LEFT JOIN AS yeni sütun?

Question

Birden çok LEFT JOIN'i bir tablodaki aynı sütunda yapmaya çalışıyorum. "Table1.color" ve "table2.words" ile birlikte "table2.words" LEFT JOIN "table2.words" ifadesine ihtiyacım var. Bunu nasıl yaparım? ve sola "table2.words" yeni bir sütun ekledikten sonra yapabilir miyim?

My SQL kodu:

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

Ne Im çıkışına çalışıyor:

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

Not: Ben masa yapılarını değiştirmek olamaz.

Answer 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; 

Kişisel Örnek Veri durumunda

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) 

My Query

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> 

GÜNCELLEME 2012-05-14 19:10 EDT arasında

Sonuçları

değerleri vardır Mevcut olmayan gıda veya renk için, ayarlanmış sorgu şu:

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; 

Ben tablo1 satır eklemek ve bu yeni bir sorgu

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> 

geçersiz veriler verildiğinde

, yine de ihtiyaç içinde LEFT JOIN çalışacaktır.

Answer 2

deneyin:

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 

Not: Burada

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 

Answer 3

sorgusu olan bir LEFT JOIN gerekli olmadığını Bu sizin verilerinden görünüyor. Tablo1'de renk veya yiyeceklerin boş olmadığı satırlar yoksa, LEFT'u bırakabilirsiniz.