Laravel Fluent Query Builder Join with subquery

Question

Okay after hours of research and still using DB::select I have to ask this question. Because I am about to trough my computer away ;).

I want to get the last input o

Answer 1

I think what you looking for is "joinSub". It's supported from laravel ^5.6. If you using laravel version below 5.6 you can also register it as macro in your app service provider file. like this https://github.com/teamtnt/laravel-scout-tntsearch-driver/issues/171#issuecomment-413062522

$subquery = DB::table('catch-text')
            ->select(DB::raw("user_id,MAX(created_at) as MaxDate"))
            ->groupBy('user_id');

$query = User::joinSub($subquery,'MaxDates',function($join){
          $join->on('users.id','=','MaxDates.user_id');
})->select(['users.*','MaxDates.*']);

Answer 2

Query with sub query in Laravel

$resortData = DB::table('resort')
        ->leftJoin('country', 'resort.country', '=', 'country.id')
        ->leftJoin('states', 'resort.state', '=', 'states.id')
        ->leftJoin('city', 'resort.city', '=', 'city.id')
        ->select('resort.*', 'country.name as country_name', 'states.name as state_name','city.name as city_name', DB::raw("(SELECT GROUP_CONCAT(amenities.name) from resort_amenities LEFT JOIN amenities on amenities.id= resort_amenities.amenities_id WHERE resort_amenities.resort_id=resort.id) as amenities_name"))->groupBy('resort.id')
        ->orderBy('resort.id', 'DESC')
       ->get();

Answer 3

I can't comment because my reputation is not high enough. @Franklin Rivero if you are using Laravel 5.2 you can set the bindings on the main query instead of the join using the setBindings method.

So the main query in @ph4r05's answer would look something like this:

$q = DnsResult::query()
    ->from($dnsTable . ' AS s')
    ->join(
        DB::raw('(' . $qqSql. ') AS ss'),
        function(JoinClause $join) {
            $join->on('s.watch_id', '=', 'ss.watch_id')
                 ->on('s.last_scan_at', '=', 'ss.last_scan');
        })
    ->setBindings($subq->getBindings());

Answer 4

You can use following addon to handle all subquery related function from laravel 5.5+

https://github.com/maksimru/eloquent-subquery-magic

User::selectRaw('user_id,comments_by_user.total_count')->leftJoinSubquery(
  //subquery
  Comment::selectRaw('user_id,count(*) total_count')
      ->groupBy('user_id'),
  //alias
  'comments_by_user', 
  //closure for "on" statement
  function ($join) {
      $join->on('users.id', '=', 'comments_by_user.user_id');
  }
)->get();

Answer 5

I was looking for a solution to quite a related problem: finding the newest records per group which is a specialization of a typical greatest-n-per-group with N = 1.

The solution involves the problem you are dealing with here (i.e., how to build the query in Eloquent) so I am posting it as it might be helpful for others. It demonstrates a cleaner way of sub-query construction using powerful Eloquent fluent interface with multiple join columns and where condition inside joined sub-select.

In my example I want to fetch the newest DNS scan results (table scan_dns) per group identified by watch_id. I build the sub-query separately.

The SQL I want Eloquent to generate:

SELECT * FROM `scan_dns` AS `s`
INNER JOIN (
  SELECT x.watch_id, MAX(x.last_scan_at) as last_scan
  FROM `scan_dns` AS `x`
  WHERE `x`.`watch_id` IN (1,2,3,4,5,42)
  GROUP BY `x`.`watch_id`) AS ss
ON `s`.`watch_id` = `ss`.`watch_id` AND `s`.`last_scan_at` = `ss`.`last_scan`

I did it in the following way:

// table name of the model
$dnsTable = (new DnsResult())->getTable();

// groups to select in sub-query
$ids = collect([1,2,3,4,5,42]);

// sub-select to be joined on
$subq = DnsResult::query()
    ->select('x.watch_id')
    ->selectRaw('MAX(x.last_scan_at) as last_scan')
    ->from($dnsTable . ' AS x')
    ->whereIn('x.watch_id', $ids)
    ->groupBy('x.watch_id');
$qqSql = $subq->toSql();  // compiles to SQL

// the main query
$q = DnsResult::query()
    ->from($dnsTable . ' AS s')
    ->join(
        DB::raw('(' . $qqSql. ') AS ss'),
        function(JoinClause $join) use ($subq) {
            $join->on('s.watch_id', '=', 'ss.watch_id')
                 ->on('s.last_scan_at', '=', 'ss.last_scan')
                 ->addBinding($subq->getBindings());  
                 // bindings for sub-query WHERE added
        });

$results = $q->get();

UPDATE:

Since Laravel 5.6.17 the sub-query joins were added so there is a native way to build the query.

$latestPosts = DB::table('posts')
                   ->select('user_id', DB::raw('MAX(created_at) as last_post_created_at'))
                   ->where('is_published', true)
                   ->groupBy('user_id');

$users = DB::table('users')
        ->joinSub($latestPosts, 'latest_posts', function ($join) {
            $join->on('users.id', '=', 'latest_posts.user_id');
        })->get();

Answer 6

Ok for all of you out there that arrived here in desperation searching for the same problem. I hope you will find this quicker then I did ;O.

This is how it is solved. JoostK told me at github that "the first argument to join is the table (or data) you're joining.". And he was right.

Here is the code. Different table and names but you will get the idea right? It t

DB::table('users')
        ->select('first_name', 'TotalCatches.*')

        ->join(DB::raw('(SELECT user_id, COUNT(user_id) TotalCatch,
               DATEDIFF(NOW(), MIN(created_at)) Days,
               COUNT(user_id)/DATEDIFF(NOW(), MIN(created_at))
               CatchesPerDay FROM `catch-text` GROUP BY user_id)
               TotalCatches'), 
        function($join)
        {
           $join->on('users.id', '=', 'TotalCatches.user_id');
        })
        ->orderBy('TotalCatches.CatchesPerDay', 'DESC')
        ->get();