hello,
im just learning jquery and ajax via YT, and i tried to create a form with combo box, and show records from database based on combo box
after that, i want to store the data to the session
i already created the form that showing the records and already worked, but the button that i created that for store to the session, the ajax is not working.
ajax3.php (index)
<form action="cari.php" method="POST">
<select name="kamar">
<option value="regular single">Regular Single Rp 99.000</option>
<option value="regular double">Regular Double Rp 120.000</option>
<option value="family single">Family Single Rp 127.500</option>
<option value="family double">Family Double Rp 150.000</option>
<option value="vip single">V.I.P Single Rp 170.000</option>
<option value="vip double">V.I.P Double Rp 200.000</option>
</select>
<br>
<button type="submit" name="submit">Submit</button></form><br><div id="tampil"></div>
<script type="text/javascript">$(document).ready(function(){$('form').on('submit', function(e)
{e.preventDefault();$.ajax({type: $(this).attr('method'),url: $(this).attr('action'),data: $(this).serialize(), success: function(data){$("#tampil").html(data);
},});$('.simpanData').on('submit', function(e){e.preventDefault();$.ajax({type: "get",url: $(this).attr('href'),data: $(this).serialize(),success: function(data){$("#tampil").html(data);
},});});});
});
here is cari.php (searching)
<?phpinclude "koneksi.php";$jenis = $_POST['kamar'];$sql= "select * from kamar where jkamar = '$jenis' ";$result = mysqli_query($conn, $sql);
?><table><?phpif(mysqli_num_rows($result) > 0)
{while ($data = mysqli_fetch_array($result))
{$link_simpan = "<a class='simpanData' href='simpan2.php?nokmr=".$data['nokamar']."'>Simpan</a>";?>
<tr>
<td value="<?php echo "nokamar = $data[nokamar]"; ?>"><?php echo $data['nokamar'];?></td>
<td><?php echo $link_simpan ?></td>
</tr><?php
}
}else
{echo "no more data!";
}?></table>
any help will be appreciated, thanks