我从来没有这样的
PHP / MysqL技巧来加入multitables.请谁有此领域的经验帮助：
表TICKETS中的字段：
表TICKETS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
最后的结果我想达成：
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,但在某处错了：
$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");
您应该能够使用以下内容来获得结果：
原文链接：https://www.f2er.com/php/135001.htmlselect 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
结果：
| TICKETID | DATENEW | NAME | PAYMENT | TOTAL | --------------------------------------------------- | 29 | 2012-12-03 | John | cash | 22.57 | | 76 | 2012-12-03 | Helen | debt | 33.98 |
