我有使用复杂的 Laravel 查询创建的集合,并且该查询的结果太大。所以我想我必须使用algolia。据我所知,algolia 将模型表数据作为 json
获取到自身,并从那里提供服务。
$result = User::search("UserName")->get();
它需要一些模型配置,例如 searchAs
等。所有这些都与现有模型相关,您可以使用 search
方法从模型中进行搜索(上面的示例)。我想问的是,我有复杂的查询,结果有太多来自另一个表(已连接)的属性。我想对我的自定义查询结果进行搜索。这可能吗?
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
})...
最佳答案
我相信最好的方法是使用此请求并链接 searchable()
方法。它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();
关于php - 使用 Algolia 搜索自定义查询结果集合 - Laravel,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46372068/