我正在申请人们出售物品并提供产品报价。如果用户查看其产品报价,则必须包含每个报价信息以及该报价的每个用户的详细信息。首先,通过ajax onclick函数激活要查询的数据库。 (我知道了)其次,在OFFERS表中查询状态为'o'且PID等于$_GET['pid']
(5)的商品。
OFFERS
------------------------------
| uid | pid | status | price |
------------------------------
| 5 | 5 | o | 49 |
| 7 | 5 | o | 45 |
------------------------------
然后根据每个要约查询USERS表以获取信息?此后,对于每个结果,根据UID,在users表中查询与用户有关的信息。
USERS
-------------------------------------
| uid | fname | lname | rating |
-------------------------------------
| 5 | John | Jan | 4.3 |
| 7 | Mark | Mull | 4.2 |
-------------------------------------
这是我对数据库的PHP语句的摘要。
$stmt = $conn->prepare('SELECT offerdate,price FROM offers WHERE pid=? AND status=?');
$o = 'o';
$stmt->bind_param('is', $_POST['pid'],$o);
$stmt->execute();
$stmt->bind_result($offerdate,$price);
我需要的另一条陈述
$stmt = $conn->prepare('SELECT fname,lname,rating, FROM users WHERE uid=?');
$stmt->bind_param('i', );
$stmt->execute();
$stmt->bind_result($fname,$lname,$rating);
任何帮助都会很棒,谢谢!
执行连接两个表的单个查询。
SELECT offerdate, price, fname, lname, rating
FROM offers
INNER JOIN users ON users.uid = offers.uid
WHERE offers.pid = ? AND offers.status = ?
您可以通过联接在一个查询中获得该结果:
select
o.offerdate,
o.price
u.fname,
u.lname,
u.rating
from offers o
inner join users u on u.uid = o.uid
where o.pid = ? and o.status = ?