Question

我想在sql字符串中找到信用卡数值。

例如

;

DECLARE @value1 NVARCHAR(MAX) = 'The payment is the place 1234567812345678'
DECLARE @value2 NVARCHAR(MAX) = 'The payment is the place 123456aa7812345678'
DECLARE @value3 NVARCHAR(MAX) = 'The payment1234567812345678is the place'

结果应为：

@value1Result 1234567812345678
@value2Result NULL
@value3Result 1234567812345678

16位必须在一起，没有空格。

如何在sql脚本或函数中执行此操作？

编辑：如果我想找到这2张信用卡价值。

@value4 = 'card 1 is : 4034349183539301 and the other one is 3456123485697865'

我该如何实现脚本？

Answer 1

您可以将PathIndex用作

PATINDEX('%[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]%', yourStr)

如果结果为0，则它不包含其他16位数字。

可以根据您的需要使用Where声明或Select声明

Answer 2

你可以写成：

SELECT case when Len(LEFT(subsrt, PATINDEX('%[^0-9]%', subsrt + 't') - 1)) = 16
                 then LEFT(subsrt, PATINDEX('%[^0-9]%', subsrt + 't') - 1)
                 else ''
                 end
FROM (
    SELECT subsrt = SUBSTRING(string, pos, LEN(string))
    FROM (
        SELECT string, pos = PATINDEX('%[0-9]%', string)
        FROM table1
    ) d
) t

Demo

Answer 3

DECLARE @ value1 NVARCHAR（MAX）=＆＃39;卡1是：4034349183539301，另一个是3456123485697865＆＃39; DECLARE @Lenght INT ，@ Count INT ，@ Candidate CHAR ，@ cNum INT ，@ result VARCHAR（16）

SELECT @Count = 1

SELECT @cNum = 0

SELECT @result =＆＃39;＆＃39;

SELECT @Lenght = LEN（@ value1）

WHILE @Count＆lt; = @Lenght 开始 SELECT @Candidate = SUBSTRING（@ value1，@ Count，1）

IF @Candidate != ' '
    AND ISNUMERIC(@Candidate) = 1
BEGIN
    SET @cNum = @cNum + 1
    SET @result = @result + @Candidate
END
ELSE
BEGIN
    SET @cNum = 1
    SET @result = ''
END

IF @cNum > 16
BEGIN
    SELECT @result 'Credit Number'
END

SET @Count = @Count + 1

END

Answer 4

先生，你先去了。

 DECLARE 
    @value3 NVARCHAR(MAX) = 'The payment1234567812345678is the place',
    @MaxCount int,
    @Count int,
    @Numbers NVARCHAR(100)

SELECT @Count = 1
SELECT @Numbers = ''
SELECT @MaxCount = LEN(@value3)


WHILE @Count <= @MaxCount
BEGIN

    IF (UNICODE(SUBSTRING(@value3,@Count,1)) >= 48 AND UNICODE(SUBSTRING(@value3,@Count,1)) <=57)
        SELECT @Numbers = @Numbers + SUBSTRING(@value3,@Count,1)

    SELECT @Count = @Count + 1


END
PRINT @Numbers

如果您计划大量使用它，可以将其作为一项功能。

搜索字符串信用卡数值

4 个答案: