Question

我有一个看起来像这样的表：

ID | FIELD_NAME   | VALUE
23 |  sign_up     |  yes
23 |  first_name  |  Fred
23 |  street      |  Barber Lane
24 |  sign_up     |  no
24 |  first_name  |  Steve
24 |  street      |  Camaro St.
25 |  sign_up     |  yes
25 |  first_name  |  Larry
25 |  street      |  Huckleberry Ave

我想运行一个查询，它将选择唯一的ID和值作为命名列，所以它看起来像这样：

ID   |   SIGN_UP   | FIRST_NAME  |  STREET           |
23   |     yes     |    Fred     |  Barber Lane      |
24   |     no      |    Steve    |  Camaro St.       |
25   |     yes     |    Larry    |  Huckleberry Ave. |

任何帮助都将非常感谢!!

Answer 1

您可以使用这个简单的解决方案：

SELECT DISTINCT
    a.id,
    b.value AS SIGN_UP,
    c.value AS FIRST_NAME,
    d.value AS STREET
FROM tbl a
LEFT JOIN tbl b ON a.id = b.id AND b.field_name = 'sign_up'
LEFT JOIN tbl c ON a.id = c.id AND c.field_name = 'first_name'
LEFT JOIN tbl d ON a.id = d.id AND d.field_name = 'street'

为了安全起见，我建立了联接LEFT JOIN，因为我不知道id是否可以有缺少字段，在这种情况下它们会显示为{{1在我们的派生列中。

SQL-Fiddle Demo

Answer 2

您还可以在分组和条件聚合的帮助下尝试旋转：

SELECT
  ID,
  MAX(CASE FIELD_NAME WHEN 'sign_up'    THEN VALUE END) AS SIGN_UP,
  MAX(CASE FIELD_NAME WHEN 'first_name' THEN VALUE END) AS FIRST_NAME,
  MAX(CASE FIELD_NAME WHEN 'street'     THEN VALUE END) AS STREET
FROM atable
GROUP BY
  ID
;

Answer 3

改编自another answer by me：

SELECT ids.ID AS ID,
       sign_up.VALUE AS SIGN_UP,
       first_name.VALUE AS FIRST_NAME,
       street.VALUE AS STREET
FROM (SELECT DISTINCT ID FROM tableName) AS ids
     LEFT JOIN tableName AS sign_up
            ON (sign_up.ID = ids.ID AND
                sign_up.FIELD_NAME = 'sign_up')
     LEFT JOIN tableName AS first_name
            ON (first_name.ID = ids.ID AND
                first_name.FIELD_NAME = 'first_name')
     LEFT JOIN tableName AS street
            ON (street.ID = ids.ID AND
                street.FIELD_NAME = 'street')

左连接将确保缺少的值将导致NULL单元格，而不是省略整行。不确定这在您的应用程序中是否重要。如果不是，则可以使用内部联接，特别是删除子查询以选择所有唯一ID。请参阅我从中导出的original answer。

Answer 4

我不确定MySQL中是否存在Pivot / Unpivot功能。试试这个：

SELECT a.ID, 
       c.FIELD_NAME AS SIGN_UP,
       a.FIELD_NAME AS FIRST_NAME,
       b.FIELD_NAME AS STREET

  FROM <YOUR-TABLE> a LEFT JOIN <YOUR-TABLE> b
        ON a.ID = b.ID 
     AND a.FIELD_NAME = 'first_name'
     AND b.FIELD_NAME = 'street' LEFT JOIN <YOUR-TABLE> c
        ON c.ID = a.ID
     AND c.FIELD_NAME = 'sign_up'

Answer 5

一种方法是使用相关子查询将每个字段值作为列返回，

SELECT t.id
     , (SELECT f1.value FROM mytable f1 
         WHERE f1.id = t.id AND f1.field_name = 'sign_up' 
         ORDER BY f1.value LIMIT 1
       ) AS SIGN_UP
     , (SELECT f2.value FROM mytable f2 
         WHERE f2.id = t.id AND f2.field_name = 'first_name' 
         ORDER BY f2.value LIMIT 1
       ) AS FIRST_NAME
     , (SELECT f3.value FROM mytable f3 
         WHERE f3.id = t.id AND f3.field_name = 'street'
         ORDER BY f3.value LIMIT 1
       ) AS STREET
  FROM (SELECT s.id
          FROM mytable s
         GROUP BY s.id
         ORDER BY s.id
       ) t

这不是唯一的方法，但这是一种可行的方法，特别是如果您担心返回的确有四列，并且它们将按特定顺序返回。

请注意，当特定ID的特定field_name“缺失”时，此方法有效（它将返回NULL代替值）。如果特定ID出现多次相同的field_name，它也可以工作。（此查询将仅返回其中一个，而忽略另一个。）

使用如下所示的查询也可以获得相同的结果集：

SELECT t.id          AS ID
     , f1.sign_up    AS SIGN_UP
     , f2.first_name AS FIRST_NAME
     , f3.street     AS STREET       
  FROM (SELECT s.id
          FROM mytable s
         GROUP BY s.id
         ORDER BY s.id
       ) t
   LEFT      
   JOIN (SELECT s1.id
              , MIN(s1.value) AS sign_up
           FROM mytable s1
          WHERE s1.field_name = 'sign_up'
            AND s1.value IS NOT NULL 
          GROUP BY s1.id
        ) f1
     ON f1.id = t.id   
   LEFT
   JOIN (SELECT s2.id
              , MIN(s2.value) AS first_name
           FROM mytable s2
          WHERE s2.field_name = 'first_name'
            AND s2.value IS NOT NULL
          GROUP BY s2.id
        ) f2
     ON f2.id = t.id
   LEFT
   JOIN (SELECT s3.id
              , MIN(s3.value) AS street
           FROM mytable s3
          WHERE s3.field_name = 'street'
            AND s3.value IS NOT NULL
          GROUP BY s3.id
        ) f3
     ON f3.id = t.id

使用其他查询，确保在给定ID的field_name“缺失”时，或者对于给定ID存在重复field_name时，或者当有其他查询时，您将获得所需的行为表格中的field_name值，您不关心。

为具有相同ID的多行选择不同的值

5 个答案: