Question

我有两张桌子，X和Y：

mysql> select * from X;
+------+------+
| id   | colX |
+------+------+
| 2001 |    0 |
| 2002 |    0 |
| 2003 |    0 |
| 2004 |    0 |
| 2005 |    0 |
| 2006 |   10 |
| 2007 |   10 |
+------+------+

mysql> SELECT * FROM Y;
+------+------+-------+
| id   | colY | score |
+------+------+-------+
| 2001 |   10 |     3 |
| 2004 |    0 |    12 |
| 2005 |    0 |    15 |
| 2007 |    0 |     1 |
+------+------+-------+

我需要以下结果，但没有子查询：

mysql> SELECT * FROM X LEFT JOIN Y ON X.id = Y.id WHERE colX=0 AND X.id NOT IN (SELECT id FROM Y WHERE colY > 0);
+------+------+------+------+-------+
| id   | colX | id   | colY | score |
+------+------+------+------+-------+
| 2002 |    0 | NULL | NULL |  NULL |
| 2003 |    0 | NULL | NULL |  NULL |
| 2004 |    0 | 2004 |    0 |    12 |
| 2005 |    0 | 2005 |    0 |    15 |
+------+------+------+------+-------+

colX必须为0
如果表Y中存在id且colY = 0则给出相应的分数
如果表Y中没有id，则给出分数= NULL

我尝试了以下查询但是id 2001显示：

mysql> SELECT * FROM X LEFT JOIN Y ON X.id = Y.id AND Y.colY = 0 WHERE colX=0;
+------+------+------+------+-------+
| id   | colX | id   | colY | score |
+------+------+------+------+-------+
| 2001 |    0 | NULL | NULL |  NULL |
| 2002 |    0 | NULL | NULL |  NULL |
| 2003 |    0 | NULL | NULL |  NULL |
| 2004 |    0 | 2004 |    0 |    12 |
| 2005 |    0 | 2005 |    0 |    15 |
+------+------+------+------+-------+

Answer 1

我认为这是你想要的逻辑：

h264_nvenc

无子查询的条件左连接MySQL

1 个答案:

Demo