MySQL从两个表中选择联接

Question

我有两个表“员工”和“家属”，

  Employees |employee_No| Employee_name | |1558 | Bean | |1557 | Juliet | |1556 | Zeke | Dependents |employee_No| dependent_name | relationship| |1558 | Kelvin | Son | |1558 | Mary | Daughter | |1556 | Janet | Spouse | 

有没有一种方法可以在一个MySQL语句中获取此数据并使用php显示，即循环雇员和该雇员下的从属，然后移至下一个雇员。

当前的PHP代码是

$employees = select_all_employees()
foreach ($employees as $covered){
  echo $covered['Employee_name'].'<br/>';
$get_dependent = $select_dependent($covered["employee_No "]);
 if($get_dependent != 0){
     foreach($get_dependent as $details){
      echo $details['dependent_name '].' '.$details['relationship'].'<br/>';
     }
  }
}

当有成千上万的员工和家属时，这会花费太多时间

预期结果

|employee_No| dependent_name | relationship|
--------------------------------------------
|1558       | Bean           | principal   |
|1558       | Kelvin         | Son         |
|1558       | Mary           | Daughter    |
|1557       | Juliet         | principal   |
|1556       | Zeke           | principal   |
|1556       | Janet          | Spouse      |

Answer 1

得到你想要的结果，最简单的方法是使用UNION中的行Employees表与JOIN的的Employees与他们的Dependents 。 我们将此UNION做为派生表，以便随后可以按employee_No排序结果，也可以将principal放在每个employee_No前面。 通过这种方式，您的PHP代码将成为所有结果的简单循环。

SELECT employee_No, Employee_name AS dependent_name, 'principal' AS relationship
FROM Employees
UNION ALL
SELECT e.employee_No, d.dependent_name, d.relationship
FROM Employees e
JOIN Dependents d on d.employee_No = e.employee_No
ORDER BY employee_No DESC, relationship = 'principal' DESC

输出：

employee_No dependent_name  relationship
1558        Bean            principal
1558        Mary            Daughter
1558        Kelvin          Son
1557        Juliet          principal
1556        Zeke            principal
1556        Janet           Spouse

dbfiddle上的演示

Answer 2

该查询如何：

SELECT a.*, "principal" as 'relationship' FROM Employees a UNION SELECT b.* FROM Dependents b ORDER BY employee_no DESC

DBFIDDLE在这里

Answer 3

您可以使用联接或映射列。 检查两个表中的索引以确保查询速度慢。 两个表中的索引都应为employee_No 。 然后循环查询输出​​。

<?php
    $qryOutput = array();
    $query = "SELECT A.*, B.* FROM Employees A LEFT JOIN Dependents B ON   A.employee_No=B.employee_No" // query

    $qryOutput = execute_Query($query); // Check this syntax. For Execute query


    // Loop through your Query Output
    foreach ($qryOutput as $key => $value) 
    {   echo $value["employee_No"]." ".$value["dependent_name"]." ".$value["relationship"]; }

 ?>

Answer 4

您可以使用INNER JOIN或LEFT JOIN来实现它。

尝试这样的事情。 虽然我没有测试代码。

您的表需要使用外键引用进行创建才能正常工作。 请参阅下面的示例。 我已经测试过了

create table Employees(employee_No int primary key,Employee_name varchar(100));

create table Dependents(employee_No int primary key,dependent_name varchar(100), relationship varchar(100)
 foreign key (employee_No) references Employees(employee_No));

插入进行测试

insert into Employees(employee_No,Employee_name) values(1558,'Bean');

insert into Employees(employee_No,Employee_name) values(1557,'Juliet');

insert into Dependents(employee_No,dependent_name,relationship) values(1558,'kevin','son');

码

<?php
$host = "localhost"; /* Host name */
$user = "root"; /* User */
$password = ""; /* Password */
$dbname = "your-db"; /* Database name */

$con = mysqli_connect($host, $user, $password,$dbname);
// Check connection
if (!$con) {
    die("Connection failed: " . mysqli_connect_error());
}

$query = 'SELECT Employees.employee_No, Employees.Employee_name, Dependents.employee_No, Dependents.dependent_name, 
Dependents.relationship FROM Employees
LEFT JOIN Dependents ON Employees.employee_No = Dependents.employee_No
ORDER BY Employees.Employee_name';

    $result = mysqli_query($con,$query);
    while($row = mysqli_fetch_assoc($result)){ 
  $empno = $row['employee_No'];
  $empname = $row['Employee_name'];
  $relation = $row['relationship'];
 $dependant_name = $row['dependent_name'];

//you can now echo

echo $dependant_name.' '.$relation.'<br/>';
}
?>