Php 联接表不返回任何结果

我在代码的另一部分成功地连接了表，所以我遵循了相同的模式，但这次它没有返回任何结果，我不明白为什么

以下是我尝试加入的表：

-角色

| id |  type  |
+----+--------+
|  1 | admin  |
|  2 | author |
|  3 | member |
+----+--------+

+----+----------+-------+----------+-------------------+--------+
| id | username | email | password |      role_id      | status |
+----+----------+-------+----------+-------------------+--------+
|    |          |       |          | 3 (default value) |        |
+----+----------+-------+----------+-------------------+--------+

-用户

| id |  type  |
+----+--------+
|  1 | admin  |
|  2 | author |
|  3 | member |
+----+--------+

+----+----------+-------+----------+-------------------+--------+
| id | username | email | password |      role_id      | status |
+----+----------+-------+----------+-------------------+--------+
|    |          |       |          | 3 (default value) |        |
+----+----------+-------+----------+-------------------+--------+

以下是请求：

public function get_users()
{
    $users_list = $this->dbh->query('SELECT users.id, roles.id AS roleid, type, role_id, id, username FROM users LEFT JOIN roles ON users.role_id = roles.id ORDER BY id ASC');
    return $users_list;
} 

SQL应该是

SELECT  
    u.id, 
    r.id AS roleid, 
    u.type,
    r.username
FROM users u
    LEFT JOIN roles r ON r.id = u.role_id
ORDER BY u.id ASC 

我省略了sql中定义的一些列，因为我根本看不出它们的意义。不仅如此，它们可能看起来模棱两可

例如：

SELECT 
    users.id,             // User ID
    roles.id AS roleid,   // Role ID
    type,                 // User type
    role_id,              // Role ID (We already have it, no need for it again)
    id,                   // What ID? This is ambiguous and will fail 
    username              // User Username

---

由于这两个表都有id列，因此需要指定

.id

public function get_users()
{
    $users_list = $this->dbh->query('SELECT users.id AS userid, roles.id AS roleid, type, role_id, username FROM users LEFT JOIN roles ON users.role_id = roles.id ORDER BY users.id ASC');
    return $users_list;
} 

像这样试试

public function get_users()
    {
        $users_list = $this->dbh->query('SELECT u.id AS userid, r.id AS roleid, r.type, u.role_id, u.username FROM users as u LEFT JOIN roles as r ON u.role_id = r.id ORDER BY u.id ASC');
        return $users_list;
    } 

---

尝试：

$users\u list=$this->dbh->query（'SELECT*FROM users AS users LEFT JOIN roles AS roles ON users.id=roles.id ORDER BY id ASC'）它仍然没有显示任何结果。下面是我遵循的模式
从表1中选择列名称左连接表2在表1上。列名称=表2。列名称但我猜我输入了一些错误的内容，我会检查againI是否尝试了您的解决方案，但它仍然返回一个空白页。我理解您对模棱两可的列的意思，我将对此进行返工first@CharlèneC从联接中的多个表中进行选择时，您应该始终使用表为所需的列添加前缀，因为这样可以避免此问题。不仅如此，您还可以确切地知道数据来自于哪个表更改了什么？答案是什么？对不起，我出人意料地离开了。你的解决方案有效！谢谢你的帮助，我以后会记住这一点。@CharlèneC，不客气。如果你不介意的话，请你也投票支持我的答案好吗？谢谢^_^