PHP删除按钮上的JavaScript确认
当用户单击删除按钮时,如何添加JavaScript警报以确认(是或否)?我尝试将类添加到警报:PHP删除按钮上的JavaScript确认,javascript,php,html,alert,Javascript,Php,Html,Alert,当用户单击删除按钮时,如何添加JavaScript警报以确认(是或否)?我尝试将类添加到警报: <?php //$con = mysqli_connect("localhost", "root", "root", "db"); $sql = "SELECT * FROM `uploads` where userId = " . $_SESSION['user']; $qry = mysqli_query($conn,$sql) or die(mysqli_error($conn)); $
<?php
//$con = mysqli_connect("localhost", "root", "root", "db");
$sql = "SELECT * FROM `uploads` where userId = " . $_SESSION['user'];
$qry = mysqli_query($conn,$sql) or die(mysqli_error($conn));
$table_content = "";
while($row = mysqli_fetch_assoc($qry)){
$id = $row['id'];
$name = $row['name'];
$table_content .= "<tr>
<td>
<a href='listen.php?id=$id' target='_new'>$name </a>
</td>
<td>
<a href='delete.php?id=$id' type='button' class='btn btn-danger'>delete</a>
</td>
</tr>";
}
echo "<table>".$table_content."</table>";
?>
如果您正在使用引导,您可以尝试一下 如果您不使用引导,请尝试一下!一些伟大的功能准备摇滚一旦你包括css和js!这个jQuery删除确认代码使用引导模式,请参见这里的示例
$(文档).ready(函数(){
$('.delete confirm').deleteConfirm();
});
请参见此处的工作示例,我认为这就是您要寻找的
<a href='delete.php?id=$id' onclick="return myFunction()" type='button' class='btn btn-danger'>delete</a>
<script>
function myFunction() {
var r = confirm("OK to delete?");
if (r == false) {
return false;
}
}
</script>
函数myFunction(){
var r=确认(“确定删除吗?”);
如果(r==false){
返回false;
}
}
您可以使用模式,单击删除按钮,显示该模式,
在模式中,保留两个按钮“是”和“否”,并分别为这两个按钮分配id,如“是”和“否”。单击“是”,执行删除操作;单击“否”,保持行不变
<td><a href='listen.php?id=$id' target='_new'>$name </a></td>
<td><a href='delete.php?id=$id' type='button' id="delete"
class='btn btn-danger'>delete</a></td>
</tr>";
}
echo "<table>".$table_content."</table>"
<script>
$('#delete).on('click',function(){
$('#modal').show();
$('#yes').on('click') {
// perform delete action.
}
}
</script>
";
}
回显“.$table\u内容”
$('#删除).on('单击',函数()){
$('模态').show();
$('yes')。在('click')上{
//执行删除操作。
}
}
为了简单起见,请查看此项
<a onClick="return confirm('Are you sure you want to delete?')" href='delete.php?id=$id' type='button' class='btn btn-danger'>delete</a>
return false实际上在调用它时做了三件非常独立的事情:
- event.preventDefault()
- event.stopPropagation()
- 停止回调执行并在调用时立即返回
请按照
<a href="javascript:void(0)" onclick="return deleteContent('<?php echo $id; ?>');" type="button" class="btn btn-danger">Delete</a>
脚本
<script>
function deleteContent(id) {
if(confirm('Are you sure you want to delete this ?')) {
window.location='delete.php?id='+id;
}
return false;
}
</script>
函数deleteContent(id){
如果(确认('是否确实要删除此?')){
window.location='delete.php?id='+id;
}
返回false;
}
已更新
<?php
$con = mysqli_connect("localhost", "root", "root", "db");
$sql = "SELECT * FROM `uploads` where userId = " . $_SESSION['user'];
$qry = mysqli_query($conn,$sql) or die(mysqli_error($conn));
echo "<table>";
while($row = mysqli_fetch_assoc($qry)){
$id = $row['id'];
$name = $row['name'];
?>
<tr>
<td><a href="javascript:void(0)" onclick="return deleteContent('<?php echo $id; ?>');" type="button" class="btn btn-danger">Delete</a></td>
<td><a href='listen.php?id=<?php echo $id; ?>' target='_new'><?php echo $name; ?> </a></td>
</tr>
<?php } echo "</table>"; ?>
您尝试了什么?出现了什么问题?您必须使用按钮上的单击事件。使用按钮的id
,并基于此,您可以显示提示或模式
(推荐)。并基于该响应,使用ajax启动删除请求。请参阅:尝试向按钮添加确认类,然后添加Javascript单击确认函数(在链接上工作),但不在PHP代码上,如图所示加载页面失败这是代码的基本结构,如果您向我显示问题,我可以帮助您解决无法加载页面的问题:(请明确一点,让它更清楚解释无法加载页面我从代码中删除了?>,因为之后还有更多的php。加载资源失败:服务器响应状态为500(内部服务器错误)尝试使用和不使用?>您看到我添加到顶部问题的完整代码了吗?是的,我看到了,它仍然不工作吗?错误是什么?
<?php
$con = mysqli_connect("localhost", "root", "root", "db");
$sql = "SELECT * FROM `uploads` where userId = " . $_SESSION['user'];
$qry = mysqli_query($conn,$sql) or die(mysqli_error($conn));
echo "<table>";
while($row = mysqli_fetch_assoc($qry)){
$id = $row['id'];
$name = $row['name'];
?>
<tr>
<td><a href="javascript:void(0)" onclick="return deleteContent('<?php echo $id; ?>');" type="button" class="btn btn-danger">Delete</a></td>
<td><a href='listen.php?id=<?php echo $id; ?>' target='_new'><?php echo $name; ?> </a></td>
</tr>
<?php } echo "</table>"; ?>