在Algolia上自定义查询结果集合的搜索 - Laravel
问题描述:
我有用复杂的laravel查询创建的集合,并且此查询的结果太大。所以我认为我必须使用algolia。据我所知,algolia将json
作为模型表格数据自身并从那里投放。在Algolia上自定义查询结果集合的搜索 - Laravel
$result = User::search("UserName")->get();
它需要像searchAs
等一些型号配置。所有与现有模型相关,您可以从模型搜索与search
方法(上面的例子)。我想问的是,我有复杂的查询和结果有太多来自其他表(加入)的属性。我想搜索我的自定义查询结果。可能吗 ?
我的示例查询:
$friendShips = Friend::
join("vp_users as users","users.id","=","friendships.friendID")
->leftJoin("vp_friendships as friendshipsForFriend",function($join) use ($request)
{
$join->on("friendships.friendID","=","friendshipsForFriend.userID");
$join->on("friendshipsForFriend.friendID","=",DB::raw($request->userID));
})
->leftJoin("vp_videos_friends as videosFromFriendMedias",function($join)
{
$join->on("videosFromFriendMedias.userID","=","friendships.friendID");
$join->on("videosFromFriendMedias.friendID", "=" ,"friendships.userID");
$join->on("videosFromFriendMedias.isCalled", "=" , DB::raw(self::CALLED));
})
->leftJoin("vp_videos_friends as videosToFriendMedias",function($join)
{
$join->on("videosToFriendMedias.userID", '=', "friendships.userID");
$join->on("videosToFriendMedias.friendID", '=', "friendships.friendID");
$join->on(function($join){
$join->on("videosToFriendMedias.isCalled", '=', DB::raw(self::CALLED));
$join->orOn("videosToFriendMedias.isActive", '=', DB::raw(self::ACTIVE));
});
})
->leftJoin("vp_videos_friends as
//some join rules too
})...
答
我认为最好的办法是使用这个请求,并链searchable()
方法。它会将查询返回的集合索引到Algolia。
$friendShips = Friend::
join("vp_users as users","users.id","=","friendships.friendID")
->leftJoin("vp_friendships as friendshipsForFriend",function($join) use ($request) {
$join->on("friendships.friendID","=","friendshipsForFriend.userID");
$join->on("friendshipsForFriend.friendID","=",DB::raw($request->userID));
})
->searchable();
+0
感谢您的回答。当我尝试成功时,我会回到你身边。 – fthopkins
+0
不要犹豫,以标志我的答案,如果这对你有用;) –
您是否编制了Friend模型的索引,并且希望为此查询创建单独的索引? –