Question

我正在尝试了解JOIN和LEFT JOIN的区别。我观看了几个youtube视频并研究了Google。但是，我无法理解为什么我的JOIN查询不起作用。如果我运行sql，我会得到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'];
     }

Answer 1

由于sales和users都有一个名为user_id的列，因此您必须在SELECT列表中明确说明您想要的内容：

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

在ON子句中，您还有向后的表和列名称。它是table.column，而不是column.table

Answer 2

我认为您对LEFT JOIN的使用缺乏理解，试试这个.. 我已经创建了你想要的表，如果我使用你的查询并解决它，那么我有错误，所以像这样的查询

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

Answer 3

关注“试图了解JOIN和LEFT JOIN”部分的区别，当您想要查看一个表中的所有行时，外部联接非常有用，即使是那些没有相应行的行也是如此。另一张相关表格。

以下是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，包括一行标识没有任何成绩的学生（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)

使用外部联接的一种非常有用的方法是，当您想要查看仅行而不在相关表中匹配行时。您也可以使用子查询执行此操作，但使用外部联接并不是一个糟糕的选择：

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)

Answer 4

如果您要加入的两个表中都有给定的ID，则使用JOIN。

当给定的ID在一个表中并且可能在第二个

中时，使用LEFT JOIN

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，您可以指定从哪个表中选择哪个列。

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

Answer 5

引入表别名，以便在引用字段时引用它。您的问题是两个表中都有user_id，并且您没有定义从哪个表读取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

加入这些表的正确方法？

5 个答案: