Question

我有两个表，其中包含有关道路建设活动的记录：

table_a是主列表。
table_b是旧列表。

对于每年的每条道路，我都希望从table_b中选择table_a中不存在的记录。

此外，记录应该不沿道路在空间上重叠。更具体地说，from_m中记录的to_m和table_b不应与from_m中的to_m和table_a重叠。

我该怎么做？我没有Oracle Spatial。

Excel中的数据（为了便于查看）：

这是Excel中数据的样子：

activeTab

绿色中的记录应由查询选择；红色中的记录不应该。

DDL：

表A：

  create table table_a 
   (
    id number(4,0), 
    road_id number(4,0), 
    year number(4,0), 
    from_m number(4,0), 
    to_m number(4,0)
   );

insert into table_a (id,road_id,year,from_m,to_m) values (1,1,2000,0,100);
insert into table_a (id,road_id,year,from_m,to_m) values (2,1,2005,0,25);
insert into table_a (id,road_id,year,from_m,to_m) values (3,1,2005,50,75);
insert into table_a (id,road_id,year,from_m,to_m) values (4,1,2005,75,100);
insert into table_a (id,road_id,year,from_m,to_m) values (5,1,2010,10,50);
insert into table_a (id,road_id,year,from_m,to_m) values (6,1,2010,50,90);
insert into table_a (id,road_id,year,from_m,to_m) values (7,1,2015,40,100);
insert into table_a (id,road_id,year,from_m,to_m) values (8,2,2020,0,40);
insert into table_a (id,road_id,year,from_m,to_m) values (9,2,2020,0,40);
insert into table_a (id,road_id,year,from_m,to_m) values (10,3,2025,90,150);
commit;

select * from table_a;

        ID    ROAD_ID       YEAR     FROM_M       TO_M
---------- ---------- ---------- ---------- ----------
         1          1       2000          0        100
         2          1       2005          0         25
         3          1       2005         50         75
         4          1       2005         75        100
         5          1       2010         10         50
         6          1       2010         50         90
         7          1       2015         40        100
         8          2       2020          0         40
         9          2       2020          0         40
        10          3       2025         90        150

表B：

  create table table_b 
   (
   id number(4,0), 
    road_id number(4,0), 
    year number(4,0), 
    from_m number(4,0), 
    to_m number(4,0)
   );

insert into table_b (id,road_id,year,from_m,to_m) values (1,1,1995,0,100);
insert into table_b (id,road_id,year,from_m,to_m) values (2,1,2001,0,50);
insert into table_b (id,road_id,year,from_m,to_m) values (3,1,2005,20,80);
insert into table_b (id,road_id,year,from_m,to_m) values (4,1,2005,0,100);
insert into table_b (id,road_id,year,from_m,to_m) values (5,1,2010,0,10);
insert into table_b (id,road_id,year,from_m,to_m) values (6,1,2010,90,100);
insert into table_b (id,road_id,year,from_m,to_m) values (7,1,2010,5,85);
insert into table_b (id,road_id,year,from_m,to_m) values (8,1,2015,40,100);
insert into table_b (id,road_id,year,from_m,to_m) values (9,1,2015,0,40);
insert into table_b (id,road_id,year,from_m,to_m) values (10,2,2020,0,41);
insert into table_b (id,road_id,year,from_m,to_m) values (11,3,2025,155,200);
insert into table_b (id,road_id,year,from_m,to_m) values (12,3,2025,199,300);
insert into table_b (id,road_id,year,from_m,to_m) values (13,4,2024,5,355);
commit;

select * from table_b;

        ID    ROAD_ID       YEAR     FROM_M       TO_M
---------- ---------- ---------- ---------- ----------
         1          1       1995          0        100
         2          1       2001          0         50
         3          1       2005         20         80
         4          1       2005          0        100
         5          1       2010          0         10
         6          1       2010         90        100
         7          1       2010          5         85
         8          1       2015         40        100
         9          1       2015          0         40
        10          2       2020          0         41
        11          3       2025        155        200
        12          3       2025        199        300
        13          4       2024          5        355

Answer 1

“不存在”子选择可以在此处提供帮助

SELECT *
FROM table_b b
WHERE
    NOT EXISTS (SELECT *
                FROM table_a a
                WHERE
                    a.road_id = b.road_id AND
                    a.year = b.year AND
                    a.to_m > b.from_m AND
                    a.from_m < b.to_m)

让我们看一下重叠的范围（f = from，t = to）

a   -------------------f=======================t-----------------

b1a -----f=============t-----------------------------------------
b1b --f=============t--------------------------------------------

b2a -------------------------------------------f======t----------
b2b -----------------------------------------------f======t------

b3  ---------------f=========t-----------------------------------
b4  ------------------------f===========t------------------------
b5  ---------------------------------------f===========t---------

范围b3，b4和b5重叠。对他们所有人来说，以下都是正确的

a.to > b.from && a.from < b.to

对于不重叠此条件的b1a，b1b和b2a，b2b，则为false。因此，对于b1a a.from == b.to，对于b1b a.from > b.to，条件a.from < b.to为false。

对于b2a a.to == b.from，对于b2b a.to < b.from，因此条件a.to > b.from为false。

诀窍是比较一个范围的from与另一个范围的to，反之亦然。

请参阅：http://sqlfiddle.com/#!4/85883/3/0

Answer 2

这将起作用：

SELECT * FROM table_b b
WHERE
    EXISTS (SELECT *
                FROM table_a a
                WHERE
                 a.year=b.year and a.from_m<>b.from_m and  a.to_m <>b.to_m );

选择数字范围不重叠的地方

2 个答案: