我有使用复杂的laravel查询创建的集合,这个查询的结果太大了。所以我想我必须使用algolia。据我所知,algolia将模型表数据作为json
自行获取并从那里开始服务。
$result = User::search("UserName")->get();
它需要某些模型配置,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(上例)。我想问的是,我有复杂的查询和结果有太多来自另一个表(加入)的属性。我想在我的自定义查询结果上进行搜索。有可能吗?
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
})...
答案 0 :(得分:0)
我认为最好的方法是使用此请求并链接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();