SQL将单词拆分为单独的字符

我需要更改一个应用程序，我需要做的第一件事是更改数据库表中的一个字段。 在这个表中，我现在有1到6个单个字符，即“abcdef” 我需要将此更改为“[a][b][c][d][e][f]”

[编辑]这意味着保持在同一字段中。所以在字段前面='abcdef'和字段后面='a][b][c][d][e][f]'

这样做的好方法是什么

rg。

---

Eric

您可以使用以下函数将字符串拆分为单独的字符：

create function ftStringCharacters
(
    @str varchar(100)
)
returns table as
return
    with v1(N) as (
        select 1 union all select 1 union all select 1 union all select 1 union all select 1
        union all
        select 1 union all select 1 union all select 1 union all select 1 union all select 1
    ),
    v2(N) as (select 1 from v1 a, v1 b),
    v3(N) as (select top (isnull(datalength(@str), 0)) row_number() over (order by @@spid) from v2)
    select N, substring(@str, N, 1) as C
    from v3
GO

然后将其应用为：

update t
set t.FieldName = p.FieldModified
from TableName t
    cross apply (
        select (select quotename(s.C)
        from ftStringCharacters(t.FieldName) s
        order by s.N
        for xml path(''), type).value('text()[1]', 'varchar(20)')
    ) p(FieldModified)

---

您可以使用以下函数将字符串拆分为单独的字符：

create function ftStringCharacters
(
    @str varchar(100)
)
returns table as
return
    with v1(N) as (
        select 1 union all select 1 union all select 1 union all select 1 union all select 1
        union all
        select 1 union all select 1 union all select 1 union all select 1 union all select 1
    ),
    v2(N) as (select 1 from v1 a, v1 b),
    v3(N) as (select top (isnull(datalength(@str), 0)) row_number() over (order by @@spid) from v2)
    select N, substring(@str, N, 1) as C
    from v3
GO

然后将其应用为：

update t
set t.FieldName = p.FieldModified
from TableName t
    cross apply (
        select (select quotename(s.C)
        from ftStringCharacters(t.FieldName) s
        order by s.N
        for xml path(''), type).value('text()[1]', 'varchar(20)')
    ) p(FieldModified)

---

不使用函数：

declare @t table(C varchar(18))
insert @t values('abc'), ('1234'), (' 1234a')

;with CTE as
(
select C, '[' + substring(c, a.n, 1) + ']' v, rn   from
(select 1 n union all 
select 2 union all 
select 3 union all 
select 4 union all 
select 5 union all 
select 6) a
cross apply
(select c, row_number() over (order by C) rn from @t group by c) b
where a.n <= len(C)
)
update t3
set C = t4.[value] 
FROM @t t3
JOIN
(
select C,
    ( 
        select v 
        from CTE t1 
        where t1.rn = t2.rn 
        for xml path(''), type 
    ).value('.', 'varchar(18)') [value] 
from CTE t2 
group by t2.rn, C
) t4
ON t3.C = t4.C

SELECT * FROM @t

---

不使用函数：

declare @t table(C varchar(18))
insert @t values('abc'), ('1234'), (' 1234a')

;with CTE as
(
select C, '[' + substring(c, a.n, 1) + ']' v, rn   from
(select 1 n union all 
select 2 union all 
select 3 union all 
select 4 union all 
select 5 union all 
select 6) a
cross apply
(select c, row_number() over (order by C) rn from @t group by c) b
where a.n <= len(C)
)
update t3
set C = t4.[value] 
FROM @t t3
JOIN
(
select C,
    ( 
        select v 
        from CTE t1 
        where t1.rn = t2.rn 
        for xml path(''), type 
    ).value('.', 'varchar(18)') [value] 
from CTE t2 
group by t2.rn, C
) t4
ON t3.C = t4.C

SELECT * FROM @t

---

谢谢，但我的意思是留在同一个领域。改变了我的问题以更清楚地反映这一点。谢谢，但我的意思是留在同一个领域。改变了我的问题，以便更清楚地反映这一点。