小编典典

SQL Server:返回字符串过程INITCAP

sql

这就是我所做的。

create proc INITCAP(@string varchar(30))
as
begin
    SET @string = UPPER(LEFT(@string,1)) + LOWER(RIGHT(@string, LEN(@string) -1))
end

declare @lastname varchar
set @lastname = exec INITCAP 'MILLER'

declare @firstname varchar
set @firstname = exec INITCAP 'StEvE'

UPDATE Employee SET firstname = @firstname, lastname = @lastname WHERE empID = 7934

我不断收到错误:

消息156,级别15,状态1,过程INITCAP,第97行
关键字“ exec”附近的语法错误。
消息156,级别15,状态1,过程INITCAP,第100行
关键字“ exec”附近的语法错误。

我该怎么办?我希望该过程INITCAP像在Oracle中一样工作:返回一个名称,例如:“ Steve”,“ Miller”


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2021-03-08

共1个答案

小编典典

解决方案1 (我不会使用此解决方案)

您可以这样使用OUTPUT参数:

create proc INITCAP(@string varchar(30) OUTPUT)
as
begin
    SET @string = UPPER(LEFT(@string,1)) + LOWER(SUBSTRING(@string, 2, 8000))
end
go

declare @lastname varchar
set @lastname = 'MILLER'
exec INITCAP @lastname OUTPUT

declare @firstname varchar
set @firstname = 'StEvE'
exec INITCAP @firstname OUTPUT

解决方案2 :相反, 我将选择 创建一个内联函数,因此:

CREATE FUNCTION dbo.Capitalize1(@string varchar(30))
RETURNS TABLE
AS
RETURN
SELECT UPPER(LEFT(@string,1)) + LOWER(SUBSTRING(@string, 2, 8000)) AS Result;

用法:

UPDATE e 
SET firstname = cap.Result
FROM Employee e
CROSS APPLY dbo.Capitalize1(e.firstname) cap;

解决方案3 :另一个选项可能是标量函数with schemabinding选项(出于性能原因):

CREATE FUNCTION dbo.Capitalize2(@string varchar(30))
RETURNS VARCHAR(30)
WITH SCHEMABINDING
AS
BEGIN
    RETURN UPPER(LEFT(@string,1)) + LOWER(SUBSTRING(@string, 2, 8000));
END;

用法:

UPDATE Employee
SET firstname = dbo.Capitalize2(firstname);
2021-03-08