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Php 单击事件打开新刀片页面{{$data->;title}_Php_Laravel_Laravel Blade - Fatal编程技术网

Php 单击事件打开新刀片页面{{$data->;title}

Php 单击事件打开新刀片页面{{$data->;title},php,laravel,laravel-blade,Php,Laravel,Laravel Blade,我正在从数据库中提取数据,我想使{{{$data->title}}可点击 “这适用于新服务器。过去,我在现有服务器上尝试过。” @foreach($ress as $key => $data) {{$data->title}} @endforeach 在routes.php中定义一个路由,并在模板中调用它 <a href="{{ route('route-name', $data->id) }}">{{ $data->title }}</a>

我正在从数据库中提取数据,我想使
{{{$data->title}}
可点击

“这适用于新服务器。过去,我在现有服务器上尝试过。”

@foreach($ress as $key => $data)

{{$data->title}}

@endforeach

routes.php
中定义一个路由,并在模板中调用它

<a href="{{ route('route-name', $data->id) }}">{{ $data->title }}</a>

这是最好的方法。您可以将变量名更改为代码中使用的名称

Route.php

Route::get('/show/{id}','DataController@show')->name('showdata);
@foreach($datas as $data)
 <a href="{{ route('showdata', $data->id) }}">{{ $data->title }}</a>
@endforeach
public function show(){
 $datas = DB::table('table_name')->get();
 return view('dataindex', ['datas'=>$datas]);
saydataindex.blade.php上的链接

Route::get('/show/{id}','DataController@show')->name('showdata);
@foreach($datas as $data)
 <a href="{{ route('showdata', $data->id) }}">{{ $data->title }}</a>
@endforeach
public function show(){
 $datas = DB::table('table_name')->get();
 return view('dataindex', ['datas'=>$datas]);

}

有一个按钮或超链接。当有人点击标题时会发生什么情况?打开包含帖子详细信息的新页面,然后你应该查看@timw answer。我很难根据标题定义可变路线,因为这将是一个slug for SEO,你能用路径定义提示我吗?我是第一次在laravelAs的初级开发人员。路径根据标题动态变化