我试图在一个表的同一列上做多个LEFT JOIN。我需要将“table1.words”与“table1.color”和“table2.words”与“table1.food”联系起来。我该怎么做呢？我可以通过使左边的“table2.words”加入一个新列吗？
我的SQL代码：
SELECT table1.id, table1.color, table2.words
FROM table1
LEFT JOIN table2 ON table1.color=table2.id
LEFT JOIN table2 ON table1.food=table2.id
表1：
--------------------------------
| id | color | food |
--------------------------------
| 1 | 1 | 3 |
| 2 | 1 | 4 |
| 3 | 1 | 3 |
| 4 | 1 | 4 |
| 5 | 2 | 3 |
| 6 | 2 | 4 |
| 7 | 2 | 3 |
| 8 | 2 | 4 |
--------------------------------
表2：
---------------------------
| id | words |
---------------------------
| 1 | yellow |
| 2 | blue |
| 3 | cookies |
| 4 | milk |
---------------------------
我想要输出的内容：
----------------------------------------
| id | colorname | foodname |
----------------------------------------
| 1 | yellow | cookies |
| 2 | yellow | milk |
| 3 | yellow | cookies |
| 4 | yellow | milk |
| 5 | blue | cookies |
| 6 | blue | milk |
| 7 | blue | cookies |
| 8 | blue | milk |
----------------------------------------
注意：我无法更改表格结构。
答案 0 :(得分：22)
SELECT
table1.id,
table2_A.words colorname,
table2_B.words foodname
FROM table1
LEFT JOIN table2 table2_A ON table1.color=table2_A.id
LEFT JOIN table2 table2_B ON table1.food=table2_B.id;
您的样本数据
mysql> drop database if exists supercoolville;
Query OK, 2 rows affected (0.06 sec)
mysql> create database supercoolville;
Query OK, 1 row affected (0.00 sec)
mysql> use supercoolville;
Database changed
mysql> create table table1
-> (
-> id int not null auto_increment,
-> color int,
-> food int,
-> primary key (id)
-> );
Query OK, 0 rows affected (0.06 sec)
mysql> insert into table1 (color,food) values
-> (1,3),(1,4),(1,3),(1,4),
-> (2,3),(2,4),(2,3),(2,4);
Query OK, 8 rows affected (0.06 sec)
Records: 8 Duplicates: 0 Warnings: 0
mysql> create table table2
-> (
-> id int not null auto_increment,
-> words varchar(20),
-> primary key (id)
-> );
Query OK, 0 rows affected (0.05 sec)
mysql> insert into table2 (words) values
-> ('yellow'),('blue'),('cookies'),('milk');
Query OK, 4 rows affected (0.07 sec)
Records: 4 Duplicates: 0 Warnings: 0
mysql> select * from table1;
+----+-------+------+
| id | color | food |
+----+-------+------+
| 1 | 1 | 3 |
| 2 | 1 | 4 |
| 3 | 1 | 3 |
| 4 | 1 | 4 |
| 5 | 2 | 3 |
| 6 | 2 | 4 |
| 7 | 2 | 3 |
| 8 | 2 | 4 |
+----+-------+------+
8 rows in set (0.01 sec)
mysql> select * from table2;
+----+---------+
| id | words |
+----+---------+
| 1 | yellow |
| 2 | blue |
| 3 | cookies |
| 4 | milk |
+----+---------+
4 rows in set (0.00 sec)
我的查询结果
mysql> SELECT
-> table1.id,
-> table2_A.words colorname,
-> table2_B.words foodname
-> FROM table1
-> LEFT JOIN table2 table2_A ON table1.color=table2_A.id
-> LEFT JOIN table2 table2_B ON table1.food=table2_B.id
-> ;
+----+-----------+----------+
| id | colorname | foodname |
+----+-----------+----------+
| 1 | yellow | cookies |
| 2 | yellow | milk |
| 3 | yellow | cookies |
| 4 | yellow | milk |
| 5 | blue | cookies |
| 6 | blue | milk |
| 7 | blue | cookies |
| 8 | blue | milk |
+----+-----------+----------+
8 rows in set (0.00 sec)
mysql>
更新2012-05-14 19:10 EDT
如果食物或颜色的值不存在，则以下是调整后的查询：
SELECT
table1.id,
IFNULL(table2_A.words,'Unknown Color') colorname,
IFNULL(table2_B.words,'Unknown Food') foodname
FROM table1
LEFT JOIN table2 table2_A ON table1.color=table2_A.id
LEFT JOIN table2 table2_B ON table1.food=table2_B.id;
我将向table1添加行并运行此新查询
mysql> drop database if exists supercoolville;
Query OK, 2 rows affected (0.13 sec)
mysql> create database supercoolville;
Query OK, 1 row affected (0.00 sec)
mysql> use supercoolville;
Database changed
mysql> create table table1
-> (
-> id int not null auto_increment,
-> color int,
-> food int,
-> primary key (id)
-> );
Query OK, 0 rows affected (0.08 sec)
mysql> insert into table1 (color,food) values
-> (1,3),(1,4),(1,3),(1,4),
-> (2,3),(2,4),(2,3),(2,4),
-> (5,3),(5,4),(2,6),(2,8);
Query OK, 12 rows affected (0.07 sec)
Records: 12 Duplicates: 0 Warnings: 0
mysql> create table table2
-> (
-> id int not null auto_increment,
-> words varchar(20),
-> primary key (id)
-> );
Query OK, 0 rows affected (0.08 sec)
mysql> insert into table2 (words) values
-> ('yellow'),('blue'),('cookies'),('milk');
Query OK, 4 rows affected (0.06 sec)
Records: 4 Duplicates: 0 Warnings: 0
mysql> select * from table1;
+----+-------+------+
| id | color | food |
+----+-------+------+
| 1 | 1 | 3 |
| 2 | 1 | 4 |
| 3 | 1 | 3 |
| 4 | 1 | 4 |
| 5 | 2 | 3 |
| 6 | 2 | 4 |
| 7 | 2 | 3 |
| 8 | 2 | 4 |
| 9 | 5 | 3 |
| 10 | 5 | 4 |
| 11 | 2 | 6 |
| 12 | 2 | 8 |
+----+-------+------+
12 rows in set (0.00 sec)
mysql> select * from table2;
+----+---------+
| id | words |
+----+---------+
| 1 | yellow |
| 2 | blue |
| 3 | cookies |
| 4 | milk |
+----+---------+
4 rows in set (0.00 sec)
mysql> SELECT
-> table1.id,
-> IFNULL(table2_A.words,'Unknown Color') colorname,
-> IFNULL(table2_B.words,'Unknown Food') foodname
-> FROM table1
-> LEFT JOIN table2 table2_A ON table1.color=table2_A.id
-> LEFT JOIN table2 table2_B ON table1.food=table2_B.id;
+----+---------------+--------------+
| id | colorname | foodname |
+----+---------------+--------------+
| 1 | yellow | cookies |
| 2 | yellow | milk |
| 3 | yellow | cookies |
| 4 | yellow | milk |
| 5 | blue | cookies |
| 6 | blue | milk |
| 7 | blue | cookies |
| 8 | blue | milk |
| 9 | Unknown Color | cookies |
| 10 | Unknown Color | milk |
| 11 | blue | Unknown Food |
| 12 | blue | Unknown Food |
+----+---------------+--------------+
12 rows in set (0.00 sec)
mysql>
鉴于任何无效数据，仍然需要LEFT JOIN。
答案 1 :(得分：3)
尝试：
SELECT table1.id, colorcodes.words, foodcodes.words
FROM table1
LEFT JOIN table2 as colorcodes
ON colorcodes.id = table1.color
LEFT JOIN table2 as foodcodes
ON foodcodes.id= table1.food
答案 2 :(得分：3)
以下是查询：
SELECT a.id as id, b.words as colorname, c.words as foodname
FROM table1 a
LEFT JOIN table2 b ON b.id = a.color
LEFT JOIN table2 c ON c.id = a.food
注意：从您的数据中可以看出不需要LEFT JOIN。如果table1中没有颜色或食物为空的行，那么您可以不使用LEFT。