【发布时间】:2018-02-14 06:54:26
【问题描述】:
我有来自 foreach 的表,在这个表中我使用 ajax (id="detail") 创建提交,但是为什么当我提交时我只得到第一行,当我点击第二行值时,我得到第一行值而不是第二行行值?
<form action="<?php echo base_url() ?>index.php/stock_opname/proses_detail" method="post">
<table class="table table-striped">
<thead>
<tr>
<th>No</th>
<th>SO No</th>
<th>Opname Date</th>
<th>Warehouse</th>
<th>Type</th>
<th>Approve</th>
<th>Close period</th>
<th>Detail</th>
</tr>
</thead>
<tbody>
<?php $no=1; foreach ($show as $key) { ?>
<tr>
<td><?php echo $no ?></td>
<td><input type="text" id="sono" value="<?php echo $key->sono ?>"></td>
<td><input type="text" id="opnamedate" value="<?php echo $key->opnamedate ?>"></td>
<td><input type="text" id="warehousecode" value="<?php echo $key->warehousecode ?>"></td>
<td><input type="text" id="stocktypeid" value="<?php echo $key->stocktypeid ?>"></td>
<td></td>
<td></td>
<td><a href="#" id="detail" class="btn btn-info">Detail</a></td>
</tr>
<?php $no++;} ?>
</tbody>
</table>
</form>
<script type="text/javascript">
//Ajax Load data from ajax
$(document).on('click', '#detail', function(){
var sono = $('#sono').val();
var opnamedate = $('#opnamedate').val();
var wh = $('#warehousecode').val();
var stocktypeid = $('#stocktypeid').val();
alert(sono);
});
【问题讨论】:
-
如何创建动态?
标签: php ajax codeigniter