从函数返回的记录具有串联的列

Question

我有一个表格，用于存储帐户随时间的变化。如果这些记录尚不存在，我需要将其与其他两个表连接起来以创建特定日期的一些记录。

为了让事情变得更简单（我希望如此），我将返回正确历史数据的查询封装到一个函数中，该函数接受帐户 ID 和日期。

如果我执行"Select * account_servicetier_for_day(20424, '2014-08-12')"，我会得到预期的结果（从函数返回的所有数据在单独的列中）。如果我在另一个查询中使用该函数，我会将所有列合并为一个：

("2014-08-12 14:20:37",hollenbeck,691,12129,20424,69.95,"2Mb/1Mb 20GB Limit",2048,1024,20.000)

我正在使用“x86_64-slackware-linux-gnu 上的 PostgreSQL 9.2.4，由 gcc (GCC) 4.7.1 编译，64 位”。

查询：

Select
    '2014-08-12' As day, 0 As inbytes, 0 As outbytes, acct.username, acct.accountid, acct.userid,
    account_servicetier_for_day(acct.accountid, '2014-08-12')
From account_tab acct
Where acct.isdsl = 1
    And acct.dslservicetypeid Is Not Null
    And acct.accountid Not In (Select accountid From dailyaccounting_tab Where Day = '2014-08-12')
Order By acct.username

功能：

CREATE OR REPLACE FUNCTION account_servicetier_for_day(_accountid integer, _day timestamp without time zone) RETURNS setof account_dsl_history_info AS
$BODY$
DECLARE _accountingrow record;
BEGIN
  Return Query
  Select * From account_dsl_history_info
  Where accountid = _accountid And timestamp <= _day + interval '1 day - 1 millisecond'
  Order By timestamp Desc 
  Limit 1;
END;
$BODY$ LANGUAGE plpgsql;

Answer 1

通常，分解从函数返回的行并获取单个列：

<b>SELECT * FROM</b> account_servicetier_for_day(20424, '2014-08-12');

---

关于查询：

Postgres 9.3 或更新版本

清理器JOIN LATERAL:

SELECT '2014-08-12' AS day, 0 AS inbytes, 0 AS outbytes
     , a.username, a.accountid, a.userid
     , f.*   -- but avoid duplicate column names!
FROM   account_tab a
     , account_servicetier_for_day(a.accountid, '2014-08-12') f  -- <-- HERE
WHERE  a.isdsl = 1
AND    a.dslservicetypeid IS NOT NULL
AND    NOT EXISTS (
   SELECT 1
   FROM   dailyaccounting_tab
   WHERE  day = '2014-08-12'
   AND    accountid = a.accountid
   )
ORDER  BY a.username;

LATERAL 关键字在这里是隐式的，函数总是可以引用更早的 FROM 项。 The manual:

LATERAL 也可以在函数调用 FROM 项之前，但在此 case 是噪声词，因为函数表达式可以参考 无论如何，之前的 FROM 项目。

相关：

• Insert multiple rows in one table based on number in another table

FROM 列表中带有逗号的短符号（大部分）等同于 CROSS JOIN LATERAL（与 [INNER] JOIN LATERAL ... ON TRUE 相同），因此会从函数调用不返回任何行的结果中删除行。要保留此类行，请使用 LEFT JOIN LATERAL ... ON TRUE：

...
FROM  account_tab a
LEFT  JOIN LATERAL account_servicetier_for_day(a.accountid, '2014-08-12') f ON TRUE
...

另外，如果可以避免，请不要使用NOT IN (subquery)。这是几种方法中最慢和最棘手的：

• Select rows which are not present in other table

我建议改为NOT EXISTS。

Postgres 9.2 或更早版本

您可以在SELECT 列表中调用集合返回函数（这是标准 SQL 的 Postgres 扩展）。出于性能原因，这最好在子查询中完成。分解外部查询中的（众所周知的！）行类型以避免重复评估函数：

SELECT '2014-08-12' AS day, 0 AS inbytes, 0 AS outbytes
     , a.username, a.accountid, a.userid
     , (a.rec).*   -- but avoid duplicate column names!
FROM  (
   SELECT *, account_servicetier_for_day(a.accountid, '2014-08-12') AS rec
   FROM   account_tab a
   WHERE  a.isdsl = 1
   AND    a.dslservicetypeid Is Not Null
   AND    NOT EXISTS (
       SELECT 1
       FROM   dailyaccounting_tab
       WHERE  day = '2014-08-12'
       AND    accountid = a.accountid
      )
   ) a
ORDER  BY a.username;

克雷格·林格的相关回答和解释，为什么我们最好在外部查询中分解：

• How to avoid multiple function evals with the (func()).* syntax in an SQL query?

Postgres 10 移除了 SELECT 中集合返回函数行为中的异常：

• What is the expected behaviour for multiple set-returning functions in SELECT clause?

Answer 2

使用from子句中的函数

Select
    '2014-08-12' As day,
    0 As inbytes,
    0 As outbytes,
    acct.username,
    acct.accountid,
    acct.userid,
    asfd.*
From
    account_tab acct
    cross join lateral
    account_servicetier_for_day(acct.accountid, '2014-08-12') asfd
Where acct.isdsl = 1
    And acct.dslservicetypeid Is Not Null
    And acct.accountid Not In (Select accountid From dailyaccounting_tab Where Day = '2014-08-12')
Order By acct.username