Question

我想做的是根据命名的路由返回“ post type”。我不想使用像MATCH (L:Location) MATCH p1 = (C1:Crime {type: 'Bicycle theft'})-[:OCCURRED_AT]->(L) MATCH p2 = (C2:Crime {type: 'Burglary'})-[:OCCURRED_AT]->(L) MATCH p3 = (C3:Crime {type: 'Drugs'})-[:OCCURRED_AT]->(L) WITH p1, p2, p3, C1, C2, C3, split(C1.date, '/') as tmp1, split(C2.date, '/') as tmp2, split(C3.date, '/') as tmp3 WITH p1, p2, p3, C1, C2, C3, datetime(tmp1[2] + '-' + tmp1[1] + '-' + tmp1[0]) as dt1, datetime(tmp2[2] + '-' + tmp2[1] + '-' + tmp2[0]) as dt2, datetime(tmp3[2] + '-' + tmp3[1] + '-' + tmp3[0]) as dt3 WHERE dt1 < dt2 AND dt1 > dt3 RETURN p1, p2, p3这样的路由参数。

web.php

Route::get('/{post-type}');

PostController

Route::prefix('/category/{category}')->group(function () { 
    Route::get('/article','PostController@topPostType')->name('articlePostType');

    //Route::get('/video','PostController@topPostType')->name('videoPostType');
});

在PostController中，假设public function topPostType(Category $category, Request $request) { return $this->getPostType($request); } protected function getPostType(Request $request) { if ($request->route()->name('videoPostType')) { return 'video'; } elseif ($request->route()->name('articlePostType')) { return 'article'; } else{ return 'Not working'; } }方法根据路由名称返回一个字符串类型的字符串。问题是，即使我注释掉名为route的videoPostType或使用名为route的articlePostType，我仍然在getPostType()方法中返回“视频”。

Answer 1

您应该使用named()而不是name()

if ($request->route()->named('profile')) {
    //
}

命名路由即使不存在也总是返回true

1 个答案: