如何在 SQL 服务器中显示日期名称？

Question

我在 SQL 服务器上有一个查询

我要显示如下：

CDATE      | CDAY
2019-04-01 | Monday
2019-04-02 | Tuesday
...        | ......
2019-04-30 | Tuesday

但是我发现错误如下：

Conversion failed when converting date and/or time from character string.

如果有人可以帮助，请

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

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))

可以直接使用datename函数获取。

  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 2

您可以很快使用 datename() 功能（从 v.2008 开始使用）

  select datename( weekday, getdate() ) as day

  day
  ------
  Friday  -- > "for today(2019-04-26)"

Demo

或如您的情况：

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

Answer 3

您的问题是：

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<>fiddle.

你没有描述你想要代码做什么。它有不必要的字符串转换，并且对于您想要执行的操作来说可能不必要地复杂化。

Answer 4

无需修复 CTE 中的日期名称，仅使用它来生成日期。

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

不需要CONVERT约会。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

如何在 SQL 服务器中显示日期名称？

How to show day name in SQL Server?

sql

sql-server

date

sql-server-2017