Question

需要帮助才能创建一个返回TRUE或FALSE的函数。 TRUE - 如果输入1或3个单词（如'__ hello _'，'_ hello'，'_ hello my frend' - 空格应该被删除），如果条件未满足则为FALSE

CREATE FUNCTION dbo.nazvFac(@f nvarchar(30))    
RETURNS bit
AS
BEGIN
DECLARE @l int = 1, @s nvarchar(30), @i int = 0, @b bit
WHILE LTRIM(RTRIM(LEN(@f))) >= @l  --error here, but I do not know how to fix it
BEGIN                                   
    SET @s = SUBSTRING(@f, @l, 1)
    IF @s BETWEEN 'А' AND 'я' 
        SET @l += 1
    ELSE IF @s = ' '
    BEGIN
        SET @l -= 1
        SET @s = SUBSTRING(@f, @l, 1)
        SET @s = RTRIM(@s) 
        SET @l += 2
        SET @i += 1
    END
    ELSE
        BREAK
END
IF @i = 0 OR @i = 2 
    SET @b = 'TRUE'
ELSE
    SET @b = 'FALSE'
RETURN @b
END
GO

Answer 1

WHILE LTRIM(RTRIM(LEN(@f))) >= @l --error here, but I do not know how to fix it

LEN()返回一个int，然后将其传递给字符串函数：RTRIM()。

Answer 2

你应修剪字符串，然后检查长度。

CREATE FUNCTION dbo.nazvFac(@f NVARCHAR(30))    
RETURNS BIT
AS
BEGIN
    DECLARE @l INT = 1, @s NVARCHAR(30), @i INT = 0, @b BIT

    WHILE LEN(LTRIM(RTRIM(@f))) >= @l  --I think youn need to trim the string and then check length
        BEGIN                                   
            SET @s = SUBSTRING(@f, @l, 1)
            IF @s BETWEEN 'А' AND 'я' 
                SET @l += 1
            ELSE IF @s = ' '
            BEGIN
                SET @l -= 1
                SET @s = SUBSTRING(@f, @l, 1)
                SET @s = RTRIM(@s) 
                SET @l += 2
                SET @i += 1
            END
            ELSE
                BREAK
        END
    IF @i = 0 OR @i = 2 
        SET @b = 'TRUE'
    ELSE
        SET @b = 'FALSE'
    RETURN @b
END
GO

Answer 3

只有在有一个或三个单词时才想返回TRUE？这应该这样做：

CREATE FUNCTION dbo.nazvFac(@f NVARCHAR(30))    
RETURNS BIT
AS
BEGIN
    DECLARE @l INT, @b BIT

    SET @l = LEN(@f) - LEN(REPLACE(@f, ' ', '')) + 1
    IF @l == 1 OR @l == 3
        SET @b = 'TRUE'
    ELSE
        SET @b = 'FALSE'

    RETURN @b
END

另外，JC。关于len（）错误是正确的。

检查输入字数的功能

3 个答案: