的Android SQLite的查询返回多个相同的列

Question

下面这个查询在Android的SQLite是返回同一列的倍数。例如，它返回“meet_id”（3）次。第一个（2）“meet_id”用正确的数字是正确的，而第三个是空的。我的“athlete_id”和“event_id”也是如此。我假定这是因为我的左手的加入，因为第三个表是空的，截至目前，但不应该只是如果返回一个列每列属性（1次相遇列多行？）：

Cursor cursorTemp = db.rawQuery("SELECT * FROM " + TABLE_MEETS + " LEFT JOIN " + TABLE_MEETS_ATHLETES + 
       " ON " + TABLE_MEETS + "." + COL_MEET_ID + " = " + TABLE_MEETS_ATHLETES + "." + COL_MEET_ID + 
       " LEFT JOIN " + TABLE_MEETS_ATHLETES_SPLITS + " ON " + TABLE_MEETS_ATHLETES + "." + COL_MEET_ID + 
       " = " + TABLE_MEETS_ATHLETES_SPLITS + "." + COL_MEET_ID + " AND " + TABLE_MEETS_ATHLETES + 
       "." + COL_ATHLETE_ID + " = " + TABLE_MEETS_ATHLETES_SPLITS + "." + COL_ATHLETE_ID + " AND " + 
       TABLE_MEETS_ATHLETES + "." + COL_EVENT_ID + " = " + TABLE_MEETS_ATHLETES_SPLITS + "." + 
       COL_EVENT_ID + " WHERE " + TABLE_MEETS + "." + COL_MEET_ID + " = ?", new String[]{String.valueOf(i)}); 

我理解这有点混乱，但我喜欢多个原始查询

而且，满足ID是在所述第一表和一个主键和复合在第二第三

Answer 1

在指令SELECT * FROM...，所述*符号包括所有列成果。由于多个连接的结果，一些列被重复。我的建议是用您感兴趣的列名替换*。

Answer 2

看来罗伯托是正确的。我在DB2中嘲笑了这一点，只要我能理解你正在尝试的东西

create table TABLE_MEETS (COL_MEET_ID int); 

insert into table_meets (COL_MEET_ID) values (1),(2),(3); 

create table TABLE_MEETS_ATHLETES (
COL_MEET_ID int, 
COL_ATHLETE_ID int, 
COL_EVENT_ID int); 

insert into TABLE_MEETS_ATHLETES (COL_MEET_ID,COL_ATHLETE_ID,COL_EVENT_ID) values 
(1,10,100), 
(2,10,101), 
(1,20,100), 
(2,20,101); 

create table TABLE_MEETS_ATHLETES_SPLITS (
COL_MEET_ID int, 
COL_ATHLETE_ID int, 
COL_EVENT_ID int); 

insert into TABLE_MEETS_ATHLETES_SPLITS (COL_MEET_ID,COL_ATHLETE_ID,COL_EVENT_ID) values 
(1,10,100), 
(2,10,101), 
(1,20,100), 
(2,20,101); 

SELECT *将列出所有表中的列。

运行针对DB2：

SELECT * FROM TABLE_MEETS LEFT JOIN TABLE_MEETS_ATHLETES ON TABLE_MEETS.COL_MEET_ID=TABLE_MEETS_ATHLETES.COL_MEET_ID LEFT JOIN TABLE_MEETS_ATHLETES_SPLITS ON TABLE_MEETS_ATHLETES.COL_MEET_ID=TABLE_MEETS_ATHLETES_SPLITS.COL_MEET_ID AND TABLE_MEETS_ATHLETES.COL_ATHLETE_ID=TABLE_MEETS_ATHLETES_SPLITS.COL_ATHLETE_ID AND TABLE_MEETS_ATHLETES.COL_EVENT_ID=TABLE_MEETS_ATHLETES_SPLITS.COL_EVENT_ID WHERE TABLE_MEETS.COL_MEET_ID=1 

COL_MEET_ID COL_MEET_ID COL_ATHLETE_ID COL_EVENT_ID COL_MEET_ID COL_ATHLETE_ID选择COL_EVENT_ID

 1   1    10   100   1    10   100 
     1   1    20   100   1    20   100 

2结果。

但你probabaly只想从一个表上报给定列。如果你真的期望NULLS，你需要考虑使用哪一个。

SELECT MEETS.COL_MEET_ID,ATHLETES.COL_ATHLETE_ID,ATHLETES.COL_EVENT_ID 
FROM TABLE_MEETS MEETS LEFT JOIN TABLE_MEETS_ATHLETES ATHLETES 
ON MEETS.COL_MEET_ID=ATHLETES.COL_MEET_ID 
LEFT JOIN TABLE_MEETS_ATHLETES_SPLITS SPLITS 
ON ATHLETES.COL_MEET_ID=SPLITS.COL_MEET_ID AND 
ATHLETES.COL_ATHLETE_ID=SPLITS.COL_ATHLETE_ID AND 
ATHLETES.COL_EVENT_ID=SPLITS.COL_EVENT_ID WHERE MEETS.COL_MEET_ID=1 

COL_MEET_ID COL_ATHLETE_ID COL_EVENT_ID

 1    10   100 
     1    20   100 

2选择结果。