Laravel - Union + Paginate at the same time?

问题:

Brief:

I am trying to union 2 tables recipes and posts then add ->paginate(5) to the queries.

But for some reason I get this error:

Cardinality violation: 1222 The used SELECT statements have a different number of columns (SQL: (select count(*) as aggregate from posts

Code:

$recipes = DB::table("recipes")->select("id", "title", "user_id", "description", "created_at")
                    ->where("user_id", "=", $id);

$items = DB::table("posts")->select("id", "title", "user_id", "content", "created_at")
                ->where("user_id", "=", $id)
                ->union($recipes)
                ->paginate(5)->get();

Am i doing something wrong?

Without ->paginate(5) the query works fine.

回答1:

You're right, pagination cause problem. Right now, you can create a view and query the view instead of the actual tables, or create your Paginator manually:

$page = Input::get('page', 1);
$paginate = 5;

$recipes = DB::table("recipes")->select("id", "title", "user_id", "description", "created_at")
            ->where("user_id", "=", $id);
$items = DB::table("posts")->select("id", "title", "user_id", "content", "created_at")
            ->where("user_id", "=", $id)
            ->union($recipes)
            ->get();

$slice = array_slice($items->toArray(), $paginate * ($page - 1), $paginate);
$result = Paginator::make($slice, count($items), $paginate);

return View::make('yourView',compact('result'));

回答2:

I faced this kind of issue already. I found a thread also not about pagination but about unions.

Please see this link : Sorting UNION queries with Laravel 4.1

@Mohamed Azher has shared a nice trick and it works on my issue.

$query = $query1->union($query2);
$querySql = $query->toSql();
$query = DB::table(DB::raw("($querySql order by foo desc) as a"))->mergeBindings($query);

This creates an sql like below:

select * from (
  (select a as foo from foo)
  union
  (select b as foo from bar)
) as a order by foo desc;

And you can already utilize Laravel's paginate same as usual like $query->paginate(5). (but you have to fork it a bit to fit to your problem)

回答3:

order by

 $page = Input::get('page', 1);

 $paginate = 5;

 $recipes = DB::table("recipes")->select("id", "title", "user_id", "description", "created_at")
                ->where("user_id", "=", $id);
$items = DB::table("posts")->select("id", "title", "user_id", "content", "created_at")
            ->where("user_id", "=", $id)
            ->union($recipes)
            ->orderBy('created_at','desc')
            ->get();

$slice = array_slice($items, $paginate * ($page - 1), $paginate);
$result = Paginator::make($slice, count($items), $paginate);

return View::make('yourView',compact('result'))->with( 'result', $result );

View page :

   @foreach($result as $data)
  {{ $data->your_column_name;}}
 @endforeach 

  {{$result->links();}}   //for pagination

its help to more peoples.. because nobody cant understand show data in view page union with pagination and orderby .. thank u

回答4:

I had this same problem, and unfortunately I couldn't get the page links with {{ $result->links() }}, but I found another way to write the pagination part and the page links appears

Custom data pagination with Laravel 5

//Create a new Laravel collection from the array data
$collection = new Collection($searchResults);

//Define how many items we want to be visible in each page
$perPage = 5;

//Slice the collection to get the items to display in current page
$currentPageSearchResults = $collection->slice($currentPage * $perPage, $perPage)->all();

//Create our paginator and pass it to the view
$paginatedSearchResults= new LengthAwarePaginator($currentPageSearchResults, count($collection), $perPage);

return view('search', ['results' => $paginatedSearchResults]);

回答5:

Reiterating jdme's answer with a more elegant method from Illuminate\Database\Query\Builder.

$recipes = DB::table("recipes") ..
$items = DB::table("posts")->union($recipes) ..

$query = DB::query()
    ->fromSub($items, "some_query_name");

// Let's paginate!
$query->paginate(5);

I hope this helps!

回答6:

I know this answer is too late. But I want to share my problems and my solution.

My problems:

1. Join with many tables at the same time
2. UNION
3. Paginate (Must use, because I have to use a common theme to show pagination. If I made own custom for pagination, it will not match to current. And in the future, a common theme may be changed.)
4. Big data: view took 4 seconds, page load took 4 seconds => total is 8 seconds. (But if I set condition inside that view, it was least than 1 second for total.)

Query

※This is the sample. MariaDB, about 146,000 records.

SELECT A.a_id
     , A.a_name
     , B.organization_id
     , B.organization_name
  FROM customers A 
    LEFT JOIN organizations B ON (A.organization_id = B.organization_id)

UNION ALL

SELECT A.a_id
     , A.a_name
     , B.organization_id
     , B.organization_name
  FROM employees A 
    LEFT JOIN organizations B ON (A.organization_id = B.organization_id)

Solution

Reference from www.tech-corgi.com (やり方２), I updated my PHP code to filter inside my query, and then call paginate normally.

I must add a condition (filter) before getting large records. In this example is organization_id.

$query = "
    SELECT A.a_id
         , A.a_name
         , B.organization_id
         , B.organization_name
      FROM customers A 
        LEFT JOIN organizations B ON (A.organization_id = B.organization_id)
     WHERE 1 = 1
       AND B.organization_id = {ORGANIZATION_ID}

    UNION ALL

    SELECT A.a_id
         , A.a_name
         , B.organization_id
         , B.organization_name
      FROM employees A 
        LEFT JOIN organizations B ON (A.organization_id = B.organization_id)

     WHERE 1 = 1
       AND B.organization_id = {ORGANIZATION_ID}
";

$organization_id = request()->organization_id;
$query = str_replace("{ORGANIZATION_ID}", $organization_id, $query);

But it still cannot be used in paginate(). There is a trick to solve this problem. See below.

Final code

Trick: put query inside (). For example: (SELECT * FROM TABLE_A).

Reason: paginage() will generate and run Count query SELECT count(*) FROM (SELECT * FROM TABLE_A), if we did not put inside brackets, Count query would not be a correct query.

$query = "
    ( SELECT A.a_id
         , A.a_name
         , B.organization_id
         , B.organization_name
      FROM customers A 
        LEFT JOIN organizations B ON (A.organization_id = B.organization_id)
     WHERE 1 = 1
       AND B.organization_id = {ORGANIZATION_ID}

    UNION ALL

    SELECT A.a_id
         , A.a_name
         , B.organization_id
         , B.organization_name
      FROM employees A 
        LEFT JOIN organizations B ON (A.organization_id = B.organization_id)

     WHERE 1 = 1
       AND B.organization_id = {ORGANIZATION_ID}
    ) AS VIEW_RESULT
";

$organization_id = request()->organization_id;
$query = str_replace("{ORGANIZATION_ID}", $organization_id, $query);

$resultSet = DB::table(DB::raw($query))->paginate(20);

Now I can use it normally:

1. SELECT, JOIN, UNION
2. paginate
3. High performance: Filter data before getting

Hope it help!!!

回答7:

The accepted answer works great for Query Builder.

But here's my approach for Laravel Eloquent Builder.

Assume that we're referring to same Model

$q1 = Model::createByMe();       // some condition
$q2 = Model::createByMyFriend(); // another condition

$q2->union($q1);
$querySql = $q2->toSql();

$query = Model::from(DB::raw("($querySql) as a"))->select('a.*')->addBinding($q2->getBindings());

$paginated_data = $query->paginate();

I'm using Laravel 5.6

回答8:

Getting the total count for pagination is the problem here. This is the error I got when used $builder->paginate()

"SQLSTATE[21000]: Cardinality violation: 1222 The used SELECT statements have a different number of columns (SQL: (select count(*) as aggregate from `institute_category_places` where `status` = approved and (`category_id` in (3) or `name` LIKE %dancing class% or `description` LIKE %dancing class% or `address_line1` LIKE %dancing class% or `address_line2` LIKE %dancing class% or `city` LIKE %dancing class% or `province` LIKE %dancing class% or `country` LIKE %dancing class%) and `institute_category_places`.`deleted_at` is null) union (select * from `institute_category_places` where `status` = approved and (`category_id` in (3, 4) or `name` LIKE %dancing% or `description` LIKE %dancing% or `address_line1` LIKE %dancing% or `address_line2` LIKE %dancing% or `city` LIKE %dancing% or `province` LIKE %dancing% or `country` LIKE %dancing% or `name` LIKE %class% or `description` LIKE %class% or `address_line1` LIKE %class% or `address_line2` LIKE %class% or `city` LIKE %class% or `province` LIKE %class% or `country` LIKE %class%) and `institute_category_places`.`deleted_at` is null))"

If you want to paginate without total count you can use

$builder->limit($per_page)->offset($per_page * ($page - 1))->get();

to get only set of rows in the page.

Getting all the rows and counting total is memory inefficient. So I used following approach to get total count.

    $bindings = $query_builder->getBindings();
    $sql = $query_builder->toSql();
    foreach ($bindings as $binding) {
        $value = is_numeric($binding) ? $binding : "'" . $binding . "'";
        $sql = preg_replace('/\?/', $value, $sql, 1);
    }
    $sql = str_replace('\\', '\\\\', $sql);

    $total = DB::select(DB::raw("select count(*) as total_count from ($sql) as count_table"));

Then we have to paginate the result manually.

    $page = Input::get('page', 1);
    $per_page = 15;

    $search_results = $query_builder->limit($per_page)->offset($per_page * ($page - 1))->get();

    $result = new LengthAwarePaginator($search_results, $total[0]->total_count, $per_page, $page, ['path' => $request->url()]);

If you can use raw sql queries, it is much more CPU and memory efficient.

回答9:

$page = Input::get('page', 1);
$paginate = 5;
$recipes = DB::table("recipes")->select("id", "title", "user_id", "description", "created_at")
            ->where("user_id", "=", $id);
$items = DB::table("posts")->select("id", "title", "user_id", "content", "created_at") ->where("user_id", "=", $id)->union($recipes)->get()->toArray();
$slice = array_slice($items, $paginate * ($page - 1), $paginate);
$result = new Paginator($slice , $paginate);