确定。我认为，我引用了THIS与我的情况类似的其他问题。但我会补充我的情况以及到目前为止我所尝试的内容。
Ticket Table
*-------------------------------*
|t_id | repair_id | customer_id |
| 1 | 10 | 11 |
| 2 | 11,12 | 12 |
| 3 | 13,14 | 13 |
| 4 | 15,16 | 14 |
*-------------------------------*
Repair Table
*-----------------------------------------------------*
|repair_id | ticket_id | device_type | repair_status |
| 10 | 1 | laptop | open |
| 11 | 2 | tablet | open |
| 12 | 2 | phone | open |
| 13 | 3 | phone | open |
| 14 | 3 | tablet | open |
| 15 | 4 | laptop | open |
| 16 | 4 | laptop | open |
*-----------------------------------------------------*
Customer Table
*-----------------------------*
|customer_id | f_name | l_name |
| 11 | John | Hughes |
| 12 | Julia | Brown |
| 13 | Tim | Duncan |
| 14 | Tony | Price |
*-----------------------------*
我希望结果表看起来像这样：
Results Table
*----------------------------------------------------------*
|customer_name | Repair_# | device_type | repair_status |
| John Hughes | 10 | laptop | open |
| Julia Brown | 11,12 | tablet,phone | open |
| Tim Duncan | 13,14 | phone, tablet | open |
| Tony Price | 15,16 | laptop, laptop | open |
*----------------------------------------------------------*
这是我尝试的查询：
$query = "SELECT customers.id, customers.f_name, customers.l_name, device_repairs.repair_id, device_repairs.ticket_id, ";
$query .= "device_repairs.device_type,device_repairs.repair_type, device_repairs.ticket_status, ";
$query .= "FROM tickets ";
$query .= "LEFT JOIN customers ON tickets.customer_id = customers.id ORDER BY device_repairs.ticket_id DESC";
我知道我的查询不对，但我不太确定如何设置INNER JOIN。
REVISION
所以经过一些审查。我已经添加了修改后的表格结构。
Ticket Table
*--------------------------------------------------------------------*
|t_id|repair_id|cust_id|cust_fname|cust_lname|device_type|repair_stat|
| 1 | 10 | 11 | John | Hughes | Laptop | open |
| 2 | 11 | 12 | Julia | Brown | Tablet | open |
| 2 | 12 | 12 | Julia | Brown | Phone | open |
*--------------------------------------------------------------------*
Repair Table
*-----------------------------------------------------*
|repair_id | ticket_id | device_type | repair_status |
| 10 | 1 | laptop | open |
| 11 | 2 | tablet | open |
| 12 | 2 | phone | open |
| 13 | 3 | phone | open |
| 14 | 3 | tablet | open |
| 15 | 4 | laptop | open |
| 16 | 4 | laptop | open |
*-----------------------------------------------------*
Customer Table
*-----------------------------*
|customer_id | f_name | l_name |
| 11 | John | Hughes |
| 12 | Julia | Brown |
| 13 | Tim | Duncan |
| 14 | Tony | Price |
*-----------------------------*
并且修订结果表看起来像这样：
Results Table
*----------------------------------------------------------*
|customer_name | Repair_# | device_type | repair_status |
| John Hughes | 10 | laptop | open |
| Julia Brown | 11 | tablet | open |
| Julia Brown | 12 | | phone | open |
*----------------------------------------------------------*
希望这有用并且更有意义。
我创建了这个查询，以便结果表输出正确。感谢您的所有投入。
$sql = "SELECT customers.id, customers.f_name, customers.l_name, device_repairs.repair_id, device_repairs.device_type, ";
$sql .="device_repairs.repair_type, device_repairs.ticket_status, device_repairs.ticketopen ";
$sql .="FROM customers, device_repairs WHERE customers.id = device_repairs.customer_id ORDER BY device_repairs.ticketopen DESC";
答案 0 :(得分：0)
票表设计不佳。我不会在一个字段中放置两个修复ID。拥有ID（主键）和外键的整个想法都将丢失。
您的票证表应该有7个而不是4个，其中的修复ID应该只有一个值。此外，repair_id应该被描述为外键并链接到Repair表。
在此之后，您可以在customer_id上将customer表连接到ticket表，并进一步加入ticket表上的修复表。