Question

有3张桌子。

items_table:
iid       
ititle  
itext   

tag_list_table
tid      
tname;  

tag_ref_table
iid      
tid      //foreign key for tag_list_table

tag_list_table中的每条记录都有一个或多个标记。

for example:
tag_list_table:
iid=1,itltle="this is title";itext="this is full text"; 
iid=2,itltle="this is title2";itext="this is full text2"; 

tag_list_table
tid=1 tname="red"
tid=2 tname="green"
tid=3 tname="yellow"
tid=4 tname="orange"

tag_ref_table
iid=1 tid=1
iid=1 tid=2
iid=1 tid=3
iid=2 tid=4

我希望得到这样的结果：

ROW1： 1- 这是标题这是全文红，绿，黄

ROW2： 2- 这是title2 这是全文2 橙

我试过这些：

SELECT i.ititle,i.itext,t.tname 
FROM tag_ref_table as i
LEFT JOIN tag_ref_table t WHERE tid=iid
LEFT JOIN tag_list_table r WHERE tid=??????....

Answer 1

您可以在选择中使用 CONCAT() 和 GROUP_CONCAT() ：

SELECT CONCAT(t1.iid,'- ',t1.ititle,' ',t1.itext,' ',t1.tname)  FROM
(SELECT i.iid,i.ititle,i.itext,GROUP_CONCAT(r.tname) tname
FROM items_table i
INNER JOIN tag_ref_table t ON i.iid = t.iid
INNER JOIN tag_list_table r ON t.tid = r.tid
GROUP BY i.ititle,i.itext,i.iid)t1;

您可以在以下网址看到完整的答案：

SQL Fiddle

Answer 2

您正在寻找GROUP_CONCAT

SELECT it.ititle, it.itext, GROUP_CONCAT(tlt.tname) 
FROM tag_ref_table as trt
INNER JOIN items_table it ON trt.iid = it.iid
INNER JOIN tag_list_table tlt ON trt.tid = tlt.tid
GROUP BY it.ititle, it.itext;

SqlFiddle here

除非您的tag_ref_table Junction表允许使用NULLS（在这样的表中不常见），否则我建议切换到INNER JOIN。

Answer 3

如果将“红色，绿色，黄色”值连接成一个字符串，在MySQL中你可以像这样使用GROUP_CONCAT函数：

SELECT i.*, group_concat(l.tname SEPARATOR ',')  as names
FROM items_table as i
JOIN tag_ref_table r ON r.iid=i.iid
JOIN tag_list_table l ON r.tid= l.tid
GROUP BY i.iid

文档：http://dev.mysql.com/doc/refman/5.7/en/group-by-functions.html#function_group-concat

更多示例：http://www.giombetti.com/2013/06/06/mysql-group_concat/

Answer 4

DROP TABLE items;

CREATE TABLE items
(item_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,item_title VARCHAR(20)
,item_text VARCHAR(20) NOT NULL
);

DROP TABLE tags;

CREATE TABLE tags
(tag_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,tag_name VARCHAR(12) NOT NULL
);  

DROP TABLE items_tags;

CREATE TABLE items_tags
(item_id INT NOT NULL
,tag_id INT NOT NULL
,PRIMARY KEY(item_id,tag_id)
);

INSERT INTO items VALUES
(1,"this is title1","this is full text1"),
(2,"this is title2","this is full text2"); 

INSERT INTO tags VALUES
(1,"red"),
(2,"green"),
(3,"yellow"),
(4,"orange");

INSERT INTO items_tags VALUES
(1,1),
(1,2),
(1,3),
(2,4);

mysql> SELECT * FROM items;
+---------+----------------+--------------------+
| item_id | item_title     | item_text          |
+---------+----------------+--------------------+
|       1 | this is title1 | this is full text1 |
|       2 | this is title2 | this is full text2 |
+---------+----------------+--------------------+
2 rows in set (0.00 sec)

SELECT * FROM tags;
+--------+----------+
| tag_id | tag_name |
+--------+----------+
|      1 | red      |
|      2 | green    |
|      3 | yellow   |
|      4 | orange   |
+--------+----------+

SELECT * FROM items_tags;
+---------+--------+
| item_id | tag_id |
+---------+--------+
|       1 |      1 |
|       1 |      2 |
|       1 |      3 |
|       2 |      4 |
+---------+--------+

SELECT i.*
     , GROUP_CONCAT(t.tag_name ORDER BY t.tag_id) tags 
  FROM items i 
  JOIN items_tags it 
    ON it.item_id = i.item_id 
  JOIN tags t 
    ON t.tag_id = it.tag_id 
 GROUP 
    BY i.item_id;
+---------+----------------+--------------------+------------------+
| item_id | item_title     | item_text          | tags             |
+---------+----------------+--------------------+------------------+
|       1 | this is title1 | this is full text1 | red,green,yellow |
|       2 | this is title2 | this is full text2 | orange           |
+---------+----------------+--------------------+------------------+

请注意，GROUP / GROUP_CONCAT部分在应用程序级别也可以执行（如果不是更好）（例如，一个简单的PHP循环作用于有序数组）

嵌套选择以在SQL中获取日期

4 个答案:

SQL Fiddle