pl/sql：我该如何写这个查询？

Question

我有两列（初级）[PLAYER_ID] [LEAUGE_ID]

是这样的：

Player_id  League_id 
2139   8 
2153   8 
2302   10 
2441   8 
2441   10 

我试图找到谁在两个同盟

据起到同样的球员上面的表格，我试图只发现：

Player_id  League_id_1  League_id_2 
2441   8    10 

在此先感谢！

Answer 1

如果你不介意它在行：

SELECT t.* 
FROM myTable t 
INNER JOIN 
(
    SELECT Player_id 
    FROM myTable 
    GROUP BY Player_id 
    HAVING COUNT(*) = (SELECT COUNT(DISTINCT(League_id)) FROM myTable) 
) p ON t.Player_id = p.Player_id 

这应返回：

 
Player_id League_id 
2441   8 
2441   10 

SQL Fiddle

Answer 2

看来，你在一排想要这个。所以，你可以使用下面的使用PIVOT得到它在一排：

select player_id, 
    league_id_1, 
    league_id_2 
from 
(
    select t1.player_id, t1.league_id, 
    row_number() over(partition by t1.player_id order by t1.league_id) rn 
    from table1 t1 
    inner join 
    (
    select player_id 
    from table1 
    group by player_id 
    having count(distinct league_id) > 1 
) t2 
    on t1.player_id = t2.player_id 
) x 
pivot 
(
    max(league_id) 
    for rn in (1 as league_id_1, 2 as league_id_2) 
) p; 

见SQL Fiddle with Demo

，或者如果你没有访问PIVOT功能，您可以使用CASE语句汇总：

select player_id, 
    max(case when rn = 1 then league_id end) league_id_1, 
    max(case when rn = 2 then league_id end) league_id_2 
from 
(
    select t1.player_id, t1.league_id, 
    row_number() over(partition by t1.player_id order by t1.league_id) rn 
    from table1 t1 
    inner join 
    (
    select player_id 
    from table1 
    group by player_id 
    having count(distinct league_id) > 1 
) t2 
    on t1.player_id = t2.player_id 
) x 
group by player_id; 

见SQL Fiddle with Demo

如果你不希望它在一排，那么你可以使用内部子查询：

select t1.player_id, t1.league_id 
    from table1 t1 
    inner join 
    (
    select player_id 
    from table1 
    group by player_id 
    having count(distinct league_id) > 1 
) t2 
    on t1.player_id = t2.player_id 

见SQL Fiddle with Demo

Answer 3

如果您无法使用PIVOT（仅11克），并且要输出在一排所有League_id，试试这个：

  DECLARE 
       v_owner VARCHAR2 (40); 
       v_player_id VARCHAR2 (40); 
       v_league_id VARCHAR2 (100); 
       v_league_id_total VARCHAR2 (1000); 

       /* First cursor */ 
       CURSOR get_player_id 
       IS 
        SELECT DISTINCT player_id 
        FROM   table1 
        ORDER BY  player_id; 

       /* Second cursor */ 
       CURSOR get_league_id 
       IS 
        SELECT league_id 
        FROM  table1 
        WHERE player_id = v_player_id 
        ORDER BY league_id; 
      BEGIN 
       -- Open first cursor 
       OPEN get_player_id; 

       LOOP 
        FETCH get_player_id 
        INTO v_player_id; 

        v_league_id_total := ''; 
        EXIT WHEN get_player_id%NOTFOUND; 

        -- Open 2nd cursor 
        OPEN get_league_id; 

        LOOP 
        FETCH get_league_id 
        INTO v_league_id; 

        EXIT WHEN get_league_id%NOTFOUND; 
        v_league_id_total := v_league_id_total || ' , ' || v_league_id; 
        END LOOP; 

        DBMS_OUTPUT.put_line (RPAD (v_player_id, 
               26, 
               ' ' 
              ) || 
             RPAD (v_league_id_total, 
               26, 
               ' ' 
              )); 

        CLOSE get_league_id; 
       END LOOP; 

       CLOSE get_player_id; 
      EXCEPTION 
       WHEN OTHERS 
       THEN 
        raise_application_error (-20001, 
              'An error was encountered - ' || 
              SQLCODE || 
              ' -ERROR- ' || 
              SQLERRM); 
      END; 

输出将是这样的：

2139  , 8      
2153  , 8      
2303  , 8 , 10 (I added one more record.)    
2441  , 8 , 10 , 11 , 12 (I added 2 more records)