我越陷入mysql,我失去的越多,现在我已经完成了失败。 所以我有一些表格:
MainTable
id|message|name
-----------------
1 |test |OP
2 |test2 |jim
3 |test3 |ted
表1
id|likes
---------
2 | 1
3 | 0
表2
id|likes
---------
2 | 1
表3
id|likes
---------
1 | 1
2 | 1
3 | 0
我想要做的是为一列中的每个id获取likes
的总数(其中a等于1),这样我就可以计算一条消息的总喜欢数(具有各自的id)
直到现在我已经设法加入我的表格,所以我最后得到一个likes
列:
SELECT id,Table1.likes,Table2.likes,Table3.likes
FROM MainTable
LEFT JOIN Table1.id ON MainTable.id = Table1.id LEFT JOIN Table2.id ON MainTable.id = Table2.id LEFT JOIN Table3.id ON MainTable.id = Table3.id
首先,有可能吗?我知道我的代码不是很好,但至少它是一个开始!
谢谢!
答案 0 :(得分:2)
我猜你正在寻找其中一个...
SELECT
id
,SUM(Table1.likes)
,SUM(Table2.likes)
,SUM(Table3.likes)
FROM MainTable
LEFT JOIN Table1 ON MainTable.id = Table1.id
LEFT JOIN Table2 ON MainTable.id = Table2.id
LEFT JOIN Table3 ON MainTable.id = Table3.id
GROUP BY MainTable.id
SELECT
id
,SUM(Table1.likes)+SUM(Table2.likes)+SUM(Table3.likes)
FROM MainTable
LEFT JOIN Table1 ON MainTable.id = Table1.id
LEFT JOIN Table2 ON MainTable.id = Table2.id
LEFT JOIN Table3 ON MainTable.id = Table3.id
GROUP BY MainTable.id
答案 1 :(得分:1)
以下是查询,计算MainTable中每行的非零喜欢:
SELECT
MainTable.id,
MainTable.name,
MainTable.message,
COUNT(Table1.likes) + COUNT(Table2.likes)
+ COUNT(Table3.likes) AS n_likes
FROM
MainTable
LEFT JOIN
Table1 ON MainTable.id = Table1.id
AND
Table1.likes=1
LEFT JOIN
Table2 ON MainTable.id = Table2.id
AND
Table2.likes=1
LEFT JOIN
Table3
ON
MainTable.id = Table3.id
AND
Table3.likes=1
GROUP BY
MainTable.id;
请注意,您的SQL语法有错误:
LEFT JOIN Table1.id
在加入时,您必须编写没有列的表名:
LEFT JOIN Table1
答案 2 :(得分:0)
您可以使用UNION ALL解决此问题。
见下面的测试。
CREATE TABLE `test`.`items` (
`id` INT NOT NULL ,
`message` VARCHAR(45) NULL ,
`name` VARCHAR(45) NULL ,
PRIMARY KEY (`id`) );
insert into items values
(1, "Message 1", "OP"),
(1, "Message 2", "jim"),
(1, "Message 3", "ted");
CREATE TABLE `test`.`table1` (
`id` INT NOT NULL ,
`likes` INT(10),
PRIMARY KEY (`id`) );
CREATE TABLE `test`.`table2` (
`id` INT NOT NULL ,
`likes` INT(10),
PRIMARY KEY (`id`) );
CREATE TABLE `test`.`table3` (
`id` INT NOT NULL ,
`likes` INT(10),
PRIMARY KEY (`id`) );
insert into table1 values
( 2, 1),
( 3, 0);
insert into table2 values
( 2, 1),
( 3, 0);
insert into table3 values
(1,1),
(2,1),
(3,0);
select source.id, sum(likes)
from (
select id, likes from table1 as t1
UNION ALL
select id, likes from table2 as t2
UNION ALL
select id, likes from table3 as t3
) as source group by id;
输出
1, 1
2, 3
3, 0