0

我正在尝试学习 和 的JOIN区别LEFT JOIN。我看了几个 youtube 视频并研究了谷歌。但是，我不太明白为什么我的 JOIN 查询不起作用。如果我运行 sql 我得到Column 'user_id' in field list is ambiguous.

销售表

sales_id   user_id   amount    date         status
3          1         1258.32   2013-07-02   S

用户表

user_id   fname   lname
1         John    Doe

我的代码：

$top = mysqli_query($mysqli, "SELECT user_id, fname, lname, SUM(amount) as total 
FROM sales
LEFT JOIN users ON
user_id.sales = user_id.users
WHERE status = 'S' AND MONTH(date) = MONTH(CURDATE()) AND YEAR(date) = YEAR(CURDATE()) GROUP BY user_id ORDER BY total DESC LIMIT 5");

     while($row = mysqli_fetch_assoc($top)) {
           $topamount[]     = $row['total'];
           $topfnameuser[]  = $row['fname'];
           $toplnameuser[]  = $row['lname'];
     }
4

5 回答 5

2

由于两者sales和users都有一个名为 的列user_id，因此您必须在SELECT列表中具体说明您想要的内容：

SELECT sales.user_id as user_id, fname, lname, SUM(amount) as total 
FROM sales
LEFT JOIN users
ON sales.user_id = users.user_id
WHERE status = 'S' AND MONTH(date) = MONTH(CURDATE()) AND YEAR(date) = YEAR(CURDATE()) 
GROUP BY user_id
ORDER BY total DESC
LIMIT 5

您还在ON子句中将表名和列名放在后面。是table.column，不是column.table

于 2013-07-05T23:05:40.127 回答
2

我认为您对使用 LEFT JOIN 有误解，试试这个。

SELECT fname,lname, SUM(amount) AS total
  FROM users LEFT JOIN sales
  ON users.user_id = sales.user_id
    WHERE sales.status = 'S' AND MONTH(sales.date) = MONTH(CURDATE()) AND YEAR(sales.date) = YEAR(CURDATE()) GROUP BY users.user_id
ORDER BY total DESC LIMIT 5
于 2013-07-05T23:26:34.323 回答
1

专注于“尝试学习JOIN和LEFT JOIN”部分的差异，当您想要查看一个表中的所有行时，外连接非常有用，即使是在另一个相关表中没有对应行的行。

这是一个使用学生和考试成绩的简单示例。

create table students (student_id serial primary key, student_name text);
create table tests (test_id serial primary key, test_name text, test_date date);
create table grades (
    student_id int,
    test_id int,
    grade char,
    primary key (student_id, test_id),
    foreign key (student_id) references students (student_id),
    foreign key (test_id) references tests (test_id));

insert into students (student_name) values ('joe');
insert into students (student_name) values ('amber');
insert into students (student_name) values ('steve');

insert into tests (test_name, test_date) values ('test 1', '2013-01-20');
insert into tests (test_name, test_date) values ('test 2', '2013-02-10');

insert into grades (student_id, test_id, grade) values (1, 1, 'A');
insert into grades (student_id, test_id, grade) values (1, 2, 'B');
insert into grades (student_id, test_id, grade) values (2, 1, 'B');
insert into grades (student_id, test_id, grade) values (2, 2, 'A');

以下查询将返回学生及其成绩的列表，其中包括一行标识尚无任何成绩的学生（在本例中为 Steve）：

select s.student_name, t.test_name, t.test_date, g.grade
from students as s
left join grades as g on s.student_id = g.student_id
left join tests as t on g.test_id = t.test_id;

student_name | test_name | test_date  | grade 
--------------+-----------+------------+-------
joe          | test 1    | 2013-01-20 | A
joe          | test 2    | 2013-02-10 | B
amber        | test 1    | 2013-01-20 | B
amber        | test 1    | 2013-01-20 | A
steve        |           |            | 
(5 rows)

使用外连接的一种非常有用的方法是，当您只想查看相关表中没有匹配行的行时。您也可以使用子查询来执行此操作，但使用外连接并不是一个坏选择：

select s.student_name                                   
from students as s
left join grades as g on s.student_id = g.student_id
where g.student_id is null;
student_name 
--------------
steve
(1 row)

...在功能上等同于：

select student_name from students where student_id not in (select student_id from grades); 
student_name 
--------------
steve
(1 row)
于 2013-07-05T23:55:32.390 回答
0

JOIN当给定的 id 在您要加入的两个表中时使用。

LEFT JOIN当给定的 id 在一个表中并且可能在第二个表中时使用

在您的情况下，您应该从用户中选择并离开加入销售

SELECT u.user_id, u.fname, u.lname, SUM(s.amount) as total 
FROM users u LEFT JOIN sales s ON s.user_id = u.user_id
WHERE s.status = 'S' AND MONTH(s.date) = MONTH(CURDATE()) AND YEAR(s.date) = YEAR(CURDATE())
GROUP BY u.user_id
ORDER BY total DESC LIMIT 5

注意表命名中的别名（用户 u）。然后使用该别名，您可以指定要从哪个表中选择哪一列。

此外，我建议不要将列命名为“日期”，因为它是 MySQL 的保留字。见：http ://dev.mysql.com/doc/refman/5.6/en/reserved-words.html

于 2013-07-05T23:06:07.967 回答
0

引入表别名，以便在引用字段时可以引用它。您的问题是您user_id在两个表中都有，并且您尚未定义从哪个表读取 user_id。

"SELECT s.user_id, u.fname, u.lname, SUM(s.amount) as total 
FROM sales AS s
LEFT JOIN users AS u ON
s.user_id = u.user_id
于 2013-07-05T23:07:26.433 回答