我试图根据错误代码的数量及其在SQL查询的特定时间段内出现的条件值,但我写的这个查询返回所有记录的第一个满足条件。
我做错了什么
SELECT ERROR_CODE,
CASE
WHEN (SELECT Count(ERROR_CODE)
FROM ERROR_LOG
WHERE CRET_DTIM > Trunc(SYSDATE - 2)
AND ERROR_CODE = '50001') > 50 THEN 'RED'
WHEN (SELECT Count(ERROR_CODE)
FROM ERROR_LOG
WHERE CRET_DTIM > Trunc(SYSDATE - 2)
AND ERROR_CODE = '50001') < 50 THEN 'GREEN'
WHEN (SELECT Count(ERROR_CODE)
FROM ERROR_LOG
WHERE CRET_DTIM > Trunc(SYSDATE - 1)
AND ERROR_CODE IS NULL) > 100 THEN 'RED'
WHEN (SELECT Count(ERROR_CODE)
FROM ERROR_LOG
WHERE CRET_DTIM > Trunc(SYSDATE - 1)
AND ERROR_CODE IS NULL) < 100 THEN 'GREEN'
END AS ALERT_COLOR
FROM ERROR_LOG
GROUP BY ERROR_CODE
示例数据
50005 05-JAN-15 11.05.51.570000000 AM
50001 05-JAN-15 02.39.57.840000000 PM
50001 05-JAN-15 02.31.06.700000000 PM
50001 05-JAN-15 02.21.49.350000000 PM
50001 05-JAN-15 01.59.13.910000000 PM
50001 05-JAN-15 01.50.05.900000000 PM
50001 05-JAN-15 01.30.19.270000000 PM
50001 05-JAN-15 01.11.10.510000000 PM
50001 05-JAN-15 12.00.00.720000000 PM
05-JAN-15 09.42.10.670000000 AM
05-JAN-15 09.37.31.590000000 AM
DESIRED OUPUT(更改计数条件以查看所需的输出) 的 COUNT
2
50001 8
50005 1
SQL
SELECT ERROR_CODE,
CASE
WHEN (SELECT Count(ERROR_CODE)
FROM ERROR_LOG
WHERE CRET_DTIM > Trunc(SYSDATE - 2)
AND ERROR_CODE = '50001') between 5 and 100 THEN 'RED'
WHEN (SELECT Count(ERROR_CODE)
FROM ERROR_LOG
WHERE CRET_DTIM > Trunc(SYSDATE - 2)
AND ERROR_CODE = '50001') < 5 THEN 'GREEN'
WHEN (SELECT Count(ERROR_CODE)
FROM ERROR_LOG
WHERE CRET_DTIM > Trunc(SYSDATE - 1)
AND ERROR_CODE IS NULL) > 100 THEN 'RED'
WHEN (SELECT Count(ERROR_CODE)
FROM ERROR_LOG
WHERE CRET_DTIM > Trunc(SYSDATE - 1)
AND ERROR_CODE IS NULL) between 5 and 100 THEN 'GREEN'
END AS ALERT_COLOR
FROM ERROR_LOG
GROUP BY ERROR_CODE
结果
Error_code Alert_color
GREEN
50005 ''
50001 RED
答案 0 :(得分:3)
您正在为表中的每一行重复每个when
子句中的子查询,没有相关性。子查询:
(SELECT Count(ERROR_CODE)
FROM ERROR_LOG
WHERE CRET_DTIM > Trunc(SYSDATE - 2)
AND ERROR_CODE = '50001')
对于表格中的每一行, ...每次评估为8;因此,无论实际在该行中的错误代码如何,案例的第一个分支将始终匹配,因此您会看到每行RED
。
重复子查询效率很低,即使您将其关联起来也是如此。 Oracle会对它进行优化/缓存,但仍然......生成计数一次会更好:
select error_code,
count(case when cret_dtim > trunc(sysdate - 2) then 1 end) as cnt_2_day,
count(case when cret_dtim > trunc(sysdate - 1) then 1 end) as cnt_1_day
from error_log
where cret_dtim > trunc(sysdate - 2)
group by error_code
order by error_code nulls first;
ERROR_CODE CNT_2_DAY CNT_1_DAY
---------- --------- ----------
(null) 2 2
50001 8 8
50005 1 1
您的所有样本数据都不到一天,因此计数相同;如果你有两天前的数据,他们会有所不同。
然后您可以使用 - 一次 - 作为内部查询(如果您愿意,也可以使用CTE),并将结果用作另一级case
的输入,以转换为您的颜色代码:
select error_code,
case
when error_code = '50001' and cnt_1_day between 5 and 100 then 'RED'
when error_code = '50001' and cnt_2_day < 5 then 'GREEN'
when error_code is null and cnt_1_day > 100 then 'RED'
when error_code is null and cnt_1_day between 5 and 100 then 'GREEN'
else null
end as alert_colour
from
(
select error_code,
count(case when cret_dtim > trunc(sysdate - 2) then 1 end) as cnt_2_day,
count(case when cret_dtim > trunc(sysdate - 1) then 1 end) as cnt_1_day
from error_log
where cret_dtim > trunc(sysdate - 2)
group by error_code
)
order by error_code nulls first;
ERROR_CODE ALERT_COLOUR
---------- ------------
(null) (null)
50001 RED
50005 (null)
这与你所说的预期结果不符,但正如我在评论中所说,我不确定你是如何从数字到你所展示的结果的。你的水桶排除了很多可能性;例如,您可能希望第一个条件为> 5
。你完全忽略了你期望看到的50005
。如果您的基础逻辑是任何代码集使用2天计数而空值使用1天计数 - 这是一个飞跃 - 那么您可以将它概括为:
select error_code,
case
when error_code is not null and cnt_not_null between 5 and 100 then 'RED'
when error_code is not null and cnt_not_null < 5 then 'GREEN'
when error_code is null and cnt_null > 100 then 'RED'
when error_code is null and cnt_null between 5 and 100 then 'GREEN'
else null
end as alert_colour
from (
select error_code,
count(case when cret_dtim > trunc(sysdate - 2)
and error_code is not null then 1 end) as cnt_not_null,
count(case when cret_dtim > trunc(sysdate - 1)
and error_code is null then 1 end) as cnt_null
from error_log
where cret_dtim > trunc(sysdate - 2)
group by error_code
)
order by error_code nulls first;
ERROR_CODE ALERT_COLOUR
---------- ------------
(null) (null)
50001 RED
50005 GREEN
这仍然不是你预期的结果,但必须要做,直到解释逻辑(和我的错误),或者你的结果发生变化......
答案 1 :(得分:0)
case
返回评估为result_expression
的第一个input_expression = when_expression
的{{1}}。试试这个。
TRUE