左加入为新列？

Posted 2023-02-24

【中文标题】左加入为新列？

【英文标题】：LEFT JOIN AS new column?

【发布时间】：2012-05-22 10:42:43

【问题描述】：

我试图在表的同一列上执行多个左连接。我需要将“table2.words”与“table1.color”和“table2.words”与“table1.food”进行左连接。我该怎么做呢？我可以通过将左连接的“table2.words”作为新列来做到这一点吗？

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 

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        |
--------------------------------

table2：

---------------------------
| id     | words          |
---------------------------
| 1      | yellow         |
| 2      | blue           |
| 3      | cookies        |
| 4      | milk           |
---------------------------

我想输出什么：

----------------------------------------
| 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：我无法更改表结构。

【问题讨论】：

从您提供的数据来看，这看起来可以通过直接的内连接来完成——真的需要左连接吗？

【参考方案1】：

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，这里是调整后的查询：

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 添加行并运行这个新查询

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。

【参考方案2】：

这里是查询：

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：从您的数据来看，LEFT JOIN 是不需要的。如果 table1 中没有 color 或 food 为 null 的行，那么您可以省略 LEFT。

【参考方案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