oracle select 今天记录但排除前几天可能存在的记录

Question

oracle 中的table 中应该存在两条记录，一条来自请求，一条来自响应。

我想 select 所有今天的记录，但问题是另一对可能存在于前两天或更长时间。怎么保证返回的记录只有1条，前几天不存在

select A.referenceNum, A.datetime, A.Type from table A where A.datetime >= sysdate - 1

为了形象化，我只想select reference num 789ef.

ReferenceNum DateTime Type

123ab （今天到今天）请求

123ab （日期到今天）响应

456cd （日期到今天）请求

456cd （datetoday-2）响应

789ef （日期到今天）请求

Answer 1

使用NOT EXISTS运算符

SELECT A.referenceNum, A.datetime, A.Type 
from table A 
   where A.datetime >= sysdate - 1
  AND NOT EXISTS (
     SELECT null FROM table B
     WHERE A.referenceNum = B.referenceNum
       AND b.datetime < a.datetime
  )

Answer 2

SELECT referenceNum
FROM A A1
WHERE TRUNC(A1.datetime) > TRUNC(sysdate)-1
AND A1.Type              = 'Request'
AND NOT EXISTS
  (SELECT 1
  FROM A
  WHERE TRUNC(A.datetime) >= TRUNC(sysdate)-1
  AND A.Type               = 'Response'
  AND A.referenceNum       = A1.referenceNum
  )

Answer 3

您可以使用分析函数在一次 table 扫描中完成（与使用 NOT EXISTS 相比，后者将使用两次 table 扫描）： SQL Fiddle

Oracle 11g R2 模式设置:

CREATE TABLE table_name ( ReferenceNum,  DateTime,  Type ) AS
SELECT '123ab', SYSDATE, 'Request' FROM DUAL UNION ALL
SELECT '123ab', SYSDATE, 'Response' FROM DUAL UNION ALL
SELECT '456cd', SYSDATE, 'Request' FROM DUAL UNION ALL
SELECT '456cd', SYSDATE - 2, 'Response' FROM DUAL UNION ALL
SELECT '789ef', SYSDATE, 'Request' FROM DUAL;

查询 1:

SELECT ReferenceNum
FROM   ( 
  SELECT ReferenceNum,
         COUNT( CASE WHEN TRUNC( DateTime ) =  TRUNC( SYSDATE ) THEN 1 END )
           OVER ( PARTITION BY ReferenceNum ) AS num_today,
         COUNT( CASE WHEN TRUNC( DateTime ) <> TRUNC( SYSDATE ) THEN 1 END )
           OVER ( PARTITION BY ReferenceNum ) AS num_other_day
  FROM   table_name t
)
WHERE  num_today = 1
AND    num_other_day = 0

Results:

| REFERENCENUM |
|--------------|
|        789ef |

oracle select 今天记录但排除前几天可能存在的记录

oracle select records today but exclude those which may exist in the previous days

oracle

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