i have created drop down and i want to filter my html tab
Posted: Tue Sep 29, 2015 3:04 am
i have created drop down and i want to filter my html table by selecting the dropdown , but when i select the drop down it create 2 html table ,, but i need one html table
Code: Select all
<!DOCTYPE html>
<!--
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To change this template file, choose Tools | Templates
and open the template in the editor.
-->
<html>
[inline] <script src="https://code.jquery.com/jquery-1.10.2.js"></script>
<div id="txtHint"></div>
<script>
function showUser(str) {
// alert (str);
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
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", "new.php?q=" + str, true);
xmlhttp.send();
}
}
</script>[/inline]
<body>
<form>
<select onchange="showUser(this.value)">
<option value="1" >
1
</option>
<option value="3" >
3
</option>
</select>
<?php
$con = mysqli_connect("localhost", "root", "");
mysqli_select_db($con, "crud_tutorial");
if (isset($_REQUEST['q'])) {
$q = intval($_GET['q']);
//echo "$q";
$sql = "SELECT * FROM customers WHERE id = '" . $q . "'";
$result = mysqli_query($con, $sql);
} else {
$sql = "SELECT * FROM customers ";
$result = mysqli_query($con, $sql);
}
echo "<table>
<tr>
<th>id</th>
<th>name</th>
</tr>";
while ($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
</form>
</body>
</html>