Question

我正在尝试创建一个带有八个不同数字的SQL语句，并从一个数据库中搜索八个不同的表以获取相应的值。我一直得到无效的语法，但我无法弄清楚为什么。这是我的代码：

SELECT cable_length.ID, drawing.ID, optional.ID, output_type.ID, pressure_range.ID, pressure_type.ID, series.ID, termination.ID 
FROM
(
    SELECT value AS cable_length FROM A1 WHERE A1.id = %s
    JOIN SELECT value AS drawing FROM A2 WHERE A2.id = %s,
    JOIN SELECT value AS optional FROM A3 WHERE A3.id = %s,
    JOIN SELECT value AS output_type FROM A4 WHERE A4.id = %s,
    JOIN SELECT value AS pressure_range FROM A5 WHERE A5.id = %s,
    JOIN SELECT value AS pressure_type FROM A6 WHERE A6.id = %s,
    JOIN SELECT value AS series FROM A7 WHERE A7.id = %s, 
    JOIN SELECT value AS termination FROM A8 WHERE A8.id = %s
);

％s将仅更改为数字。每个表中的列名是“ID”和“值”。我想按“ID”搜索并返回“Value”。表名是cable_length等。非常感谢任何帮助。

Answer 1

如果你知道所有8个值都存在，你可以像这样加入：

SELECT  A1.value as cable_length
  , A2.value as drawing
  , A3.value as optional
  , A4.value as output_type
  , A5.value as pressure_range
  , A6.value as pressure_type
  , A7.value as series
  , A7.value as termination
FROM A1, A2, A3, A4, A5, A6, A7, A8
WHERE A1.ID = %s
  AND A2.ID = %s
  AND A3.ID = %s
  AND A4.ID = %s
  AND A5.ID = %s
  AND A6.ID = %s
  AND A7.ID = %s
  AND A8.ID = %s

如果您不知道这一点，但保证第一个值存在，那么您的查询必须变得更加丑陋，因为您需要左连接。

SELECT  A1.value as cable_length
  , A2.value as drawing
  , A3.value as optional
  , A4.value as output_type
  , A5.value as pressure_range
  , A6.value as pressure_type
  , A7.value as series
  , A7.value as termination
FROM A1
  LEFT JOIN A2
    ON A2.ID = %s
  LEFT JOIN A3
    ON A3.ID = %s
  LEFT JOIN A4
    ON A4.ID = %s
  LEFT JOIN A5
    ON A5.ID = %s
  LEFT JOIN A6
    ON A6.ID = %s
  LEFT JOIN A7
    ON A7.ID = %s
  LEFT JOIN A8
    ON A8.ID = %s
WHERE A1.ID = %s

如果没有任何值保证存在，那么hack的子查询是8个查询的UNION ALL，返回8列，只有一个填充，然后选择每个MAX列作为该列名。

这是我过去用过的一个非常糟糕的技巧，可以避免许多只偶尔加入的左连接的性能问题。

顺便说一下，名为A1..A8的8个表格表明，您需要具有数据库经验的人来查看您的设计并为您提供更好的布局。

Answer 2

也许这就是你想要的。也许不是，我只是重写了你的问题：

SELECT
  cable_length.ID,
  drawing.ID,
  optional.ID,
  output_type.ID,
  pressure_range.ID,
  pressure_type.ID,
  series.ID,
  termination.ID 
FROM (SELECT ID FROM A1 WHERE A1.id = %s) AS cable_length
JOIN (SELECT ID FROM A2 WHERE A2.id = %s) AS drawing
JOIN (SELECT ID FROM A3 WHERE A3.id = %s) AS optional
JOIN (SELECT ID FROM A4 WHERE A4.id = %s) AS output_type
JOIN (SELECT ID FROM A5 WHERE A5.id = %s) AS pressure_range
JOIN (SELECT ID FROM A6 WHERE A6.id = %s) AS pressure_type
JOIN (SELECT ID FROM A7 WHERE A7.id = %s) AS series
JOIN (SELECT ID FROM A8 WHERE A8.id = %s) AS termination;

Answer 3

UNION ALL将在这种情况下工作，但它可能不是最好的解决方案，更不用说乱七八糟了（参见@btilly提供的答案）。另请注意，这是SQL Server语法，可能与MySql略有不同

select max(cable_length), max(drawing), max(optional), max(output_type), max(pressure_range), max(pressure_type), max(series), max(termination)

from
(
    SELECT value as cable_length, null, null, null, null, null, null, null, null 
    FROM cable_length WHERE id = %s

    UNION ALL

    SELECT null, value as drawing, null, null, null, null, null, null, null 
    FROM drawing WHERE id = %s

    UNION ALL

    SELECT null, null, value as optional, null, null, null, null, null, null 
    FROM optional WHERE id = %s

    UNION ALL

    SELECT null, null, null, value as output_type, null, null, null, null, null 
    FROM output_type WHERE id = %s

    UNION ALL

    SELECT null, null, null, null, value as pressure_range, null, null, null 
    FROM pressure_range WHERE id = %s

    UNION ALL

    SELECT null, null, null, null, null, value as pressure_type, null, null 
    FROM pressure_type WHERE id = %s

    UNION ALL

    SELECT null, null, null, null, null, null, value as series, null 
    FROM series WHERE id = %s

    UNION ALL

    SELECT null, null, null, null, null, null, null, value as termination 
    FROM termination WHERE id = %s
);

来自八个表的多个JOIN和SELECT语句

3 个答案: