我的代码正在从另一个代码获取 ID，在获取该 ID 后，我会将其插入到另一个表中。问题是它不起作用，知道为什么吗？
<?php
session_start();
include("Connection.php");
if (isset($_POST['submit'])){
$name = $_POST['customerName'];
mysql_query("INSERT INTO `starbucks`.`orders` (
`ID` ,
`NAME` ,
`TOTAL_PRICE` ,
`TOTAL_ITEMS` ,
`TIME`
)
VALUES (
'' , '$name', '', '',NOW())");
$_SESSION['user'] = $name;
}
$dTime = time();
$myValue = isset($_REQUEST['dValue']) ?$_REQUEST['dValue'] : '';
echo "The time is: {$dTime}<br/>
The choice is {$myValue} ";
$sql = "Select * from product where NAME = '{$myValue}'";
$result = mysql_query($sql);
while ($row = mysql_fetch_assoc($result)){
$price = $row['PRICE'];
$id = $row['ID'];
echo $id;
$sql2 ="INSERT INTO starbucks`.order_details (ID, ORDER_ID, PRODUCT_ID, QTY) VALUES ('', '', '$id', '1')";
$result2 = mysql_query($sql2);
}
?>
最佳答案
插入中额外的反勾号，添加另一个或删除
关于php - 数据未存储在数据库中，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/7382109/