我有 3 个表，其结构如下:
福内西多
+-------------+--------------+----------+--------+
| codmaterial | idfornecedor | idmodelo | preco |
+-------------+--------------+----------+--------+
| 8978998 | 0 | 234 | 2.00 |
| 8978998 | 718 | 0 | 22.00 |
| 8978998 | 578 | 201 | 222.00 |
+-------------+--------------+----------+--------+
fornecedor
+--------------+---------------------------+---------+
| idfornecedor | nome | contact |
+--------------+---------------------------+---------+
| 578 | ACAR-IMPORT.EXPORT.LD. | NULL |
| 580 | ACCESS OFFICE VB | NULL |
| 581 | ACCIOP-CONST.OBR.PUB.,LD. | NULL |
| 582 | ACCUSONIC TECHNOLOGIES | NULL |
| 583 | ACE NAVILECTRA TOME | NULL |
| 584 | ACE-ART.CIRURG.EUROPA,LD. | NULL |
| 718 | ADIHOTEL-MAQ.EQUIP.HOT.LD | NULL |
+--------------+---------------------------+---------+
模型
+----------+-----------------+
| idmodelo | nome |
+----------+-----------------+
| 224 | 24TDMEJ50 |
| 140 | 259118 |
| 201 | 122AVP |
| 234 | 1YMB531002M0005 |
+----------+-----------------+
我想用 fornecedor 和 modelo 的名称而不是 ids 显示结果。
SELECT
fornecedor.nome, modelo.nome, preco
FROM
fornecido, fornecedor, modelo
WHERE
fornecido.codmaterial=8978998 AND condition
我尝试过的每个条件或导致所有可能的组合或唯一同时具有 fornecedor 和 modelo 的组合:s
我假装是这样的:
null, 1YMB531002M0005, 2.00
ADIHOTEL-MAQ.EQUIP.HOT.LD, null, 22.00
ACAR-IMPORT.EXPORT.LD., 122AVP, 222.00
谢谢:)
最佳答案
SELECT
fn.nome as FNOME, m.nome as MNOME, f.preco
FROM fornecido f left join fornecedor fn
on f.idfornecedor = fn.idfornecedor
left join modelo m on f.idmodelo =m.idmodelo
WHERE f.codmaterial=8978998
在此处检查 SQlFiddle http://sqlfiddle.com/#!2/251cc/5/0
供引用:
http://placeisimportant.files.wordpress.com/2013/02/sql-joins-vis-rep-1.png http://placeisimportant.files.wordpress.com/2013/02/sql-joins-vis-rep-2.png
关于mysql - 查询显示其中一个值为 0 的关系的结果，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/18334022/