PHP-DataTables为mysql\u查询从表行发布用户名?
我正在尝试向我的DataTables表中添加一个delete按钮,但我不知道在按下delete按钮时如何发布所选行的用户名 以下是数据表的打印方式:PHP-DataTables为mysql\u查询从表行发布用户名?,php,mysql,post,datatable,datatables,Php,Mysql,Post,Datatable,Datatables,我正在尝试向我的DataTables表中添加一个delete按钮,但我不知道在按下delete按钮时如何发布所选行的用户名 以下是数据表的打印方式: <div class="panel-body"> <div class="dataTable_wrapper"> <table class="table table-striped table-bordered table-hover" id="dataTables-example">
<div class="panel-body">
<div class="dataTable_wrapper">
<table class="table table-striped table-bordered table-hover" id="dataTables-example">
<thead>
<tr>
<th>Username</th>
<th>Email</th>
<th>Fullname</th>
<th>Class ID</th>
<th>Delete</th>
</tr>
</thead>
<tbody>
<?php
while($listDetailsRow = mysql_fetch_array($listDetails)){
?>
<tr>
<td class="success"><?=$listDetailsRow['username']?></td>
<td class="info"><?=$listDetailsRow['email']?></td>
<td class="success"><?=$listDetailsRow['fullname']?></td>
<td class="info"><?=$listDetailsRow['class_id']?></td>
<td>
<form id='delete' action='deleteUser.php' method='post'>
<button type="submit" name="submit" class="btn btn-danger" href="deleteUser.php">
Delete
</button>
</form>
</td>
</tr>
<?php
}
?>
</tbody>
</table>
</div>
我只是想让它从按下删除按钮的那一行打印用户名
这可能吗?JS或Ajax不太好,因此非常感谢所有帮助。您可以使用a元素而不是按钮元素中的表单。请参阅此代码:
<form id='delete' action='deleteUser.php' method='post'>
<input type='hidden' name='username' value="<?=$listDetailsRow['username'];?>" > <--add this line
<button type="submit" name="submit" class="btn btn-danger" href="deleteUser.php">
Delete
</button>
</form>
<tbody>
<?php
while($listDetailsRow = mysql_fetch_array($listDetails)){
?>
<tr>
<td class="success"><?=$listDetailsRow['username']?></td>
<td class="info"><?=$listDetailsRow['email']?></td>
<td class="success"><?=$listDetailsRow['fullname']?></td>
<td class="info"><?=$listDetailsRow['class_id']?></td>
<td>
<!------------- New -------------->
<?php
$params = array( 'username' => $listDetailsRow['username'] );
$href = "deleteUser.php?" . http_build_query($params);
?>
<a class="btn btn-danger" href="<?php echo $href; ?>" > Delete </a>
<!------------- End New -------------->
</td>
</tr>
<?php
}
?>
</tbody>
您可以在button元素中使用a元素而不是表单。请参阅此代码:
<tbody>
<?php
while($listDetailsRow = mysql_fetch_array($listDetails)){
?>
<tr>
<td class="success"><?=$listDetailsRow['username']?></td>
<td class="info"><?=$listDetailsRow['email']?></td>
<td class="success"><?=$listDetailsRow['fullname']?></td>
<td class="info"><?=$listDetailsRow['class_id']?></td>
<td>
<!------------- New -------------->
<?php
$params = array( 'username' => $listDetailsRow['username'] );
$href = "deleteUser.php?" . http_build_query($params);
?>
<a class="btn btn-danger" href="<?php echo $href; ?>" > Delete </a>
<!------------- End New -------------->
</td>
</tr>
<?php
}
?>
</tbody>
您从未在表单中嵌入任何标识信息,因此无论单击哪个按钮,每次都会发送完全相同的信息。另外,在每个表单标签上都会有许多重复的ID,这是非法的html。您从未在表单中嵌入任何标识信息,因此无论您单击哪个按钮,每次都会发送完全相同的信息。另外,在每个表单标记上都有许多重复的ID,这是非法的html。
<tbody>
<?php
while($listDetailsRow = mysql_fetch_array($listDetails)){
?>
<tr>
<td class="success"><?=$listDetailsRow['username']?></td>
<td class="info"><?=$listDetailsRow['email']?></td>
<td class="success"><?=$listDetailsRow['fullname']?></td>
<td class="info"><?=$listDetailsRow['class_id']?></td>
<td>
<!------------- New -------------->
<?php
$params = array( 'username' => $listDetailsRow['username'] );
$href = "deleteUser.php?" . http_build_query($params);
?>
<a class="btn btn-danger" href="<?php echo $href; ?>" > Delete </a>
<!------------- End New -------------->
</td>
</tr>
<?php
}
?>
</tbody>
$userDel = isset($_GET['username']) ? $_GET['username'] : null;