我有一张表格如下：
Order No | Customers | Amount
---------+----------------------+-------------
1 | Briant~~Luck | 23~~2122
2 | Mike~~Lee~~David | 10~~200~37
3 | Stak | 100
使用格式，每个客户在Amount中都有一个值。
我正在尝试弄清楚如何扩展~~分隔值以填充新的customer表，它应如下所示：
Order No | Customer | Amount
---------+----------------------+---------
1 | Briant | 23
1 | Luck | 2122
2 | Mike | 10
2 | Lee | 200
2 | David | 37
3 | Stak | 100
我该怎么办？
赞赏SQL查询，函数或游标中的任何解决方案。
由于
答案 0 :(得分：2)
我认为您可以将数据存储为预期的结果结构。它好多了。
顺便说一句，你可以使用拆分功能来获得输出
DECLARE @SampleData AS TABLE
(
OrderNo int,
Customers varchar(200),
Amount varchar(200)
)
INSERT INTO @SampleData
(
OrderNo,
Customers,
Amount
)
VALUES
( 1, 'Briant~~Luck','23~~2122'),
( 2, 'Mike~~Lee~~David','10~~200~~37'),
( 3, 'Stak','100')
SELECT sd.OrderNo, c.[Value] AS Customer, a.[Value] AS Amount
FROM @SampleData sd
CROSS APPLY
(
SELECT Pos, Value
FROM [dbo].[SplitString](sd.Customers,'~~')
) c
CROSS APPLY
(
SELECT Pos, Value
FROM [dbo].[SplitString](sd.Amount,'~~')
) a
WHERE c.Pos = a.Pos
ORDER BY sd.OrderNo
分割功能
CREATE FUNCTION [dbo].[SplitString] (@Text varchar(max),@Delimiter varchar(10))
Returns Table
As
Return (
Select Pos = Row_Number() over (Order By (Select null))
,Value = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(max)')))
From (Select x = Cast('<x>'+ Replace(@Text,@Delimiter,'</x><x>')+'</x>' as xml).query('.')) as A
Cross Apply x.nodes('x') AS B(i)
);
答案 1 :(得分：0)
如果您使用的是SQL2016 +，请尝试以下操作：
drop table if exists dbo.Orders;
create table dbo.Orders (
ID int
, Customers varchar(100)
, Amount varchar(100)
);
insert into dbo.Orders (ID, Customers, Amount)
values (1,'Briant~~Luck', '23~~2122')
, (2, 'Mike~~Lee~~David', '10~~200~~37')
, (3, 'Stak', '100');
select
o.ID
, t01.value as Customer
, t02.value as Amount
from dbo.Orders o
outer apply (
select
ROW_NUMBER () over (order by o.Customers ASC) as Rbr
, t.value
from string_split (replace(o.Customers, '~~', '~'), '~') t
) t01
outer apply (
select
ROW_NUMBER () over (order by o.Amount ASC) as Rbr
, t.value
from string_split (replace(o.Amount, '~~', '~'), '~') t
) t02
where t01.Rbr = t02.Rbr
答案 2 :(得分：0)
如果您使用的是2016年之前的SQL Server版本，则需要创建自己的String Splitting函数并在脚本中引用它。我使用的版本如下：
create function [dbo].[StringSplit]
(
@str nvarchar(4000) = ' ' -- String to split.
,@delimiter as nvarchar(1) = ',' -- Delimiting value to split on.
,@num as int = null -- Which value to return.
)
returns table
as
return
( -- Start tally table with 10 rows.
with n(n) as (select n from (values(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) n(n))
-- Select the same number of rows as characters in @str as incremental row numbers.
-- Cross joins increase exponentially to a max possible 10,000 rows to cover largest @str length.
,t(t) as (select top (select len(@str) a) row_number() over (order by (select null)) from n n1,n n2,n n3,n n4)
-- Return the position of every value that follows the specified delimiter.
,s(s) as (select 1 union all select t+1 from t where substring(@str,t,1) = @delimiter)
-- Return the start and length of every value, to use in the SUBSTRING function.
-- ISNULL/NULLIF combo handles the last value where there is no delimiter at the end of the string.
,l(s,l) as (select s,isnull(nullif(charindex(@delimiter,@str,s),0)-s,4000) from s)
select rn as ItemNumber
,Item
from(select row_number() over(order by s) as rn
,substring(@str,s,l) as item
from l
) a
where rn = @num -- Return a specific value where specified,
or @num is null -- Or the everything where not.
)
使用如下。请注意，我已将outer apply拆分为单独的查询以避免行重复：
declare @t table(OrderNo int,Customers nvarchar(500),Amount nvarchar(500));
insert into @t values
(1,'Briant~~Luck','23~~2122')
,(2,'Mike~~Lee~~David','10~~200~~37')
,(3,'Stak','100');
with c as
(
select t.OrderNo
,c.ItemNumber
,c.Item as Customers
from @t t
outer apply dbo.StringSplit(replace(t.Customers,'~~','|'),'|',null) c
),a as
(
select t.OrderNo
,a.ItemNumber
,a.Item as Amount
from @t t
outer apply dbo.StringSplit(replace(t.Amount,'~~','|'),'|',null) a
)
select c.OrderNo
,c.Customers
,a.Amount
from c
join a
on(c.OrderNo = a.OrderNo
and c.ItemNumber = a.ItemNumber
)
order by a.OrderNo
,c.Customers;
输出：
+---------+-----------+--------+
| OrderNo | Customers | Amount |
+---------+-----------+--------+
| 1 | Briant | 23 |
| 1 | Luck | 2122 |
| 2 | David | 37 |
| 2 | Lee | 200 |
| 2 | Mike | 10 |
| 3 | Stak | 100 |
+---------+-----------+--------+
答案 3 :(得分：0)
此解决方案为plsql
declare
type t_array_text is table of varchar2(30);
type t_array_number is table of number;
array_text t_array_text := t_array_text();
array_number t_array_number := t_array_number();
i number := 1;
cursor c1 is
select * from deneme;
begin
FOR c in c1
LOOP
i := 1;
array_text := t_array_text();
array_number := t_array_number();
for rec in (
SELECT regexp_substr(c.customer, '[[:alpha:]]+', 1, level) a frOM dual
CONNECT BY level<=regexp_count(c.customer,'~~')+1)
loop
array_text.extend();
array_text (i) := rec.a;
i := i + 1;
end loop;
i := 1;
for rec in (
SELECT regexp_substr(c.amount, '[0-9]+', 1, level) a frOM dual
CONNECT BY level<=regexp_count(c.amount,'~~')+1)
loop
array_number.extend();
array_number (i) := rec.a;
i := i + 1;
end loop;
for y in 1..array_text.count loop
dbms_output.put_line (c.order_no || ' ' || array_text(y) || ' ' || array_number(y));
end loop;
END LOOP;
end;
结果如下：
1 Briant 23
1 Luck 2122
2 Mike 10
2 Lee 200
2 David 37
3 Stak 10
答案 4 :(得分：0)
此解决方案使用XML，CROSS APPLY＆amp; ROW_NUMBER用于解构'~~'分隔字段。
它不需要SQL Server 2016中的UDF或STRING_SPLIT函数。
-- Using a table variable for the test
declare @Orders table ([Order No] int, Customers varchar(30), Amount varchar(30));
insert into @Orders ([Order No], Customers, Amount) values
(1,'Briant~~Luck','23~~2122'),
(2,'Mike~~Lee~~David','10~~200~~37'),
(3,'Stak','100');
SELECT C.[Order No], C.Customer, A.Amount
FROM
(
SELECT
[Order No],
row_number() over (partition by [Order No] order by (select 1)) as rn,
Customers.Name.value('.', 'VARCHAR(100)') AS Customer
FROM (
SELECT [Order No], CAST ('<x>' + REPLACE(Customers, '~~', '</x><x>') + '</x>' AS XML) AS XCustomers
FROM @Orders
) AS Q1
CROSS APPLY Q1.XCustomers.nodes ('/x') AS Customers(Name)
) C
JOIN (
SELECT
[Order No],
row_number() over (partition by [Order No] order by (select 1)) as rn,
Amounts.Value.value('.', 'VARCHAR(100)') AS Amount
FROM (
SELECT [Order No], CAST ('<x>' + REPLACE(Amount, '~~', '</x><x>') + '</x>' AS XML) AS XAmounts
FROM @Orders
) AS Q1
CROSS APPLY Q1.XAmounts.nodes ('/x') AS Amounts(Value)
) A
ON (C.[Order No] = A.[Order No] AND C.RN = A.RN);
或者，如果您知道这些字符串中最多有3个值 然后可以使用PARSENAME的技巧：
SELECT [Order No],
PARSENAME(REPLACE(Customers, '~~', '.'), v.n) as Customer,
PARSENAME(REPLACE(Amount, '~~', '.'), v.n) as Amount
FROM @Orders t
JOIN (VALUES (1),(2),(3)) AS v(n)
ON v.n <= (len(Customers) - len(replace(Customers, '~~', ','))+1);