Question

如何简化此查询？

我想要做的是通过子查询派生列S9_Unlock，其中我只查找从主查询返回的user_ids，但这对我来说看起来很尴尬，特别是因为这里的查询只是一个摘录。实际上，我正在做这些子查询的多个来派生不同的列......

SELECT userid, CAST(to_char(S9_unlock,'YYYY/MM/DD') AS timestamp) AS "S9_Unlock"
FROM (
  SELECT ca.user_id AS userid
  FROM shop_db.invoices AS inv
  LEFT JOIN shop_db.carts AS ca ON inv.id = ca.invoice_id
  LEFT JOIN shop_db.cart_items AS ci ON ca.id = ci.cart_id
  WHERE (inv.created BETWEEN '2014-11-13' AND '2014-11-14' OR inv.created BETWEEN '2013-11-14' AND '2013-11-15')
    AND inv.status <> 'do_not_book'
    AND inv.id IS NOT NULL
  GROUP BY user_id) AS master
LEFT JOIN (
  SELECT MIN(s3.unl) AS "S9_Unlock", s3.user_id
  FROM (
    SELECT user_id, challenge_codes.created AS unl,
           MAX /* Check if license contains Suite9 */
             (CASE WHEN substring(bundle_article_code,1,6) = 'BuSu90' THEN 1 ELSE 0 END) AS "S9_Unlock"
    FROM licensing_db.serial_numbers
    LEFT JOIN licensing_db.licenses ON licenses.id = serial_numbers.license_id 
    LEFT JOIN user_db.users ON users.id = licenses.user_id
    LEFT JOIN licensing_db.challenge_codes ON challenge_codes.serial_number_id = serial_numbers.id
    WHERE user_id IN (
      SELECT ca.user_id AS userid
      FROM shop_db.invoices AS inv
      LEFT JOIN shop_db.carts AS ca ON inv.id = ca.invoice_id
      LEFT JOIN shop_db.cart_items AS ci ON ca.id = ci.cart_id
      WHERE (inv.created BETWEEN '2014-11-13' AND '2014-11-14' OR inv.created BETWEEN '2013-11-14' AND '2013-11-15')
        AND inv.status <> 'do_not_book'
        AND inv.id IS NOT NULL
      GROUP BY user_id
    )
    GROUP BY user_id, challenge_codes.created) AS s3
  )
  WHERE "S9_Unlock" = 1
    AND s3.unl IS NOT NULL
  GROUP BY s3.user_id) AS "S9_Unlock" ON "S9_Unlock".user_id = master.userid

Answer 1

在您的查询中，您有两个相同的子查询;这是一个CTE的尖叫声。

在有关许可问题的子查询中，您可以使用GROUP BY子句在HAVING子句后过滤掉有效许可证。也可以使它成为WITH QUERY，最终会有更可读性：

WITH inv AS (
  SELECT ca.user_id AS userid
  FROM shop_db.invoices AS inv
  LEFT JOIN shop_db.carts AS ca ON ca.invoice_id = inv.id
  LEFT JOIN shop_db.cart_items AS ci ON ci.cart_id = ca.id
  WHERE (inv.created BETWEEN '2014-11-13' AND '2014-11-14' OR inv.created BETWEEN '2013-11-14' AND '2013-11-15')
    AND inv.status <> 'do_not_book'
    AND inv.id IS NOT NULL
), s3 AS (
    SELECT u.user_id, min(cc.created) AS first_unlocked, bundle_article_code
    FROM licensing_db.serial_numbers AS sn
    LEFT JOIN licensing_db.licenses AS lic ON lic.id = sn.license_id 
    LEFT JOIN user_db.users AS u ON u.id = lic.user_id
    LEFT JOIN licensing_db.challenge_codes AS cc ON cc.serial_number_id = sn.id
    WHERE u.user_id IN (SELECT userid FROM inv)
    GROUP BY u.user_id, bundle_article_code
    HAVING bundle_article_code LIKE 'BuSu90%'
       AND first_unlocked IS NOT NULL
)
SELECT userid, date_trunc('day', first_unlocked) AS "S9_Unlock"
FROM inv
LEFT JOIN s3 ON s3.user_id = inv.userid;

因此，主查询现在减少到3行，并且WITH-QUERY都执行数据库的逻辑自包含查询。您引用的其他子查询可以类似地成为WITH-QUERY，然后在主查询中汇编它们。请记住，您可以在with-queries列表中引用先前命名的查询，如上所示inv引用s3。虽然这样的CTE在语法上不提供新功能（RECURSIVE变体除外），但它们确实使复杂查询更具可读性，因此更易于维护。

另一种方法是将逻辑子组件（例如inv子查询）分解出来并从中生成VIEW。然后，您可以简单地引用主查询中的视图。如果您想使查询更灵活，那么将整个事物视为一个视图可能也是一个好主意。如果您想在2014年3月27日查询Suite9.1（'BuSu91％'）怎么办？取出这些文字，然后在视图中将它们用作WHERE子句，使您的查询更加通用;这可以是子查询，也可以是完整的CTE。

（请检查s3 with-query中的语义是否仍然正确，因为如果没有您的表结构和示例数据，我无法测试上面的代码。）

Answer 2

我不会将你的问题解决为一个巨大的单片关系sql查询，而是会认真考虑去＆＃34;程序＆＃34;通过使用内置的＆＃34; plpgsql＆＃34; postgresql的语言。这可以为您的应用程序带来很多清晰度。

如何简化这个postgres查询

2 个答案: