Question

我正在尝试根据value列中的值从name列中选择值。

例如：

mysql> select name, value from rss_feed_property_value where rss_feed_id = 31;
+-------------+---------------+
| name        | value         |
+-------------+---------------+
| High        | 45            |
| Description | Rain And Snow |
| Low         | 25            |
| Day         | Fri           |
| Date        | 29 Dec 2017   |
+-------------+---------------+
5 rows in set (0.00 sec)

我几乎通过以下查询实现了这一目标：

select

  if (rfpv.name = 'Date', rfpv.value, null) as `Date`,
  if (rfpv.name = 'High', rfpv.value, null) as `High`,
  if (rfpv.name = 'Low', rfpv.value, null) as `Low`,
  if (rfpv.name = 'Description', rfpv.value, null) as `Description`

from rss_feed rf
join rss_feed_definition rfd on rf.rss_feed_definition_id = rfd.id
join rss_feed_property_value rfpv on rfpv.rss_feed_id = rf.id

where rfd.type = 'weather'
and rf.id = 31
order by rf.id;

产生以下输出：

+-------------+------+------+---------------+
| Date        | High | Low  | Description   |
+-------------+------+------+---------------+
| NULL        | 45   | NULL | NULL          |
| NULL        | NULL | NULL | Rain And Snow |
| NULL        | NULL | 25   | NULL          |
| NULL        | NULL | NULL | NULL          |
| 29 Dec 2017 | NULL | NULL | NULL          |
+-------------+------+------+---------------+
5 rows in set (0.00 sec)

我现在如何通过忽略空值来展平上述内容，只留下如下一行：

+-------------+------+------+---------------+
| Date        | High | Low  | Description   |
+-------------+------+------+---------------+
| 29 Dec 2017 | 45   | 25   | Rain and Snow |
+-------------+------+------+---------------+

Answer 1

按rf.id聚合并使用条件聚合。我会在这里使用CASE表达式，因为这种语法在数据库中得到的支持比MySQL的IF更广泛。

SELECT
    rf.id,
    MAX(CASE WHEN rfpv.name = 'Date'        THEN rfpv.value END) AS Date,
    MAX(CASE WHEN rfpv.name = 'High'        THEN rfpv.value END) AS High,
    MAX(CASE WHEN rfpv.name = 'Low'         THEN rfpv.value END) AS Low,
    MAX(CASE WHEN rfpv.name = 'Description' THEN rfpv.value END) AS Description
FROM rss_feed rf
INNER JOIN rss_feed_definition rfd
    ON rf.rss_feed_definition_id = rfd.id
INNER JOIN rss_feed_property_value rfpv
    ON rfpv.rss_feed_id = rf.id
WHERE
   rfd.type = 'weather' AND
   rf.id = 31
GROUP BY
    rf.id
ORDER BY
    rf.id;

Answer 2

只需将MAX()放在您要折叠的所有内容（MAX()忽略NULL值），GROUP BY ID为''重新尝试将所有内容折叠起来......

SELECT
  rf.id,
  MAX(if (rfpv.name = 'Date',        rfpv.value, null))  AS `Date`,
  MAX(if (rfpv.name = 'High',        rfpv.value, null))  AS `High`,
  MAX(if (rfpv.name = 'Low',         rfpv.value, null))  AS `Low`,
  MAX(if (rfpv.name = 'Description', rfpv.value, null))  AS `Description`
FROM
  rss_feed                   AS rf
INNER JOIN
  rss_feed_definition        AS rfd
    ON rf.rss_feed_definition_id = rfd.id
INNER JOIN
  rss_feed_property_value    AS rfpv
    ON rfpv.rss_feed_id = rf.id
WHERE
      rfd.type = 'weather'
  AND rf.id = 31
GROUP BY
  rf.id
ORDER BY
  rf.id
;

我已将字段rf.id添加到SELECT，只是为了更清楚地了解正在做什么。 （还添加了关联的GROUP BY rf.id）

Answer 3

我通过以下查询实现了这一目标：

select

  rf.id as rssFeedId,

  (select value
   from rss_feed_property_value
   where name = 'Date'
   and rss_feed_id = rssFeedId) as `Date`,

   (select value
   from rss_feed_property_value
   where name = 'High'
   and rss_feed_id = rssFeedId) as `High`,

   (select value
   from rss_feed_property_value
   where name = 'Low'
   and rss_feed_id = rssFeedId) as `Low`,

   (select value
   from rss_feed_property_value
   where name = 'Description'
   and rss_feed_id = rssFeedId) as `Description`


from rss_feed rf
join rss_feed_definition rfd on rf.rss_feed_definition_id = rfd.id
join rss_feed_property_value rfpv on rfpv.rss_feed_id = rf.id

where rfd.type = 'weather'
and rf.id = 31

group by rf.id

order by rf.id;

我不确定这是否是最有效的方式。

如何根据另一列的值获取单行值？

3 个答案: