Question

我在SQLite中有一个表：

CREATE TABLE test_results(timestamp TEXT, npass INTEGER, nfails INTEGER)

我想返回每天的最后一次通过/失败信息。例如，如果表包含...

2009-08-31 23:30:19|0|24
2009-08-31 23:37:18|0|24
2009-08-31 23:40:00|0|24
2009-09-01 19:02:13|0|2
2009-09-01 19:08:24|2|0
2009-09-01 19:20:29|2|0

我想要一个select语句来返回......

2009-08-31 23:40:00|0|24
2009-09-01 19:20:29|2|0

提前致谢！

Answer 1

尝试：

SELECT t.timestamp,
       t.npass,
       t.nfails
  FROM TEST_RESULTS t
  JOIN (SELECT MAX(tt.timestamp) 'maxtimestamp'
         FROM TEST_RESULTS tt
     GROUP BY date(tt.timestamp)) m ON m.maxtimestamp = t.timestamp

Answer 2

您可以尝试此查询：

SELECT * FROM (select * from test_results ORDER BY timestamp)
GROUP BY date(timestamp)

如果您还希望每天都能获得第一次失败：

SELECT * FROM (select * from test_results ORDER BY timestamp DESC)
GROUP BY date(timestamp)

Answer 3

这就是我解决这类问题的方法：

SELECT t1.*
FROM test_results t1
LEFT OUTER JOIN test_results t2
  ON (DATE(t1.timestamp) = DATE(t2.timestamp) AND t1.timestamp < t2.timestamp)
WHERE t2.timestamp IS NULL;

这假定timestamp列对其具有唯一或主键约束。否则，每天可以获得超过一行的（小）机会。

选择每日最新条目和相应数据

3 个答案: