JOIN許多表，而GROUP BY和COUNT在GROUP BY行的MySQL

Question

我的表像這樣在我的數據庫

lp_kon

ID | lpkonName 
1 | kon1 
2 | kon2 

lp_seb

ID | ID_lpkon | lpsebName 
1 | 1  | seb1 
2 | 2  | seb3 

lp_rek

ID | ID_lpseb | lprekName 
1 | 1  | rek1 
2 | 1  | rek2 

lp_tl

ID | ID_lprek | lptlName 
1 | 1  | tl1 
2 | 1  | tl2 

我用這個查詢加入這些4桌

SELECT 
    a.id as id_lpkon, 
    b.id as id_seb, 
    c.id as id_rek, 
    d.id as id_tl 
FROM `lp_kon` a 
left OUTER JOIN lp_seb b ON a.id = b.ID_lpkon 
LEFT OUTER join lp_rek c on b.id = c.ID_lpseb 
LEFT OUTER join lp_tl d on c.id = d.ID_lprek 
ORDER by a.id ASC 

我喜歡這個

id_lpkon|id_lpseb|id_lprek|id_lptl| 
1  | 1  | 1  | 1 | 
1  | 1  | 1  | 2 | 
1  | 1  | 2  | NULL | 
1  | 2  | NULL | NULL | 
2  | Null | NULL | NUll | 

我如何算一個resut錶行，而分組。像下面

id_lpkon | count(lpkon) | id_seb | count(lpseb | id_rek | count(lprek) | id_lptl | count(lptl) 
1  |  4  | 1 | 3  | 1 | 2   |1  | 1  | 
1  |  4  | 1 | 3  | 1 | 2   |2  | 1  | 
1  |  4  | 1 | 3  | 2 | 1   |NUll  |NUll  | 
1  |  4  | 2 | 1  | NULL | NULL   |NULL  |NULL  | 
2  |  1  | NULL | NULL  | NULL | NULL   |NULL  | 

SO每一行的結果，他們有一個數列的每個關係表

更新與查詢

Lp_kon

CREATE TABLE `lp_kon` (`ID` int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT, `lpkonName` text NOT NULL) ENGINE=InnoDB DEFAULT CHARSET=latin1 
INSERT INTO `lp_kon` (`ID`, `lpkonName`) VALUES (NULL, 'kon1'), (NULL, 'kon2'); 

Lp_seb

CREATE TABLE `lp_seb` (`ID` int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,`ID_lpkon` text NOT NULL, `lpsebName` text NOT NULL) ENGINE=InnoDB DEFAULT CHARSET=latin1 
INSERT INTO `lp_seb` (`ID`,`ID_lpkon`, `lpsebName`) VALUES (NULL,`1`,'seb1'), (NULL,`2`,'seb3'); 

Lp_rek

CREATE TABLE `lp_rek` (`ID` int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,`ID_lpseb` text NOT NULL, `lprekName` text NOT NULL) ENGINE=InnoDB DEFAULT CHARSET=latin1 
INSERT INTO `lp_rek` (`ID`,`ID_lpkon`, `lprekName`) VALUES (NULL,`1`,'rek1'), (NULL,`1`,'rek2'); 

Lp_tl

CREATE TABLE `lp_tl` (`ID` int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,`ID_lprek` text NOT NULL, `lptlName` text NOT NULL) ENGINE=InnoDB DEFAULT CHARSET=latin1 
INSERT INTO `lp_tl` (`ID`,`ID_lprek`, `lpsebName`) VALUES (NULL,`1`,'tl1'), (NULL,`1`,'tl2'); 

Answer 1

我覺得這個問題是比較容易用PHP解決。

使用您的查詢並將數據提取到關聯數組中。我應該是這樣的：

$dbData = [ 
    ['id_lpkon' => 1, 'id_lpseb' => 1, 'id_lprek' => 1, 'id_lptl' => 1], 
    ['id_lpkon' => 1, 'id_lpseb' => 1, 'id_lprek' => 1, 'id_lptl' => 2], 
    ['id_lpkon' => 1, 'id_lpseb' => 1, 'id_lprek' => 2, 'id_lptl' => null], 
    ['id_lpkon' => 1, 'id_lpseb' => 2, 'id_lprek' => null, 'id_lptl' => null,], 
    ['id_lpkon' => 2, 'id_lpseb' => null, 'id_lprek' => null, 'id_lptl' => null,] 
]; 

現在使用嵌套foreach循環計算所有讀取值：

$columns = ['id_lpkon', 'id_lpseb', 'id_lprek', 'id_lptl']; 
$counts = []; 
foreach ($columns as $column) { 
    $counts[$column] = []; 
} 

foreach ($dbData as $row) { 
    foreach ($row as $column => $value) { 
     if ($value === null) { 
      continue; 
     } elseif (isset($counts[$column][$value])) { 
      $counts[$column][$value]++; 
     } else { 
      $counts[$column][$value] = 1; 
     } 
    } 
} 

你可以停在這裏，因爲你已經擁有了$counts陣列，它看起來在所需要的數據像這樣：

$counts = [ 
    'id_lpkon' => [1 => 4, 2 => 1], 
    'id_lpseb' => [1 => 3, 2 => 1], 
    'id_lprek' => [1 => 2, 2 => 1], 
    'id_lptl' => [1 => 1, 2 => 1], 
]; 

但是如果你需要在一個「表」中的數據，使用下一個嵌套的foreach循環：

foreach ($dbData as $rn => $row) { 
    foreach ($row as $column => $value) { 
     $countColumn = $column . '_count'; 
     if ($value === null) { 
      $dbData[$rn][$countColumn] = null; 
     } else { 
      $dbData[$rn][$countColumn] = $counts[$column][$value]; 
     } 
    } 
} 

演示：http://rextester.com/XQTSX69507

Answer 2

我相信sum()是你在找什麼。

Answer 3

我相信，因爲每次他需要爲每一個值計算一個值時，他都想要它。

COUNT（表）作爲標題， a.ID AS id_lpkon， COUNT（表）AS頭2， b.ID AS id_seb，

等等等等