Question

我正在使用Oracle 10g数据库。我正在试图弄清楚如何编写一个简单的SQL查询：

在86002895和86005197（含）之间的表格中找到缺失的数字，86002895和86005197之间有1955行。

例如：当前场景：table_1：

tracking_no | id_value
86002895 | 10
86002896 | 10
86002899 | 10
86002900 | 10
86002910 | 10
86005196 | 10
86005197 | 10

预期结果1：

“缺少tracking_id”，其中id_value 从table_1开始= 10;

86002897

86002898
86002900 到

86002910

86002910 到

86005196

提前致谢

Answer 1

with data as 
(
  select tracking_no from table_1 where id_value = 10
),  
data_n as 
(
  select level + (select min(tracking_no) from data) n 
  from dual 
  connect by level <= (select max(tracking_no) - min(tracking_no) from data) 
)
select * from data_n
where not exists (select 1 from data where tracking_no = n);

如果要包括86002895和86005197，请执行：

with data as 
(
  select tracking_no from table_1 
  where id_value = 10
  and   tracking_no between 86002895 and 86005197
),  
data_n as 
(
  select level + (select min(tracking_no) from data) n 
  from dual 
  connect by level <= (select max(tracking_no) - min(tracking_no) from data) 
)
select * from data_n
where not exists (select 1 from data where tracking_no = n);

Answer 2

试试光标？不是一个完整的解决方案...

declare
    V_IDX   number := 86002895;
begin
    for REC in (select   *
                from     TABLE_1
                order by TRACKING_NO asc)
    loop
        if V_IDX <> REC.TRACKING_NO then
            dbms_output.PUT_LINE('missing tracking_id '|| REC.TRACKING_NO || ' where id_value = ' || REC.ID_VALUE || ' from table_1');
        end if;
        V_IDX := V_IDX + 1;
    end loop;
end;

更新：我还不能添加评论，但除了彼得的答案，你可以动态制作数字表。例如，以下内容将返回86002895和86004849之间的所有数字：

select rownum+86002895-1
from dual
connect by level <= 1955

Answer 3

使用MINUS设置操作。

-- all numbers
SELECT ROWNUM
FROM dual
CONNECT BY level <= :SOME_LARGE_VALUE_HERE
MINUS
-- some numbers missing
SELECT id 
FROM table_1

根据需要调整。

Answer 4

使用model子句的解决方案：

select rangech
from
(
  select rangech
  from      table_1
  where  id_value = 10
  model
  dimension by (row_number() over (order by tracking_no) rn)
  measures (cast(null as varchar2(25)) rangech,tracking_no no)
  rules
  (
    rangech[any] = case
                   when no[cv()+1] is not null and no[cv()]+1 < no[cv()+1]-1
                        then to_char(no[cv()]+1)||'-'||to_char(no[cv()+1]-1)
                   when no[cv()+1] is not null and no[cv()]+1 = no[cv()+1]-1
                        then to_char(no[cv()]+1)
                   else
                        'X'
                   end
  )
)
where rangech <> 'X'
order by rangech;

输出：

RANGECH
-------------------------
86002897-86002898
86002901-86002909
86002911-86005195

Answer 5

如果你有一个数字表，你会这样做：

SELECT t.min_no+n.Number-1 AS missing_no
FROM Numbers n
INNER JOIN (
  SELECT MIN(tracking_no) AS min_no, MAX(tracking_no) AS max_no 
  FROM TABLE WHERE id_value = 10
  ) t ON n.Number BETWEEN 1 AND t.max_no-t.min_no+1
WHERE n.Number+t.min_no-1 NOT IN (
  SELECT tracking_no FROM TABLE
  WHERE id_value = 10
  );

数字表是一个包含一个整数列的表，数字从0或1到你需要的高度。

Answer 6

不是最优雅的解决方案，但它有效（在MySQL中 - 我不使用Oracle，所以我希望它适合你！）：

SELECT tracking_no
FROM yourtable
WHERE id_value = 10
AND tracking_no-1 NOT IN (SELECT tracking_no FROM yourtable WHERE id_value=10)

UNION

SELECT tracking_no
FROM yourtable
WHERE id_value = 10
AND tracking_no+1 NOT IN (SELECT tracking_no FROM yourtable WHERE id_value=10)

ORDER BY tracking_no

Answer 7

以下方法为您提供了所有缺失跟踪号的列表，但没有列出范围：

步骤1：找到tracking_no的最大值和最小值：M1和M2

步骤2：创建包含单列tracking_no

的临时表 TempNumbers

步骤3：将（M2 - M1）+ 1个唯一行插入 TempNumbers ，其值从M1到M2包含

第4步：

SELECT tracking_no FROM TempNumbers 
 WHERE NOT EXISTS (SELECT 'not found' FROM table_1 
                    WHERE TempNumbers.tracking_no = table_1.tracking_no)

Answer 8

这将返回每个缺失范围的起始端列表：

select s, e from
(select s, rownum sr
from
(
select tracking_no + 1 s
from table_1
where id_value = 10
MINUS
select tracking_no
from table_1
where id_value = 10
order by s
)),
(
select e, rownum er
from
(
select tracking_no - 1 e
from table_1
where id_value = 10
MINUS
select tracking_no
from table_1
where id_value = 10
order by e
))
where er-1 = sr;

Answer 9

 select  next_in_sequence missing_range_begin, 
    next_actual -1  missing_range_end
    from
    (
    select
    tracking_no, 
    tracking_no + 1 next_in_sequence,
    lead(tracking_no, 1) over (order by tracking_no) next_actual
    from test
    where id_value = 10
    order by tracking_no
    )
    where next_in_sequence <> next_actual
    ;

Answer 10

基本上使用86002894作为偏移量。然后它变成一个简单的查询。如果你的表非常大，你可以添加一个where子句。

我已经使用了ALL_OBJECTS，但是你可以使用任何有足够行数的表。


SELECT rownum+86002894
FROM All_Objects
WHERE rownum between 1 AND 86005197-86002895
MINUS
SELECT tracking_no
FROM your_table

查找表中缺少的序列

10 个答案: