PHP / MYSQL连接多个表

Question

我从来没有做过这样的PHP / MYSQL技巧来加入多表。请具有该领域经验的人员帮助： TICKETS 表中的字段：

ID TICKETID CUSTOMER
234   29      9798797
235   76      7887878

表 RECEPTS中的 字段：

ID   DATENEW    TOTAL
234 2012-12-03   22.57
235 2012-12-03   33.98

表 PAYMENTS中的 字段：

RECEIPT   PAYMENT
234       cash
235       debt

表 CUSTOMERS中的 字段：

ID            NAME
9798797       John
7887878       Helen

表之间的关系很容易理解： TICKETS.CUSTOMER=CUSTOMERS.ID;
PAYMENTS.RECEIPT=RECEIPTS.ID=TICKETS.ID

我希望达到的最终结果：

TICKETID DATENEW      NAME    PAYMENT TOTAL
29       2012-12-03   John     cash   22.57
76       2012-12-03   Helen    debt   33.98

我试图做这样的事情，但在某个地方出错：

$qry = mysql_query("Select TICKETS.TICKETID, RECEIPTS.DATENEW, PAYMENTS.TOTAL,  CUSTOMERS.NAME, PAYMENTS.PAYMENT FROM PEOPLE, RECEIPTS 
INNER JOIN TICKETS ON RECEIPTS.ID = TICKETS.ID
INNER JOIN CUSTOMERS ON TICKETS.CUSTOMER = CUSTOMERS.ID
ORDER BY RECEIPTS.DATENEW");

Answer 1

您应该可以使用以下方法获得结果：

select t.ticketid,
  date_format(r.datenew, '%Y-%m-%d') datenew,
  c.name,
  p.payment,
  r.total
from tickets t
left join RECEPTS r
  on t.id = r.id
left join CUSTOMERS c
  on t.customer = c.id
left join payments p 
  on t.id = p.RECEIPT
  and r.id = p.RECEIPT

参见带有演示的SQL Fiddle

结果：

| TICKETID |    DATENEW |  NAME | PAYMENT | TOTAL |
---------------------------------------------------
|       29 | 2012-12-03 |  John |    cash | 22.57 |
|       76 | 2012-12-03 | Helen |    debt | 33.98 |