Laravel Route::使用where条件获取
我正在开发Laravel 5.1,我有以下代码:Laravel Route::使用where条件获取,laravel,routes,laravel-5.1,Laravel,Routes,Laravel 5.1,我正在开发Laravel 5.1,我有以下代码: //Routes.php Route::get('/listagem/{campo}','ControllerParque@listar'); //ControllerParque.php public function listar($campo) { $listagem = DB::SELECT("SELECT COUNT(x.".$campo.") AS nreg, x.* FROM (SELECT
//Routes.php
Route::get('/listagem/{campo}','ControllerParque@listar');
//ControllerParque.php
public function listar($campo) {
$listagem = DB::SELECT("SELECT COUNT(x.".$campo.") AS nreg, x.* FROM (SELECT * FROM computadores ORDER BY id DESC) AS x GROUP BY ".$campo." ORDER BY ".$campo." ASC");
$dados = array('listagem' => $listagem, 'campo' => $campo);
return view("ViewListagem")->with($dados);
}
我想限制“listagem/usuario”或“listagem/tag”的任何不同之处。我的意思是,如果用户在导航器(URL栏)中键入“listagem/crap”,它将被拒绝,因为它不是“listagem/usuario”或“listagem/tag”
我试过“where”子句:
Routes.php
Route::get('/listagem/{campo}','ControllerParque@listar')
->where('campo', 'usuario');
但这样我只限制了一个条件,这个例子是'listagem/usuario'
有什么想法吗?我不确定是否可以通过where完成。相反,您可以在控制器功能中检查它
public function listar($campo) {
if(!in_array($campo,['usuario','tag'])) {
// throw an error, return to previous page
return Redirect::back()->withErrors('error message here');
}
// Continue on with your function
}
我不确定是否可以通过where完成。相反,您可以在控制器功能中检查它
public function listar($campo) {
if(!in_array($campo,['usuario','tag'])) {
// throw an error, return to previous page
return Redirect::back()->withErrors('error message here');
}
// Continue on with your function
}
可以在“where”子句中使用正则表达式<代码>usario |标记仅与
usario
或tag
可以在“where”子句中使用正则表达式<代码>usario |标记仅与
usario
或tag