php - 内连接不起作用

Question

大家好，请帮我解决这个从四个表中获取数据的查询。这是编写查询的正确方法吗

SELECT task.employee_id , task.user_id , task.service_id from task 
INNER JOIN employee  employee.name , employee.pic ON employee.pno =employee_id 
INNER JOIN user  user.name , user.pic ON user.pno = user_id 
INNER JOIN service  service.name , service.description ON service.service_id =service_id";

当我获取数据时，我将如何像我们一样显示它们 $a = $data['id'];

并且出现错误

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result,

现在我已经更新到此

function viewAll()
{


$this->query = "SELECT task.employee_id , task.user_id , task.service_id, employee.name , employee.pic, user.name , user.pic, service.name , service.description
FROM task  INNER JOIN employee  ON employee.pno = task.employee_id  INNER JOIN user  ON user.pno = employee.user_id  INNER JOIN service  ON service.service_id = user.service_id;";



$rd = $this->executeQuery();
$recordsArray = array(); // create a variable to hold the informationm
while (($row = mysqli_fetch_array($rd)) ){
$recordsArray[] = $row; // add the row in to the results (data) array
}

我收到此错误

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result,

Answer 1

不，这不正确。

SELECT task.employee_id , task.user_id , task.service_id, employee.name , employee.pic, user.name , user.pic, service.name , service.description
FROM
    task 
    INNER JOIN employee  ON employee.pno = task.employee_id 
    INNER JOIN user  ON user.pno = employee.user_id 
    INNER JOIN service  ON service.service_id = user.service_id;

我无法告诉你它到底应该是什么，但应该更接近。我根据原始查询对您的连接和列名称进行了最佳猜测。您想要“查看”的所有字段都需要位于 SELECT 子句中。当您加入另一张 table 时，您需要解释如何加入。例如，您要从 task 加入 employee，因此我假设 task 有一个名为 emmployee_id 的列(或者类似)这是您需要加入的领域。