我在SQL服务器上有一个查询
我想显示如下:
CDATE | CDAY
2019-04-01 | Monday
2019-04-02 | Tuesday
... | ......
2019-04-30 | Tuesday
但是我发现错误如下:
从字符转换日期和/或时间时转换失败 字符串。
请有人帮忙
DECLARE @V_DATE DATE = GETDATE()
;WITH CTE_DATE AS (
SELECT DATEADD(dd,-(DAY(@V_DATE)-1),@V_DATE) CDATE,
DATENAME(dw, CONVERT(varchar, DATEADD(dd,-(DAY(@V_DATE)-1),@V_DATE))) CDAY
UNION ALL
SELECT DATEADD(dd,1,CDATE),
DATENAME(dw, CONVERT(varchar, DATEADD(dw,1,CDAY)))
FROM CTE_DATE
WHERE DATEADD(dd,1,CDATE) <= DATEADD(dd,-(DAY(DATEADD(mm,1,CDATE))),DATEADD(mm,1,CDATE))
)
SELECT * FROM CTE_DATE
答案 0 :(得分:2)
无需转换为varchar即可获得工作日。
UNION ALL
SELECT DATEADD(dd,1,CDATE),
DATENAME(dw, CONVERT(varchar, DATEADD(dw,1,CDAY))) -- No need to convert to varchar in order to get weekday.
FROM CTE_DATE
WHERE DATEADD(dd,1,CDATE) <= DATEADD(dd,-(DAY(DATEADD(mm,1,CDATE))),DATEADD(mm,1,CDATE))
您可以使用datename函数直接获取它。
DECLARE @V_DATE DATE = GETDATE()
;WITH CTE_DATE AS (
SELECT DATEADD(dd,-(DAY(@V_DATE)-1),@V_DATE) CDATE,
DATENAME(dw, CONVERT(varchar, DATEADD(dd,-(DAY(@V_DATE)-1),@V_DATE))) CDAY
UNION ALL
SELECT DATEADD(dd,1,CDATE),
DATENAME(dw, DATEADD(dd,1,CDATE)) -- modified
FROM CTE_DATE
WHERE DATEADD(dd,1,CDATE) <= DATEADD(dd,-(DAY(DATEADD(mm,1,CDATE))),DATEADD(mm,1,CDATE))
)
SELECT * FROM CTE_DATE
答案 1 :(得分:2)
您可以很快使用datename()
函数(自2008年起使用)
select datename( weekday, getdate() ) as day
day
------
Friday -- > "for today(2019-04-26)"
或根据您的情况:
with t(cdate) as
(
select '2019-04-01' union all
select '2019-04-02' union all
select '2019-04-30'
)
select cdate, datename( weekday, cdate ) as cday
from t;
+----------+-------+
| cdate | cday |
+----------+-------+
|2019-04-01|Monday |
|2019-04-02|Tuesday|
|2019-04-30|Tuesday|
+----------+-------+
答案 2 :(得分:2)
您的问题是:
DATENAME(dw, DATEADD(dw, 1, CDAY))
我认为您打算这样做
DATENAME(dw, DATEADD(dw, 1, CDATE))
我将CTE写为:
WITH CTE_DATE AS (
SELECT DATEADD(day ,-(DAY(@V_DATE)-1),@V_DATE) as CDATE,
DATENAME(dw, DATEADD(day, -(DAY(@V_DATE) - 1), @V_DATE)) as CDAY
UNION ALL
SELECT DATEADD(day, 1, CDATE),
DATENAME(dw, DATEADD(dw, 1, CDATE))
FROM CTE_DATE
WHERE DATEADD(day, 1, CDATE) <= DATEADD(day, -(DAY(DATEADD(month, 1, CDATE))), DATEADD(month, 1, CDATE))
)
SELECT *
FROM CTE_DATE;
Here是db <>小提琴。
您没有描述您想要代码执行的代码。它具有不必要的字符串转换功能,对于您想做的事情可能会不必要地复杂。
答案 3 :(得分:2)
无需在CTE中固定日期名称,只需将其用于生成日期即可。
DECLARE @V_DATE DATE = GETDATE()
WITH CTE_DATE AS
(
SELECT DATEADD(day ,-(DAY(@V_DATE)-1),@V_DATE) as CDATE
UNION ALL
SELECT DATEADD(day, 1, CDATE)
FROM CTE_DATE
WHERE DATEADD(day, 1, CDATE) <= DATEADD(day, -(DAY(DATEADD(month, 1, CDATE))), DATEADD(month, 1, CDATE))
)
SELECT CDATE, DATENAME(dw, CDATE) FROM CTE_DATE
答案 4 :(得分:2)
不需要CONVERT
日期。使用FORMAT
函数。并使用EOMONTH
函数:
DECLARE @V_DATE DATE = DATEADD(DAY, 1, EOMONTH(GETDATE(), -1));
WITH CTE_DATE AS (
SELECT @V_DATE CDATE
UNION ALL
SELECT DATEADD(dd, 1, CDATE)
FROM CTE_DATE
WHERE DATEADD(dd, 1, CDATE) <= EOMONTH(@V_DATE)
)
SELECT CDATE, FORMAT(CDATE, 'dddd') AS CDAY, FORMAT(CDATE, 'ddd') AS CDAYSHORT
FROM CTE_DATE