mysql，连接2个表，并检查2个条件

MySQL

table1:

+--------+------+
| listid | type |
+--------+------+
|    1   |  a   |
+--------+------+
|    2   |  a   |
+--------+------+
|    3   |  b   |
+--------+------+

table2:

+----+--------+------+
| id | listid | code |
+----+--------+------+
|  1 |    1   |  ax  |
+----+--------+------+
|  2 |    1   |  bx  |
+----+--------+------+
|  3 |    2   |  ax  |
+----+--------+------+
|  4 |    2   |  bx  |
+----+--------+------+
|  5 |    2   |  cx  |
+----+--------+------+
|  6 |    3   |  ax  |
+----+--------+------+
|  7 |    3   |  bx  |
+----+--------+------+

Task

In one query, I want to check if:

1) in the table table2 only "ax" & "bx" are listed as code

2) the type of the listid I get in 1) is "a" in the table table1

PHP

$a = mysql_query("SELECT t1.listid FROM table1 AS t1, table2 AS t2......");
$b = mysql_fetch_assoc($a);

if($b['listid'])
{
    echo $b['listid'];
}
else
{
    echo 'nothing found';
}

Output

listid = 1

listid = 2 is false, because "cx" is also included in table2

listid = 3 is false, because it has type "b" in table1

I hope this makes sense :)

SELECT  t1.listid
FROM    t1
WHERE   type = 'a'
        AND id NOT IN
        (
        SELECT  listid
        FROM    t2
        WHERE   code NOT IN ('ax', 'bx')
        )

This will also match records from t1 which have no corresponding records in t2 at all.