Question

我正在尝试对四张桌子进行一些计数，我正在努力做到这一点！

这是我的表脚本（4个简单的表，相同的表）

CREATE TABLE `TableA` (
  `id` int(45) NOT NULL,
  `zone` varchar(45) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `TableA` (`id`, `zone`) VALUES
(1, 'ca'),(2, 'ca'),(3, 'fr'),(4, 'ca'),(5, 'ca'),(6, 'fr');

CREATE TABLE `TableB` (
  `id` int(45) NOT NULL,
  `zone` varchar(45) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `TableB` (`id`, `zone`) VALUES
(1, 'ca'),(2, 'it'),(3, 'de'),(4, 'ca'),(5, 'it'),(6, 'fr');

CREATE TABLE `TableC` (
  `id` int(45) NOT NULL,
  `zone` varchar(45) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `TableC` (`id`, `zone`) VALUES
(1, 'ca'),(2, 'ma'),(3, 'fr'),(4, 'pl'),(5, 'usa'),(6, 'fr');

CREATE TABLE `TableD` (
  `id` int(45) NOT NULL,
  `zone` varchar(45) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `TableD` (`id`, `zone`) VALUES
(1, 'ca'),(2, 'pl'),(3, 'it'),(4, 'pl'),(5, 'ca'),(6, 'it');

这是我到目前为止所尝试的内容

SELECT DISTINCT Zone,
    TableAa,
    TableBb,
    TableCc,
    TableDd 
    FROM
    (  
            SELECT DISTINCT Ta.zone AS Zone,
                   COUNT(  Ta.id  ) AS TableAa,                     
                   COUNT(  Tb.id  ) AS TableBb,                    
                   COUNT(  Tc.id  ) AS TableCc,                    
                   COUNT(  Td.id  ) AS TableDd
              FROM TableA Ta
            LEFT JOIN TableB Tb ON Ta.zone = Tb.zone
            LEFT JOIN TableC Tc ON Ta.zone = Tc.zone
            LEFT JOIN TableD Td ON Ta.zone = Td.zone
            GROUP BY Ta.zone          
      UNION              
            SELECT DISTINCT Tb.zone AS Zone,
                   COUNT(  Ta.id  ) AS TableAa,                     
                   COUNT(  Tb.id  ) AS TableBb,                    
                   COUNT(  Tc.id  ) AS TableCc,                    
                   COUNT(  Td.id  ) AS TableDd
              FROM TableB Tb
            LEFT JOIN TableA Ta ON Tb.zone = Ta.zone
            LEFT JOIN TableC Tc ON Tb.zone = Tc.zone
            LEFT JOIN TableD Td ON Tb.zone = Td.zone
            GROUP BY Tb.zone          
      UNION              
            SELECT DISTINCT Tc.zone AS Zone,
                   COUNT(  Ta.id  ) AS TableAa,                     
                   COUNT(  Tb.id  ) AS TableBb,                    
                   COUNT(  Tc.id  ) AS TableCc,                    
                   COUNT(  Td.id  ) AS TableDd
              FROM TableC Tc
            LEFT JOIN TableB Tb ON Tc.zone = Tb.zone
            LEFT JOIN TableA Ta ON Tc.zone = Ta.zone
            LEFT JOIN TableD Td ON Tc.zone = Td.zone
            GROUP BY Tc.zone          
      UNION              
            SELECT DISTINCT Td.zone AS Zone,
                   COUNT(  Ta.id  ) AS TableAa,                     
                   COUNT(  Tb.id  ) AS TableBb,                    
                   COUNT(  Tc.id  ) AS TableCc,                    
                   COUNT(  Td.id  ) AS TableDd
              FROM TableA Td
            LEFT JOIN TableB Tb ON Td.zone = Tb.zone
            LEFT JOIN TableC Tc ON Td.zone = Tc.zone
            LEFT JOIN TableA Ta ON Td.zone = Ta.zone
            GROUP BY Td.zone  
          ) A
ORDER BY Zone              
    ;

我有一些结果，但结果不连贯。

我应该

  zone   |   TableAa   |   TableBb   |   TableCc   |   TableDd
  fr     |   2         |   1         |   2         |   0
  ca     |   4         |   2         |   1         |   2
  it     |   0         |   2         |   0         |   2
  de     |   0         |   1         |   0         |   0
  ma     |   0         |   0         |   1         |   0
  pl     |   0         |   0         |   1         |   2
  usa    |   0         |   0         |   1         |   0

相反，我有

  zone   |   TableAa   |   TableBb   |   TableCc   |   TableDd
  ca     |   16        |   16        |   16        |   16
  ca     |   32        |   32        |   32        |   32
  de     |   0         |   1         |   0         |   0
  fr     |   4         |   4         |   4         |   0
  fr     |   8         |   8         |   8         |   8
  it     |   0         |   4         |   0         |   4
  ma     |   0         |   0         |   1         |   0
  pl     |   0         |   0         |   2         |   2
  usa    |   0         |   0         |   1         |   0

结果完全错误，我真的不知道为什么！这是一个用于测试的SQL小提琴：http://sqlfiddle.com/#!9/1cc0ab/3

需要任何帮助。

感谢。

PS：我正在使用mysql

Answer 1

您需要稍微分开代码并分别计算每个总和，然后加入区域列表。

 SELECT T.ZONe,IFNULL(A.TableAa,0) AS TableAa,IFNULL(B.TableBb,0) AS TableBb,IFNULL(C.TableCc,0) AS TableCc,IFNULL(D.TableDd,0) AS TableDd
 FROM (
            SELECT DISTINCT Ta.zone AS Zone
            FROM TableA Ta
            UNION
            SELECT DISTINCT Tb.zone AS Zone
            FROM TableB Tb
            UNION
            SELECT DISTINCT Tc.zone AS Zone
            FROM TableC Tc
            UNION
            SELECT DISTINCT Td.zone AS Zone
            FROM TableD Td
     ) T       
     LEFT JOIN (
       SELECT Zone, COUNT(*) AS TableAa FROM TableA GROUP BY Zone
       ) A 
       ON A.Zone=T.ZOne
     LEFT JOIN (
       SELECT Zone, COUNT(*) AS TableBb FROM TableB GROUP BY Zone
       ) B 
       ON B.Zone=T.ZOne
       LEFT JOIN (
       SELECT Zone, COUNT(*) AS TableCc FROM TableC GROUP BY Zone
       ) C 
       ON C.Zone=T.ZOne
       LEFT JOIN (
       SELECT Zone, COUNT(*) AS TableDd FROM TableD GROUP BY Zone
       ) D 
       ON D.Zone=T.ZOne
    ;

SQL Fiddle

依靠具有相同标识符的多个表mysql

1 个答案: