Построить массив из присоединиться выберите

Question

У меня есть это соединение выбрать с доктриной:

->select("concat(c.name, ' ', c.second_name) as name_client","c.id as id_client", "p.id as id_project", "s.id as id_status","p.renew_date","p.domain") 
       ->from("clients", "c") 
       ->leftJoin("c", "projects", "p", "c.id = p.id_client") 
       ->leftJoin("c", "todos", "t", "t.id = c.id") 
       ->orderBy('p.renew_date', 'ASC'); 

мне нужно иметь выход JSon так:

[ 
{ 
    "id":1, 
    "name_client" : "Tim", 
    "projects" : [ 
     { 
     "id" : 1 
     "domain" : "www.test.com" 
     }, 
     { 
     "id" : 1 
     "domain" : "www.test.com" 
     } 
    ], 
    "todos": [ 
     { 
     "id" : 1 
     "text" : "do something" 
     }, 
     { 
     "id" : 1 
     "text" : "do something" 
     } 
    ] 
} 
] 

Как я петля мои данные, чтобы получить этот массив состав? Это мой массив данных:

Array 
(
    [0] => Array 
     (
      [name_client] => Aaaaaaa 
      [id_todo] => 
      [text] => 
      [id_client] => 9 
      [id_project] => 13 
      [renew_date] => 
      [domain] => xxx.it 
     ) 

    [1] => Array 
     (
      [name_client] => Bbbbbb 
      [id_todo] => 
      [text] => 
      [id_client] => 8 
      [id_project] => 12 
      [renew_date] => 2016-01-23 
      [domain] => vvvv.it 
     ) 

    [2] => Array 
     (
      [name_client] => Dddddddd 
      [id_todo] => 
      [text] => 
      [id_client] => 1 
      [id_project] => 1 
      [renew_date] => 2016-03-19 
      [domain] => eeeeee.it 
     ) 

    [3] => Array 
     (
      [name_client] => Assssass 
      [id_todo] => 
      [text] => 
      [id_client] => 5 
      [id_project] => 8 
      [renew_date] => 2016-03-26 
      [domain] => erreerr.net 
     ) 

    [4] => Array 
     (
      [name_client] => Eeeeeeee 
      [id_todo] => 
      [text] => 
      [id_client] => 3 
      [id_project] => 5 
      [renew_date] => 2016-04-27 
      [domain] => rrrrrrr.it 
     ) 

    [5] => Array 
     (
      [name_client] => Edsdee 
      [id_todo] => 
      [text] => 
      [id_client] => 7 
      [id_project] => 10 
      [renew_date] => 2016-07-19 
      [domain] => rrrrrrr.com 
     ) 

    [6] => Array 
     (
      [name_client] => Dddddddd 
      [id_todo] => 
      [text] => 
      [id_client] => 6 
      [id_project] => 9 
      [renew_date] => 2016-07-27 
      [domain] => rrrrrrrrr.it 
     ) 

    [7] => Array 
     (
      [name_client] => wwwwwwwww 
      [id_todo] => 
      [text] => 
      [id_client] => 4 
      [id_project] => 6 
      [renew_date] => 2016-09-24 
      [domain] => rrttttrrr.it 
     ) 

    [8] => Array 
     (
      [name_client] => yyyyyyyy 
      [id_todo] => 1 
      [text] => Todo!!!!! 
      [id_client] => 2 
      [id_project] => 4 
      [renew_date] => 2016-09-29 
      [domain] => uuuuuuuuu.it 
     ) 

) 

Спасибо.

Answer 1

не могли бы вы показать, что ваш «выбор» возвращает whith print_r, пожалуйста, чтобы иметь структуру данных?

Попробуйте это:

$clients = array(); 
foreach ($queryResults as $entry) { 
    if (! isset($clients[$entry['id_client']])) { 
     $clients[$entry['id_client']] = array(
      'id' => $entry['id_client'], 
      'name_client' => $entry['name_client'], 
      'todos' => array(), 
      'projects' => array() 
     ); 
    } 
    if (! empty($entry['id_todo'])) { 
     $clients[$entry['id_client']]["todos"][] = array(
      'id' => $entry['id_todo'], 
      'text' => $entry['text'] 
     ); 
    } 
    if (! empty($entry['id_project'])) { 
     $clients[$entry['id_client']]["projects"][] = array(
      'id' => $entry['id_project'], 
      'domain' => $entry['text'] 
     ); 
    } 
}