Home >>Ajax Tutorial >Display Data Using Ajax
<script> function showUser(str) { if (str=="") { document.getElementById("txtHint").innerHTML=""; return; } 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) { document.getElementById("txtHint").innerHTML=xmlhttp.responseText; } } xmlhttp.open("GET","getdatafromDB.php?q="+str,true); xmlhttp.send(); } </script> <select name="users" onchange="showUser(this.value)"> <option disabled="disabled" selected="selected" value="">Select a person:</option> <option value="1">vikash</option> <option value="2">Ravi</option> <option value="3">Sandeep</option> <option value="4">Saurabh</option> </select> <div id="txtHint" style="width: 400px; border: 2px solid red;"><b>Person info will be listed here.</b></div>
<?php $con = mysql_connect('localhost', 'root', ''); mysql_select_db("test", $con); $q=$_GET["q"]; $sql="SELECT * FROM stuinfo WHERE stu_id ='$q'"; $result = mysql_query($sql); echo "<table border='1'> <tr> <th>Email </th> <th>Mobile</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['mobile'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> ?>