How to determine whether the number is float or integer in SQL Server?

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挽巷
挽巷 2021-02-20 13:11

I need to write this query in sql server:

IF isFloat(@value) = 1
BEGIN
    PRINT \'this is float number\'
END
ELSE
BEGIN
    PRINT \'this is integer number\'
END         


        
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4条回答
  • 2021-02-20 13:47

    Martin, under certain circumstances your solution gives an incorrect result if you encounter a value of 1234.0, for example. Your code determines that 1234.0 is an integer, which is incorrect.

    This is a more accurate snippet:

    if cast(cast(123456.0 as integer) as varchar(255)) <> cast(123456.0 as varchar(255)) 
    begin 
      print 'non integer' 
    end 
    else 
    begin 
      print 'integer' 
    end
    

    Regards,

    Nico

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  • 2021-02-20 13:51

    See whether the below code will help. In the below values only 9, 2147483647, 1234567 are eligible as Integer. We can create this as function and can use this.

    CREATE TABLE MY_TABLE(MY_FIELD VARCHAR(50))
    INSERT INTO MY_TABLE
    VALUES('9.123'),('1234567'),('9'),('2147483647'),('2147483647.01'),('2147483648'), ('2147483648ABCD'),('214,7483,648')
    
    SELECT *
    FROM MY_TABLE
    WHERE CHARINDEX('.',MY_FIELD) = 0 AND CHARINDEX(',',MY_FIELD) = 0       
    AND ISNUMERIC(MY_FIELD) = 1 AND CONVERT(FLOAT,MY_FIELD) / 2147483647 <= 1
    DROP TABLE MY_TABLE
    

    OR

    DECLARE @num VARCHAR(100)
    SET @num = '2147483648AS'
    IF ISNUMERIC(@num) = 1 AND @num NOT LIKE '%.%' AND @num NOT LIKE '%,%' 
    BEGIN
        IF CONVERT(FLOAT,@num) / 2147483647 <= 1
            PRINT 'INTEGER'
        ELSE
            PRINT 'NOT-INTEGER' 
    END
    ELSE
        PRINT 'NOT-INTEGER'
    
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  • 2021-02-20 13:53
    declare @value float = 1
    
    IF FLOOR(@value) <> CEILING(@value)
    BEGIN
        PRINT 'this is float number'
    END
    ELSE
    BEGIN
        PRINT 'this is integer number'
    END
    
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  • 2021-02-20 13:53
    DECLARE @value FLOAT = 1.50
    IF CONVERT(int, @value) - @value <> 0
    BEGIN
        PRINT 'this is float number'
    END
    ELSE
    BEGIN
        PRINT 'this is integer number'
    END
    
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