Php 是否使用laravel表单确认删除?
确认删除有问题,所有工作正常,但如果在确认弹出窗口中单击取消,它将从数据库中删除该项。有什么想法吗?请用代码示例说明,我是拉威尔的新手Php 是否使用laravel表单确认删除?,php,forms,laravel,confirm,Php,Forms,Laravel,Confirm,确认删除有问题,所有工作正常,但如果在确认弹出窗口中单击取消,它将从数据库中删除该项。有什么想法吗?请用代码示例说明,我是拉威尔的新手 <script> function ConfirmDelete() { var x = confirm("Are you sure you want to delete?"); if (x) return true; else return false; } </script> {!! For
<script>
function ConfirmDelete()
{
var x = confirm("Are you sure you want to delete?");
if (x)
return true;
else
return false;
}
</script>
{!! Form::open(['method' => 'DELETE', 'route' => ['path_delete_time', $time->id], 'onsubmit' => 'ConfirmDelete()']) !!}
{!! Form::hidden('case_id', $project->id, ['class' => 'form-control']) !!}
{!! Form::button('<i class="glyphicon glyphicon-trash"></i>', array('type' => 'submit', 'class' => 'specialButton')) !!}
{!! Form::close() !!}
函数ConfirmDelete()
{
var x=确认(“您确定要删除吗?”);
if(x)
返回true;
其他的
返回false;
}
{!!Form::open(['method'=>'DELETE','route'=>['path\u DELETE\u time',$time->id],'onsubmit'=>'ConfirmDelete()'))
{!!Form::hidden('case_id',$project->id,['class'=>'Form control'])
{!!表单::按钮(“”,数组('type'=>'submit','class'=>'specialButton'))
{!!Form::close()!!}
现在发生的唯一一件事是,您会看到一个弹出窗口,延迟重定向到删除页面
您可能只需要生成一个链接,该链接会在js中生成一个表单 只需将return添加到您的
onsubmit
通话中即可。因此,函数返回的值决定表单是否被提交
'onsubmit' => 'return ConfirmDelete()'
您可以为表单提供一个id,并使用jQuery:
{!! Form::open(['method' => 'DELETE', 'route' => ['path_delete_time', $time->id], 'id' => 'FormDeleteTime']) !!}
<script type="text/javascript">
$("#FormDeleteTime").submit(function (event) {
var x = confirm("Are you sure you want to delete?");
if (x) {
return true;
}
else {
event.preventDefault();
return false;
}
});
</script>
{!!Form::open(['method'=>'DELETE','route'=>['path\u DELETE\u time',$time->id],'id'=>'FormDeleteTime'])
$(“#FormDeleteTime”).submit(函数(事件){
var x=确认(“您确定要删除吗?”);
if(x){
返回true;
}
否则{
event.preventDefault();
返回false;
}
});
提交表单或删除记录时,简单地将jQuery代码放入提示框
jQuery(document).ready(function($){
$('.deleteGroup').on('submit',function(e){
if(!confirm('Do you want to delete this item?')){
e.preventDefault();
}
});
});
在我的新项目(Laravel 5.3)中,我使用了以下内容:
{!! Form::open(['method' => 'DELETE', 'route' => ['admin.user.delete', $user], 'onsubmit' => 'return confirmDelete()']) !!}
<button type="submit" name="button" class="btn btn-default btn-sm">
<i class="fa fa-trash-o"></i>
</button>
{!! Form::close() !!}
没有任何函数调用
'onsubmit' => 'return confirm("are you sure ?")'
范例
{!! Form::open(['url'=>'posts/'.$post->id.'/delete','method'=>'DELETE','class'=>'form-horizontal',
'role'=>'form','onsubmit' => 'return confirm("are you sure ?")'])!!}
使用laravel 5,我想我可以概括一下: .html
$(“.specialButton”)。单击(函数(){
返回确认(“是否要删除此项?”);
});
我知道这已经有一段时间了,但这是最顺利的方式:
'onsubmit' => 'return confirm("Do you want to delete this user?");'
在表单标签中。非常感谢,通过Laravel的“数据确认”链接很容易,但这一点很难解决。只需返回
结果
,而这段代码片段可能会解决问题,真正有助于提高您文章的质量。请记住,您将在将来回答读者的问题,这些人可能不知道您的代码建议的原因。
<form method="post" action="..." class="has-confirm" data-message="Delete this Thing?">...</form>
$("form.has-confirm").submit(function (e) {
var $message = $(this).data('message');
if(!confirm($message)){
e.preventDefault();
}
});
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script>
$(".specialButton").click(function(){
return confirm("Do you want to delete this ?");
});
</script>
'onsubmit' => 'return confirm("Do you want to delete this user?");'