Question

我正在搜索一个问题的评论。

我有一个4位数字，如3521，问题是我需要找到前两位数大于后两位的数字。

Answer 1

假设你的号码是lnMyNumber：

lnMyNumber = 3521
? (Val(Left(Str(lnMyNumber, 4, 0),2)) > Val(right(Str(lnMyNumber, 4, 0),2)))

Answer 2

命令窗口示例：

LOCAL unknownType
unknownType = 3521
? CompareSumsOfTwoBytePairs(unknownType) && returns true
unknownType = "3521"
? CompareSumsOfTwoBytePairs(unknownType) && returns true
unknownType = "YZAB"
? CompareSumsOfTwoBytePairs(unknownType) && returns true
unknownType = 35210
? CompareSumsOfTwoBytePairs(unknownType) && too long, returns false
* The following unexpected types return all false:
unknownType = .F.
? CompareSumsOfTwoBytePairs(unknownType)
unknownType = .Null.
? CompareSumsOfTwoBytePairs(unknownType)
unknownType = CREATEOBJECT("Form")
? CompareSumsOfTwoBytePairs(unknownType)
unknownType = DATE(2015, 2, 7)
? CompareSumsOfTwoBytePairs(unknownType)

FUNCTION CompareSumsOfTwoBytePairs(u)
    LOCAL c
    c = TRANSFORM(u)
    RETURN LEN(c) == 4 AND (SUBSTR(c, 1, 2) > SUBSTR(c, 3, 2))
ENDFUNC

Answer 3

另一种选择是将数字保持为数字与强制字符串转换。

n = 3521
? INT( n/100 ) > n-(INT(n/100)*100)

n/100 = 35.21... the int leaves just 35

同样的基础在另一半......

n-(INT(n/100)*100)
3521 - ( INT( 3521/100) * 100 )
= 3521 - (  35 * 100 )
= 3521 - 3500 = 21

so 35 > 21

Answer 4

这是一个完整的代码，而不是你可以放入一个函数

    local lnVal, llReturn

    lnVal=35821

    llReturn=left(trans(lnVal),2)>right(trans(lnVal),2)

    return llReturn

Visual Fox Pro 9 4位数字

4 个答案: