出于某种原因，当我添加到我的数据库时，这一行
$query = "INSERT INTO customers (id, name, address) VALUES (NULL,'$name', '$address')"; 添加客户两次。所以一个客户将同时拥有 id 1 和 id 2。随后获取 $last_id 将始终获得第二个 id 号，因此对于输入 id 1 和 2 的客户，$last_id == 2。我通过删除第一个副本来消除这个问题，但是我如何首先防止它发生呢？我正在使用 Chrome，但它也发生在 safari 中。
<?php
// Include the ShoppingCart class. Since the session contains a
// ShoppingCard object, this must be done before session_start().
require "../application/cart.php";
session_start();
?>
<!DOCTYPE html>
<?php
$orderErr = $nameErr = $addressErr = "";
$name = $address = "";
// If this session is just beginning, store an empty ShoppingCart in it.
if (!isset($_SESSION['cart'])) {
$_SESSION['cart'] = new ShoppingCart();
}
if (($_SESSION['cart']->count_order()) == 0) {
$orderErr = "There is nothing in the order, cannot checkout";
}
// sanitizing function
function test_input($data) {
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
if ($_SERVER["REQUEST_METHOD"] == "POST") {
if (empty($_POST["name"])) {
$nameErr = "Name is required";
}
else {
$name = test_input($_POST["name"]);
}
if (empty($_POST["address"])) {
$addressErr = "Address is required";
}
else {
$address = test_input($_POST["address"]);
}
//if there is an order, and name and address exist after sanaitizing,
// add to db
if ($name != "" && $address != "" && ($_SESSION['cart']->count_order()) != 0) {
require_once 'login.php';
$conn = new mysqli($hn, $un, $pw, $db);
if ($conn->connect_error) die($conn->connect_error);
//adding current customer
$query = "INSERT INTO customers (id, name, address) VALUES (NULL, '$name', '$address')";
echo "ok<br>";
$result = $conn->query($query);
if ($conn->query($query) === TRUE) {
echo "New record created successfully" . "<br>";
$last_id = $conn->insert_id;
echo $last_id . "<br>";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$delete_id = $last_id - 1;
// deleting duplicate
$query = "DELETE FROM customers WHERE id = '$delete_id'";
if ($conn->query($query) === TRUE) {
echo "duplicate deleted successfully" . "<br>";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
//send orders to db
$_SESSION['cart']->send_order_to_db($last_id, $conn);
$conn->close();
session_unset(); // remove all session variables
session_destroy();
}
}
?>
<html lang="en">
<head>
<title>Checkout</title>
<style>
.error {color: #FF0000;}
</style>
</head>
<body>
<h2>Checkout</h2>
<p>Here is your order: <?php
// Poor man's display of shopping cart
if (!isset($_SESSION['cart'])) {
$_SESSION['cart'] = new ShoppingCart();
}
$_SESSION['cart']->table();
?></p>
<span class="error"><?php echo $orderErr;?></span>
<h3> Checkout Form: </h3>
<p><span class="error">* required field.</span></p>
<form method = "post" >
Name: <input type="text" name="name" value="<?php echo $name;?>">
<span class="error">* <?php echo $nameErr;?></span>
<br><br>
Address: <input type="text" name="address" value="<?php echo $address;?>">
<span class="error">* <?php echo $nameErr;?></span>
<br><br>
<input type="submit" name="submit" value="Submit">
</form>
<p>Your credit card will be billed. Thanks for the order!</p>
<p><a href="index4.php">Shop some more!</a></p>
</body>
</html>
第二个问题是函数SOLVE。请参阅下面的编辑
public function send_order_to_db($last_id, $conn) {
foreach ($this->order as $variety => $quantity)
$query = "INSERT INTO orders (id, variety, quantity) VALUES" .
"('$last_id', '$variety', '$quantity')";
echo 'added ' . $quantity . ' ' . $variety;
$result = $conn->query($query);
if (!$result) echo "INSERT failed: $query<br>" .
$conn->error . "<br><br>";
}
我在输入客户信息后调用它，只添加数组 $this->order 中的最后一个元素。是因为 foreach 不像循环那样工作吗？那么我会使用 for 循环吗？还是跟数据库有关系？
编辑 我意识到我在 foreach 语句周围没有括号。现在我这样做了，它有效。添加所有具有相同 ID 的订单。
这是我的数据库:
/* To start with a fresh new database named store, we will delete one
* if one already exists:
*/
DROP DATABASE IF EXISTS store;
CREATE DATABASE store;
GRANT ALL PRIVILEGES ON store.* to user@localhost IDENTIFIED BY 'name';
USE store;
CREATE TABLE IF NOT EXISTS customers (
id int NOT NULL AUTO_INCREMENT,
name text NOT NULL,
address text NOT NULL,
PRIMARY KEY (id)
) ENGINE=InnoDB;
CREATE TABLE IF NOT EXISTS orders (
id int NOT NULL,
variety varchar(20) NOT NULL,
quantity int(11) NOT NULL,
PRIMARY KEY (id, variety),
FOREIGN KEY (id)
REFERENCES customers(id)
ON DELETE CASCADE
ON UPDATE CASCADE
) ENGINE=InnoDB;
最佳答案
问题1
您的第一个问题是记录被插入两次是因为您运行了两次查询:
$result = $conn->query($query);
if ($conn->query($query) === TRUE) {
您应该使用 $result 内容来检查它是否有效，而不是重新运行查询:
$result = $conn->query($query);
if ($result === true) {
问题2
记录只插入最后一条记录 - 这是因为你没有为你的 foreach 循环使用大括号，所以它只在循环中的控制结构之后运行第一行 - 其余的将仅在循环完成后执行。查看注释流程:
public function send_order_to_db($last_id, $conn) {
foreach ($this->order as $variety => $quantity)
// This line is run in the loop
$query = "INSERT INTO orders (id, variety, quantity) VALUES" .
"('$last_id', '$variety', '$quantity')";
// This line is only run once, after the foreach has finished
echo 'added ' . $quantity . ' ' . $variety;
$result = $conn->query($query);
if (!$result)
echo "INSERT failed: $query<br>" .
$conn->error . "<br><br>";
}
你只需要将内容用大括号括起来:
public function send_order_to_db($last_id, $conn) {
foreach ($this->order as $variety => $quantity) {
// This line is run in the loop
$query = "INSERT INTO orders (id, variety, quantity) VALUES" .
"('$last_id', '$variety', '$quantity')";
// So is all below now
echo 'added ' . $quantity . ' ' . $variety;
$result = $conn->query($query);
if (!$result) {
echo "INSERT failed: $query<br>" .
$conn->error . "<br><br>";
}
}
}
请注意，末尾 if 之后的大括号不是必需的，但最好始终使用大括号以提高可读性(并帮助您发现此类错误)。
关于php - 使用 PHP 添加到 MySQL 会添加两次并出现外部 ID 约束问题，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/29276535/