当我不尝试使用 store_ID 进行搜索时,我的查询有效。这是我所拥有的。<?php$search = $mysqli-> real_escape_string($_GET['store_ID']);$end = $mysqli-> real_escape_string($_GET['end']);$start = $mysqli-> real_escape_string($_GET['start']);$resultSet = $mysqli->query("SELECT * FROM orders INNER JOIN customers ON orders.store_ID=customers.store_ID WHERE order_date between '$start' and '$end' AND PurchaseMethod LIKE 'deferred"); if($resultSet-> num_rows>0){ while($rows = $resultSet->fetch_assoc()) { $storename= $rows['store_name']; $orderID = $rows['OrderID']; $storeID = $rows['store_ID']; $orderdate = $rows['order_date']; $purchaseMethod = $rows['PurchaseMethod']; $price = $rows['TotalPrice']; $output .= "<tr><td style='width:14.28%;'>$storename</td> <td style='width:14.28%;'>$orderID</td><td style='width:14.28%;'>$storeID</td> <td style='width:14.28%;'>$orderdate</td><td style='width:14.28%;'>$purchaseMethod</td> <td style='width:14.28%;'>$$price</td></tr>"; } }else{ echo var_dump($resultSet); $output = "No results"; } } ?> <?php echo $output; ?>这就是我想要开始工作的内容:$resultSet = $mysqli->query("SELECT * FROM orders INNER JOIN customers ON orders.store_ID=customers.store_ID WHERE order_date between '$start' and '$end' AND PurchaseMethod LIKE 'deferred' AND store_ID = '$search'");当我尝试添加AND store_ID = '$search'");它时,它给了我这个错误:注意:尝试在第 66 行的 customerreportC.php 中获取非对象的属性
1 回答
蝴蝶不菲
TA贡献1810条经验 获得超4个赞
由于 store_id 存在于两个表中,它会给出不明确的错误。尝试这个
$resultSet = $mysqli->query("SELECT * FROM orders
INNER JOIN customers
ON orders.store_ID=customers.store_ID
WHERE order_date between '$start' and '$end'
AND PurchaseMethod LIKE 'deferred' AND orders.store_ID = '$search'");
- 1 回答
- 0 关注
- 206 浏览
添加回答
举报
0/150
提交
取消