Question

这是我的查询

select 
    custid,
    Stuff(Coalesce('' + t12, '') + Coalesce('' + t11, '') + Coalesce('' + t10, '') + 
          Coalesce('' + t9, '') + Coalesce('' + t8, '') + Coalesce('' + t7, '') + 
          Coalesce('' + t6, '') + Coalesce('' + t5, '') + Coalesce('' + t4, '') + 
          Coalesce('' + t3, '') + Coalesce('' + t2, '') + Coalesce('' + t1, '') +
          Coalesce('' + id, ''), 1, 0, '') AS path
from 
    Table

这是我回来的结果：

431222294701031547005760100001411000302910000718.

需要这样：

43122229/47010315/47005760/10000141/10003029/10000718

如果我使用

SELECT columns, stuff(Coalesce('/' + t2, '') + Coalesce('/' + t1, '') , 1, 0, '') AS path  
FROM table

结果如下所示：

/43122229/47010315/47005760/10000141/10003029/10000718.

如何解决此问题？请帮助我几个小时以来一直挠头。...

Answer 1

如果知道字符串的最大长度，则可以使用CROSS APPLY提取每8个字符，然后在它们之间用“ /”连接

declare @sample table
(
    col varchar(100)
)

insert into @sample (col) select '12345678901234567890123'
insert into @sample (col) select '123456789012345678901234567890'
insert into @sample (col) select '12'


select  s.col, 
        new_col = isnull(col1, '') 
                + isnull('/' + col2, '') 
                + isnull('/' + col3, '') 
                + isnull('/' + col4, '') 
                + isnull('/' + col5, '')
from    @sample s
    cross apply
    (
        select  col1 = nullif(substring(col, 1, 8), '')
    ) p1
    cross apply
    (
        select  col2 = nullif(substring(col, 9, 8), '')
    ) p2
    cross apply
    (
        select  col3 = nullif(substring(col, 17, 8), '')
    ) p3
    cross apply
    (
        select  col4 = nullif(substring(col, 25, 8), '')
    ) p4
    cross apply
    (
        select  col5 = nullif(substring(col, 33, 8), '')
    ) p5

Answer 2

在下面的查询中尝试此

DECLARE @txt varchar(max)='/43122229/47010315/47005760/10000141/10003029/10000718'
SELECT STUFF(@txt,1,1,'') as newText

示例：

这表示从第一个1字符位置开始，将id=xxx字符替换为”

如何在SQL中每8个字符后添加斜杠

2 个答案: