SQL中返回包含相同值/字符的不同行

我有一些棘手的情況。我有一列在表中的許多行中包含一組以管爲分隔符的數字。例如：SQL中返回包含相同值/字符的不同行

Courses 
------------------- 
1|2 
1|2|3 
1|2|8 
10 
11 
11|12

我想實現的是返回數字在輸出中只出現一次的行。

理想情況下，我想嘗試使用SQL執行此操作，而不必在Web應用程序級別執行檢查。執行DISTINCT並不能達到我想要的效果。

所需的輸出將是：

Courses 
------------------- 
1 
2 
3 
8 
10 
11 
12

我會理解，如果任何人都可以引導我在正確的方向。

謝謝。

來源

2013-05-21 sbhomra

，你會想到什麼作爲輸出？只有10？ 3,8,10和12？其他？ –

@sbhomra：給定樣本輸入的期望輸出是多少？ –

@RaviSingh我已經用期望的輸出更新了我的問題。 – sbhomra

請嘗試：

declare @tbl as table(Courses nvarchar(max)) 
insert into @tbl values 
('1|2'), 
('1|2|3'), 
('1|2|8'), 
('10'), 
('11'), 
('11|12') 

select * from @tbl 

SELECT 
    DISTINCT CAST(Split.a.value('.', 'VARCHAR(100)') AS INT) AS CVS 
FROM 
(
    SELECT CAST ('<M>' + REPLACE(Courses, '|', '</M><M>') + '</M>' AS XML) AS CVS 
    FROM @tbl 
) AS A CROSS APPLY CVS.nodes ('/M') AS Split(a) 
ORDER BY 1

來源

2013-05-21 09:42:04 TechDo

偉大的解決方案。但只是爲了我的SQL知識，你能解釋一行：「CAST（Split.a.value（'。'，'VARCHAR（100）'）AS INT）」。爲什麼Split.a.value需要在整個停止狀態下分割？只是好奇。 :-) – sbhomra

不能沒有遞歸管理:(像這樣的東西可以做的trich

WITH splitNum(num, r) 
AS 
(
    SELECT 
    SUBSTRING(<field>,1, CHARINDEX('|', <field>)-1) num, 
    SUBSTRING(<field>,CHARINDEX('|', <field>)+1, len(<field>)) r 
    FROM <yourtable> as a 
    UNION ALL 
    SELECT 
    SUBSTRING(r,1, CHARINDEX('|', r)-1) num, 
    SUBSTRING(r,CHARINDEX('|', r)+1, len(r)) r 
    FROM <yourtable> b 
    WHERE CHARINDEX('|', r) > 0 
    inner join splitNum as c on <whatevertheprimarykeyis> 
) 
SELECT distinct num FROM splitNum

沒能成功運行，但它應該做的伎倆，？只需更換和正確的信息

來源

2013-05-21 09:42:22 Davide

一種方法將使用遞歸CTE：

with cte as 
(select cast(case charindex('|',courses) when 0 then courses 
        else left(courses,charindex('|',courses)-1) end as int) course, 
     case charindex('|',courses) when 0 then '' 
      else right(courses,len(courses)-charindex('|',courses)) end courses 
from courses 
union all 
select cast(case charindex('|',courses) when 0 then courses 
        else left(courses,charindex('|',courses)-1) end as int) course, 
     case charindex('|',courses) when 0 then '' 
      else right(courses,len(courses)-charindex('|',courses)) end courses 
from cte 
where len(courses)>0) 
select distinct course from cte

SQLFiddle here。

來源

2013-05-21 09:56:45

嘗試這一個 -

SET NOCOUNT ON; 

DECLARE @temp TABLE 
(
     string VARCHAR(500) 
) 

DECLARE @Separator CHAR(1) 
SELECT @Separator = '|' 

INSERT INTO @temp (string) 
VALUES 
    ('1|2'), 
    ('1|2|3'), 
    ('1|2|8'), 
    ('10'), 
    ('11'), 
    ('11|12') 

-- 1. XML 

SELECT p.value('(./s)[1]', 'VARCHAR(500)') 
FROM (
    SELECT field = CAST('<r><s>' + REPLACE(t.string, @Separator, '</s></r><r><s>') + '</s></r>' AS XML) 
    FROM @temp t 
) d 
CROSS APPLY field.nodes('/r') t(p) 

-- 2. CTE 

;WITH a AS 
(
    SELECT 
      start_pos = 1 
     , end_pos = CHARINDEX(@Separator, t.string) 
     , t.string 
    FROM @temp t 

    UNION ALL 

    SELECT 
      end_pos + 1 
     , CHARINDEX(@Separator, string, end_pos + 1) 
     , string 
    FROM a 
    WHERE end_pos > 0 
) 
SELECT d.name 
FROM (
    SELECT 
      name = SUBSTRING(
       string 
      , start_pos 
      , ABS(end_pos - start_pos) 
     ) 
    FROM a 
) d 
WHERE d.name != ''

來源

2013-05-21 10:02:16 Devart

試試這個：

create table course (courses varchar(100)) 

insert into course values('1|2') 
insert into course values('1|2|3') 
insert into course values('1|2|8') 
insert into course values('10') 
insert into course values('11') 
insert into course values('11|12') 

Declare @col varchar(200) 

SELECT 
@col=(
     SELECT DISTINCT c.courses + '|' 
     FROM course c 

     FOR XML PATH('') 
    ); 




select * from course 

;with demo as(


select cast(substring(@col,1,charindex('|',@col,1)-1) AS INT) cou,charindex('|',@col,1) pos 

    union all 
    select cast(substring(@col,pos+1,charindex('|',@col,pos+1)-pos-1)AS INT) cou,charindex('|',@col,pos+1) pos 
    from demo where pos<LEN(@col)) 
select distinct cou from demo

來源

2013-05-21 10:13:57

SQL中返回包含相同值/字符的不同行

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