【已解决】webman路由

开赛灵 更新于2021-08-20

如何做到
只设置一条路由 /api/* 所有生效
例如 /api/user/query

 3173  3 3

3个回答

葱香小油条 2021-08-20

https://github.com/walkor/webman/issues/54

开赛灵 2021-08-20

谢谢你

不败少龙 2021-08-20

能不能把api/v1,api/v2弄成自动匹配

葱香小油条 2021-08-20

根据https://github.com/walkor/webman/issues/54作者的答复把v1改成verion，在加点判断版本，类存不存在啥的，自由发挥，如下：
Route::any('/api/{verion}/{controller}/{action}', function($request, $verion, $controller, $action){
$class_name = "app\api\{$verion}\controller\" . $controller;
var_export($verion);
$controller = new $class_name;
return call_user_func([$controller, $action], $request);
});
不败少龙 2021-08-20

OK 谢谢我去试试
yzh52521 2021-08-20

Route::group('/api',function () {
Route::group("/{verion}", function () {
Route::any('/{controller}/{action}', function ($request, $verion, $controller, $action) {
$class_name = "app\api\{$verion}\controller\" . $controller;
$controller = new $class_name;
return call_user_func([$controller, $action], $request);
});
});
});
不败少龙 2021-08-20

已经写好了谢谢

不败少龙 2021-08-20

Route::group('/api/{verion}', function () {
    Route::group("/{controller}", function () {
        Route::any("/{action}", function (Request $request, $verion, $controller, $action) {
            $class_name = 'app\\api\\controller\\' . $verion . '\\' . $controller;
            if(!is_dir( 'app\\api\\controller\\' . $verion)){
                return json(['code'=>500,'msg'=>$verion.'目录不存在!']);
            }
            if(!class_exists($class_name)){
                return json(['code'=>500,'msg'=>$verion."目录下的控制器: {$controller}不存在!"]);
            }
            if (!method_exists($class_name, $action)) {
                return json(['code'=>500,'msg'=>$verion."目录下的控制器: {$controller}里面的方法: {$action}不存在"]);
            }
            $controller = new $class_name;
            $request->controller = $class_name;
            return call_user_func([$controller, $action], $request);
        });
    });
});

本地目录结构