Question

您好我对SQL查询的新手我只知道简单的查询。我的问题是SELECT skipped check number EX2001,EX2002,EX2004可能 select result will show EX2003。提前谢谢，对不起我的英语。你只能回答算法，我试着把它实现到SQL。例： SELECT * FROM SETTLEMENT WHERE checkno not in（在ex2001和ex2900之间）这样可能吗？即时通讯使用MS SQL 2008。

Answer 1

如果您在查询中需要硬编码值，那么就可以这样做（因为您没有发布任何代码，所以这里是一个简单的查询）。您可以在sql中使用IN()或NOT IN()

select * from table where check_number not in ('EX2001','EX2002','EX2004' );

Answer 2

create table sequence(st varchar(50))

 insert into sequence values('EX2001');
 insert into sequence values('EX2002');
 insert into sequence values('EX2004');
 insert into sequence values('EX2005');
 insert into sequence values('EX2008');

假设您的原始表格名称为序列且只有一个字段，您可以根据需要进行修改

尝试以下

DECLARE @all TABLE
(
st varchar(20) 

)
declare @start int
declare @end int
declare @str varchar(20)
set @start=2000 //define starting point
set @end=2010 //define end point
while(@start<@end)
    BEGIN
        SET @start=@start+1
        set @str='EX'+cast(@start as varchar(20))
        INSERT INTO @all VALUES (''+@str+'') 
    END

SELECT * from @all 
except
select * from sequence

输出

st
EX2003
EX2006
EX2007
EX2009
EX2010

Answer 3

由于未提及SQL引擎，因此该答案仅适用于 Oracle 11G

选项1：使用分层查询

See Fiddle here

CREATE TABLE TEST1 ( A VARCHAR2 ( 9 ) );

INSERT INTO
      TEST1
VALUES
      ( 'EX2001' );

INSERT INTO
      TEST1
VALUES
      ( 'EX2002' );

INSERT INTO
      TEST1
VALUES
      ( 'EX2004' );

COMMIT;

WITH TEST2
    AS (SELECT
             TO_NUMBER(SUBSTR ( A,
                    3 ))
                 AS A
        FROM
             TEST1)
SELECT
        MIN_A
      - 1
      + LEVEL
FROM
      (SELECT
            MIN ( A ) MIN_A,
            MAX ( A ) MAX_A
       FROM
            TEST2)
CONNECT BY
      LEVEL <=   MAX_A
             - MIN_A
             + 1
MINUS
SELECT A FROM TEST2;

选项2：使用Oracle分析功能

See Fiddle here

WITH T
    AS (SELECT
             TO_NUMBER(SUBSTR ( A,
                    3 ))
                 AS SNO,
                 SYSDATE AS SDATE
        FROM
             TEST1)
SELECT
      SDATE,
      SNO
      + 1
          FIRST_MISSING,
      DECODE ( NEXT_SNO
             - 1,
             SNO
             + 1, TO_NUMBER ( NULL ),
             NEXT_SNO
             - 1 )
          LAST_MISSING
FROM
      (SELECT
            SDATE,
            SNO,
            LAG ( SNO )
                OVER ( PARTITION BY  SDATE 
                      ORDER BY SNO )
                LAST_SNO,
            LEAD ( SNO )
                OVER ( PARTITION BY SDATE
                      ORDER BY SNO )
                NEXT_SNO
       FROM
            T)
WHERE
      NVL ( NEXT_SNO,
           SNO
           + 1 ) <> SNO
                  + 1;

SQL查询知道跳过的数字

3 个答案: