如何在SQL Server中显示日期名称？

Question

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

但我发现错误如下：

从字符串转换日期和/或时间时转换失败。

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

Answer 1

你的问题是：

DATENAME(dw, DATEADD(dw, 1, CDAY))

我想你打算：

DATENAME(dw, DATEADD(dw, 1, CDATE))

我会把CTE写成：

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是一个db <>小提琴。

您没有描述代码要执行的代码。它对字符串进行了不必要的转换，对于您想要做的事情可能会不必要地复杂化。

Answer 2

无需转换为varchar以获得工作日。

 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))

你可以使用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

Answer 3

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)"

Demo

或者在你的情况下：

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|
+----------+-------+

Answer 4

无需在CTE中修复日期名称，仅将其用于生成日期。

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

Answer 5

没有必要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