3

我有三个表：applications、permissions 和 applications_permissions

|------------|   |------------------------|   |-----------|
|applications|   |applications_permissions|   |permissions|
|------------|   |------------------------|   |-----------|
| id         | <-| application_id         |   | id        |
| price      |   | permission_id          |-> | name      |
|------------|   |------------------------|   |-----------|

对于应用程序有两类：免费和商业应用程序（价格 = '0' 和价格！= '0'）

现在我想知道每个权限有多少百分比的应用程序引用它；这对于这两个类别

自由：

id, percentage
1 , 20.0230
2 ,  0.0000
3 ,  0.0312
...

商业的：

id, percentage
1 , 18.0460
2 ,  0.0000
3 ,  0.0402
...

我已经制定了以下查询，但它不包括没有应用程序的权限 ID：/

SELECT (SELECT name FROM permissions WHERE id = applications_permissions.permission_id) AS "name",
        100::float * COUNT(*)/(SELECT COUNT(name) FROM applications WHERE price = \'0\') AS "percent"
  FROM applications, applications_permissions
  WHERE applications.id = applications_permissions.application_id 
    AND applications.price = \'0\'
  GROUP BY applications_permissions.permission_id
  ORDER BY percent DESC')

我该怎么做呢？我已经尝试了几个小时（那个查询，misc JOINs）但它让我无法理解：/

4

4 回答 4

4

简化。初稿非常理想。
要在一个查询中计算所有内容：

SELECT p.id
     ,(100 * sum((a.price > 0)::int)) / cc.ct AS commercial
     ,(100 * sum((a.price = 0)::int)) / cf.ct AS free
FROM  (SELECT count(*)::float AS ct FROM applications WHERE price > 0) AS cc
     ,(SELECT count(*)::float AS ct FROM applications WHERE price = 0) AS cf
      ,permissions p
LEFT   JOIN applications_permissions ap ON ap.permission_id = p.id
LEFT   JOIN applications a ON a.id = ap.application_id
GROUP  BY 1, cc.ct, cf.ct
ORDER  BY 2 DESC, 3 DESC, 1;

假设您的价格实际上是一个数字列 - 所以0而不是'0'.

这包括permissions根本没有附加applications的（LEFT JOIN）。

如果有可能applications没有附加到任何permissions列表，则列表加起来不会达到 100%。

我做了一次总计数（ct）并将其转换float为子查询。其余的计算可以用整数算术来完成，只有最终/ ct将数字转换为浮点数。这是最快和最精确的。

与 CTE 相同

如果您对更多新事物持开放态度：尝试使用CTE（通用表表达式 - WITH 查询）进行相同操作 - 自 PostgreSQL 8.4 起可用。
它更干净，可能稍微快一些，因为我在一个 CTE 中同时计算并且更便宜GROUP BY- 这两者都可以通过子查询来完成：

WITH  c AS (
    SELECT sum((a.price > 0)::int) AS cc
          ,sum((a.price = 0)::int) AS cf
    FROM   applications
    ), p AS (
    SELECT id
          ,sum((a.price > 0)::int) AS pc
          ,sum((a.price = 0)::int) AS pf
    FROM   permissions p
    LEFT   JOIN applications_permissions ap ON ap.permission_id = p.id
    LEFT   JOIN applications a ON a.id = ap.application_id
    GROUP  BY 1
    )
SELECT p.id
     ,(100 * pc) / cc::float AS commercial
     ,(100 * pf) / cf::float AS free
FROM   c, p
ORDER  BY 2 DESC, 3 DESC, 1;
于 2012-05-07T18:33:06.697 回答
3

使用LEFT OUTER JOIN：

SELECT * FROM permissions LEFT OUTER JOIN
applications_permissions as rel on permissions.id = rel.permission_id LEFT OUTER JOIN
applications on rel.application_id = applications.id
于 2012-05-07T16:18:17.630 回答
1

这行得通吗？

对于这种free情况：

SELECT p.id, (100::float * COUNT(p.id)/(SELECT COUNT(*) from Applications)) Percent
FROM Applications a, Permissions p, Applications_Permissions a_p
WHERE a.id = a_p.application_id AND p.id = a_p.permission_id AND a.price = 0
GROUP BY p.id
ORDER BY Percent DESC
于 2012-05-07T16:19:07.270 回答
1

这是一个查询的结果：

SELECT p.id
, p.name
, (CASE WHEN total.free=0 THEN NULL ELSE 100::float * sub.free::float / total.free::float END) AS percent_free
, (CASE WHEN total.comm=0 THEN NULL ELSE 100::float * sub.comm::float / total.comm::float END) AS percent_comm
FROM permissions AS p
LEFT JOIN (
  SELECT permission_id
  , SUM(CASE WHEN a.price<=0 THEN 1 ELSE 0 END) AS free
  , SUM(CASE WHEN a.price>0  THEN 1 ELSE 0 END) AS comm
  FROM applications_permissions AS pa
  JOIN applications AS a ON (pa.application_id=a.id)
  GROUP BY permission_id
) AS sub ON (p.id=sub.permission_id)
, (
  SELECT
    SUM(CASE WHEN price<=0 THEN 1 ELSE 0 END) AS free
  , SUM(CASE WHEN price>0  THEN 1 ELSE 0 END) AS comm
  FROM applications
) AS total

或者仅适用于免费应用程序的结果（分别通过更改 where 子句的商业应用程序）：

SELECT p.id
, p.name
, (CASE WHEN total.nbr=0 THEN NULL ELSE 100::float * sub.nbr::float / total.nbr::float END) AS percent
FROM permissions AS p
LEFT JOIN (
  SELECT permission_id, COUNT(*) AS nbr
  FROM applications_permissions AS pa
  JOIN applications AS a ON (pa.application_id=a.id)
  WHERE (a.price<=0)
  GROUP BY permission_id
) AS sub ON (p.id=sub.permission_id)
, (
  SELECT COUNT(*) AS nbr
  FROM applications
  WHERE (price<=0)
) AS total
于 2012-05-07T23:45:08.137 回答