5

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 ID 和值作为命名列，因此它看起来像这样：

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

任何帮助将非常感激！！

4

5 回答 5

5

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，我做了 joinsLEFT JOIN因为我不知道 id 是否可以缺少字段，在这种情况下，它们将显示NULL在我们的派生列中。

SQL-Fiddle 演示

于 2012-07-10T18:10:02.260 回答
5

You could also try pivoting with the help of grouping and conditional aggregating:

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
;
于 2012-07-10T19:31:15.947 回答
2

改编自我的另一个答案：

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')

The left joins will ensure that missing values will result in NULL cells, instead of an omission of the whole row. Not sure whether that is important in your application. If it is not, you can use an inner join and in particular get rid of the subquery to select all unique IDs. See my original answer from which I derived this.

于 2012-07-10T19:17:21.660 回答
0

我不确定 MySQL 中是否有 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'
于 2012-07-10T18:06:32.653 回答
0

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，但它是一种可行的方法，特别是如果您担心您将得到恰好返回的四列，并且它们将以特定的顺序返回。

请注意，当特定 ID“缺少”特定字段名称时，此方法有效（它将返回 NULL 代替值）。如果特定 ID 多次出现相同的 field_name，它也可以工作。（此查询将只返回其中一个，而忽略另一个。）

同样的结果集也可以通过这样的查询获得：

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

对于其他查询，请确保在field_name给定 ID“缺少”a、给定 ID 有重复项field_name或表中存在您不关心的其他 field_name 值时获得所需的行为。

于 2012-07-10T18:11:04.477 回答