比较 i 和 i+1 元素 oracle

Question

我有table

CREATE table table_x 
(
        id NUMBER,  -- ID
        NUMBER_history NUMBER, --  consecutive number
        start_date date,    -- start_date  of next id=end_date of previous
        end_date date   -- should be >=start_date 
);
 INSERT INTO table_x VALUES  (14 , 1, '13-NOV-92' ,'14-NOV-92' );
 INSERT INTO table_x VALUES  (15 , 2, '14-NOV-92' ,'16-NOV-92' );
 INSERT INTO table_x VALUES  (19 , 3, '16-NOV-92' ,'18-NOV-92' );
 INSERT INTO table_x VALUES  (11 , 4, '18-NOV-92' ,'14-NOV-93' );
 INSERT INTO table_x VALUES  (17 , 5, '15-NOV-93' ,'16-NOV-93' );
 INSERT INTO table_x VALUES  (151 , 6, '16-NOV-93' ,'17-NOV-93' );
 INSERT INTO table_x VALUES  (139 , 7, '17-NOV-93' ,'18-NOV-93' );
 INSERT INTO table_x VALUES  (121 , 8, '19-NOV-93' ,'20-DEC-93' );
 INSERT INTO table_x VALUES  (822 , 9, '20-DEC-93' ,'20-DEC-93' );

我想写查询，我可以找到 where start_date of next row > end_date of previous.他们必须相等。

我尝试使用 NUMBER_history 作为计数器来做类似的事情。我按变量 i 组织循环并比较 i 和 i+1 (NUMBER_history 和 NUMBER_history+1)

的 C 方式

select * INTO row_tblx from table_x where NUMBER_history=NUMBER_history and end_date<(select start_date from table_x where NUMBER_history=NUMBER_history+1);

但我必须按 n_counter 从 1 到最后一个 NUMBER_history 值组织循环并将数据提取到多行中。我该怎么做？
我试试

set serveroutput on
DECLARE
CURSOR cur IS
      SELECT * FROM table_x;
TYPE row_tblx_type
IS
TABLE OF cur%ROWTYPE;
row_tblx row_tblx_type;

  rowx  cur%ROWTYPE;
  nh_count NUMBER;
BEGIN
FOR NUMBER_history IN cur LOOP
select * INTO row_tblx from table_x where NUMBER_history=NUMBER_history and end_date<(select start_date from table_x where NUMBER_history=NUMBER_history+1);
DBMS_OUTPUT.PUT_LINE (row_tblx(NUMBER_history).id);
END LOOP;
END;
/

如何使用 for 或另一个循环、多条记录或 table 条记录、游标、table 行作为计数器 (NUMBER_history)？没有光标怎么办？

Answer 1

您不需要 PL/SQL 或循环：

select *
from (
   select id, number_history, start_date, end_date, 
          lead(start_date) over (order by number_history) as next_start
   from table_x
) t
where next_start > end_date;

Answer 2

例如，这是一个不需要 PL/SQL 或 LEAD/LAG 函数的解决方案

select a.*, b.*
from table_x a
join table_x b
  on a.number_history+1 = b.number_history
where b.start_date > a.end_date

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