (twice) and I ended up copy pasting. but still it doesn't show any data from my db html code <html> <head> <script type="text/javascript" src="selectuser.js"></script> </head> <body> <form> Select a User: <select name ="users" onchange="showUser(this.value)"> <option value="1">Peter Griffin</option> <option value="2">Lois Griffin</option> <option value="3">Glenn Quagmire</option> <option value="4">Joseph Swanson</option> </select> </form> <br /> <div id="txtHint"><b>Person info will be listed here.</b></div> </body> </html> the ajax code var xmlhttp; function showUser(str){ xmlhttp.GetXmlHttpObject(); if(xmlhttp == null){ alert("Your browser does not support Ajax"); return; } var url = "getuser.php"; url=url+"?q="+str; url=url+"&sid="+ Math.random(); xmlhttp.onreadystatechange = stateChanged; xmlhttp.open("GET",url,true); xmlhttp.send(null); } function stateChanged(){ if(xmlhttp.readyState == 4){ document.getElementById("txtHint").innerHTML = xmlhttp.responseText; } } function GetXmlHttpRequest(){ if(window.XMLHttpRequest){ return new XMLHttpRequest(); } if(window.ActiveXObject){ return new ActiveXObject("Microsoft.XMLHTTP"); } return null; } php code <?php $q = $_GET["q"]; $con = mysql_connect('localhost','root','mypassword'); if(!$con){ die("Could not connect: " mysql_error()); } mysql_select_db('ajaxsample', $con); $sql = "SELECT * FROM user WHERE id ='".$q."'"; $result = mysql_query($sql); echo "<table border='1'> <tr> <th>Firstname</th> <th>Lastname</th> <th>Age</th> <th>Hometown</th> <th>Job</th> </tr>"; while($row = mysql_fetch_array($result)){ echo "<tr>"; echo "<td>" . $row['FirstName'] . "</td>"; echo "<td>" . $row['LastName']. "</td>"; echo "<td>" . $row['Age']. "</td>"; echo "<td>" . $row['Hometown']."</td>"; echo "<td>" . $row['Job'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> :'( Quote Link to comment https://forums.phpfreaks.com/topic/191945-php-ajax-help/ Share on other sites More sharing options...
MRushton Posted February 13, 2010 Share Posted February 13, 2010 You've done: xmlhttp.GetXmlHttpObject(); It should be: xmlhttp = GetXmlHttpObject(); Quote Link to comment https://forums.phpfreaks.com/topic/191945-php-ajax-help/#findComment-1011708 Share on other sites More sharing options...
sasori Posted February 13, 2010 Author Share Posted February 13, 2010 arghhh...thanks for pointing that out. I got another mistake,.it now works Quote Link to comment https://forums.phpfreaks.com/topic/191945-php-ajax-help/#findComment-1011709 Share on other sites More sharing options...
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