Question

请您帮我解决以下问题：考虑汽车维修公司的这两个表：

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我试图计算处理时间，这是＆＃34;进行中＆＃34;之间的日期差异。和＆＃34;固定＆＃34;状态：

Old_Value   New_Value   Created
0           1           2016/09/14
1           2           2016/09/15
2           3           2016/09/19

Value   Description
0       Not Diagnosed Yet
1       In Queue
2       In Progress
3       Fixed

我得到的只是一个错误信息。请帮忙，谢谢。

Answer 1

在MySQL中，您不使用day。我希望这样的查询：

Select repair_id,
       datediff(max(case when old_value = 1 and new_value = 2 then created end),
                max(case when old_value = 2 and new_value = 3 then created end)
               ) as handle_time
from tb1
group by repair_id;

我发明了专栏repair_id，因为它对我有意义。如果表中只有一个修复，则可以将其从查询中删除。

Answer 2

不清楚你想要什么，但尝试这将返回4.

Select TOP 1
datediff(DAY, (SELECT CREATED FROM T1 WHERE New_Value = 2), (SELECT CREATED FROM T1 WHERE New_Value = 3)) as handle_time
from t1

我刚刚运行您的查询，但没有发现任何错误。它只返回所有3行为NULL。

CREATE TABLE T1(Old_Value int, New_Value int, Created DATE)
CREATE TABLE T2(Value int, Description varchar(50))

INSERT INTO T1 VALUES
(0, 1, '2016/09/14'),
(1, 2, '2016/09/15'),
(2, 3, '2016/09/19')

INSERT INTO T2 VALUES
(0, 'Not Diagnosed Yet'),
(1, 'In Queue'),
(2, 'In Progress'),
(3, 'Fixed')

Select 
datediff(day, (case when Old_Value='1' and New_Value='2' then Created else Null end), (case when Old_Value='2' and New_Value='3' then Created else Null end))
as Handle_Time
from T1

DROP TABLE T1
DROP TABLE T2

结果：

Handle_Time
NULL
NULL
NULL

现在只需运行以下查询

Select 
(case when Old_Value='1' and New_Value='2' then Created else Null end) AS param1, 
(case when Old_Value='2' and New_Value='3' then Created else Null end) AS param2,
datediff(day, (case when Old_Value='1' and New_Value='2' then Created else Null end), (case when Old_Value='2' and New_Value='3' then Created else Null end))
as Handle_Time
from T1

结果：

param1      param2      Handle_Time
NULL        NULL        NULL
2016-09-15  NULL        NULL
NULL        2016-09-19  NULL

因此，在datediff中，您的一个参数始终为null，返回null。

SELECT DateDiff(DAY, NULL, NULL) --return NULL
SELECT DateDiff(DAY, GETDATE(), NULL) --return NULL
SELECT DateDiff(DAY, NULL, GETDATE()) --return NULL

Datediff声明中的用例

2 个答案: