在我的应用程序中,我有一个带有顺序和编辑按钮的用户记录表。当我单击编辑按钮时,我想在该行的下面打开一个新表单,并以该表单获取该用户的数据。我使用href传递该用户ID。 这是我的用户记录表代码
<div class="panel-body">
<div class="dataTable_wrapper">
<table class="table table-striped table-bordered table-hover" id="dataTables-example">
<thead>
<tr>
<th>Candidate</th>
<th>Gender</th>
<th>Date of Birth</th>
<th>Gotra</th>
<th>Action</th>
</tr>
</thead>
<tbody>
<?php
$result=mysqli_query($conn,"select candidateid,candidatename,gender,dob,gotra from candidateinfo where clientid='$clientid' ");
while($row = mysqli_fetch_array($result))
{
echo "<tr class='odd gradeX'>";
echo "<td>" . $row[1] . "</td>";
echo "<td>" . $row[2] . "</td>";
echo "<td>" . $row[3] . "</td>";
echo "<td>" . $row[4] . "</td>";
echo "<td><a data-toggle='tooltip' data-placement='bottom' title='Order' href=reorders.php?vcandidateid=" . urlencode(base64_encode("$row[0]")) . ">Order</a>";
echo " || <a data-toggle='tooltip' data-placement='bottom' title='Edit' href=candidateedit.php?vcandidateid=" . urlencode(base64_encode("$row[0]")) . ">Edit</a></td>";
echo "</tr>";
}
?>
</tbody>
</table>
</div>
</div>
我从此表中的数据库中检索数据。请帮助我。
jquery和表单
<script>
$(function() {
$('#edit').click(function() {
$('#update_form').show();
});
});
</script>
<form id="update_form" class="form-horizontal" action="candidatedb.php" method="post">
<div class="col-xs-12 col-sm-6 col-md-6 col-lg-6">
<div class="form-group">
<label class="hidden-xs col-sm-4 control-label" for="candidateName"><small class="color-red vlarge-text">*</small>Candidate Name</label>
<div class="col-sm-8 xs-padding-l-r-none">
<input type="text" id="candidatename" name="candidatename">
<span class="help-block errmsg margin-none"></span>
</div>
</div>
</div>
<div class="col-xs-12 col-sm-12 col-md-12 col-lg-12">
<div class="form-group">
<button class="col-sm-2 btn btnDark" id="Submit" type="submit" value="" >Submit</button>
</div>
</div>
</form>