如何连接表格以获得推荐的ID&数据?

时间:2014-03-07 23:16:55

标签: mysql

表1

TYPE | ID | PRICE | SIZE
 A      1    100     4
 A      2    150     8
 B      3    400     12

参考

FIELD_ID | LONG
    A        Llamas
    B        Alpacas
    C        Cats
    D        Dogs

我想要得到结果:

TYPE   | ID | PRICE | SIZE
llamas | 1  |  100  | 4
llamas | 2  |  150  | 8
Alpacas| 3  |  400  | 12

我尝试过使用

SELECT TYPE, ID, PRICE, SIZE, Reference.FIELD_ID, Reference.LONG
FROM Table1
LEFT JOIN Reference on TYPE = Reference.FIELD_ID
WHERE (PRICE BETWEEN '0' AND '200') AND
      (SIZE BETWEEN '1' AND '10')

仅返回[“TYPE”] => “A”和[“FIELD_ID”] => NULL

0 个答案:

没有答案