我有一个问题，我的INSERT。。。ON DUPLICATE KEY UPDATE是插入一个新记录，而不是更新行，我使用的表既有主键也有唯一键。所以我很困惑为什么会发生这种事。
表
CREATE TABLE `Product` (
`Product_Id` bigint(255) NOT NULL AUTO_INCREMENT,
`Resturant_ID` bigint(255) NOT NULL,
`Product_Desc` text NOT NULL,
`Product_Name` varchar(100) NOT NULL,
`Product_Price` decimal(8,0) NOT NULL,
`Add_On_ID` int(11) NOT NULL,
PRIMARY KEY (`Product_Id`),
UNIQUE KEY `Product_Name` (`Product_Name`)
) ENGINE=InnoDB AUTO_INCREMENT=30 DEFAULT CHARSET=utf8
查询
$add_product_errors = array();
if (isset($_POST['add'])) {
$item_name = $_POST['item_name'];
$desc = $_POST['desc'];
$price = $_POST['price'];
$rest_id = mysqli_real_escape_string($dbc, $_SESSION['Resturant_ID']);
if (empty($_POST['price']) || !filter_var($_POST['price'], FILTER_VALIDATE_FLOAT) || ($_POST['price'] <= 0)) {
$add_product_errors['price'] = "Please enter a product price";
}
if (empty($_POST['item_name'])) {
$add_product_errors['item_name'] = "Please enter a name";
}
if (empty($_POST['desc'])) {
$add_product_errors['desc'] = "Please enter a product description";
}
$query = "INSERT INTO Product(Resturant_ID,Product_Name,Product_Desc,Product_Price) VALUES (?,?,?,?)
ON DUPLICATE KEY
UPDATE
Resturant_ID = VALUES(Resturant_ID)
,Product_Name = VALUES(Product_Name)
,Product_Desc = VALUES(Product_Desc)
,Product_Price = VALUES(Product_Price)";
$run_query = mysqli_prepare($dbc, $query);
if (!$run_query) {
die(mysqli_error($dbc));
}
mysqli_stmt_bind_param($run_query, 'sssd', $rest_id, $item_name, $desc, $price);
$execute = mysqli_stmt_execute($run_query);
$item_name = strip_tags($_POST['item_name']);
$desc = strip_tags($_POST['desc']);
//100 - changes the way the decimal displays in database
$price = strip_tags($_POST['price'] * 100);
if ($execute) {
echo "<script> alert('Addrrss Saved')</script>";
} else {
echo "<b>Oops! we have an issue </b>";
mysqli_stmt_close($run_query);
}
}
?>
语法对我来说很不错。也许可以先写SQL并在控制台或MySQL工作台中测试它，或者先做其他什么？试试这个：
$query = "INSERT INTO Product(Resturant_ID,Product_Name,Product_Desc,Product_Price) VALUES (?,?,?,?)
ON DUPLICATE KEY UPDATE
Resturant_ID = ?
,Product_Name = ?
,Product_Desc = ?
,Product_Price = ?";
$run_query = mysqli_prepare($dbc, $query);
if (!$run_query) {
die(mysqli_error($dbc));
}
mysqli_stmt_bind_param($run_query, 'issdissd', $rest_id, $item_name, $desc, $price, $rest_id, $item_name, $desc, $price);
或者?八次，绑定东西两次。。。不确定mysqli是否支持命名参数
根据Martin的反馈[再次]更新