Question

我是PHP和SQL的新手。对于学校作业，我需要为用户创建一个表单来更新客户数据。但是，我注意到更新功能仅使用数据集中的最后一个ID更新客户。例如，如果我在数据集中有4个客户。我使用下拉框列出了客户ID。当我选择id，1~3时，它表示0行被更新。它只适用于我选择id，4。所以我只能更新第4个。有人可以看看我的代码，并给我一些关于问题是什么以及如何解决它的提示？谢谢！

这是“updatecustomer.php”代码：

    <?php
//Turn on error reporting
ini_set('display_errors', 'On');
//Connects to the database
$mysqli = new mysql(SERVER_NAME, USERNAME,PASSWORD, DATABASE);
if($mysqli->connect_errno){
    echo "Connection error " . $mysqli->connect_errno . " " . $mysqli->connect_error;
}


if(!($stmt = $mysqli->prepare("UPDATE customer SET fName=?, lName=?, email=?, phone_number=?, address_no=?, address_street1=?, 
    address_street2=?, address_city=?, address_state=?, address_zip=? WHERE customer_id = ?"))){
    echo "Prepare failed: "  . $stmt->errno . " " . $stmt->error;
}

if(!($stmt->bind_param("isssiissssi",$_POST['customer_id'],$_POST['fName'],$_POST['lName'],$_POST['email'],$_POST['phone_number'], $_POST['address_no'],
    $_POST['address_street1'],$_POST['address_street2'],$_POST['address_city'],$_POST['address_state'], $_POST['address_zip']))){
    echo "Bind failed: "  . $stmt->errno . " " . $stmt->error;
}
if(!$stmt->execute()){
    echo "Execute failed: "  . $stmt->errno . " " . $stmt->error;
} else {
    echo "Updated " . $stmt->affected_rows . " rows to customer.";
}

?>

以下是我表单中代码的一部分：

<div>
            <form method="post" action="updatecustomer.php">
                <fieldset>
                    <legend>Update Existing Customer</legend>
                    <li>Customer ID:
                        <select name="customer_id">
        <?php
        if(!($stmt = $mysqli->prepare("SELECT customer_id, customer_id FROM customer"))){
            echo "Prepare failed: "  . $stmt->errno . " " . $stmt->error;
        }

        if(!$stmt->execute()){
            echo "Execute failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;
        }
        if(!$stmt->bind_result($customer_id, $customer_id)){
            echo "Bind failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;
        }
        while($stmt->fetch()){
            echo '<option value=" '. $customer_id . ' "> ' . $customer_id . '</option>\n';
        }
        $stmt->close();
        ?>
                        </select>
                    </li>
                    <li>First Name: <input type="text" name="fName"> Last Name: <input type="text" name="lName"</li>
                    <li>Email Address: <input type="text" name="email"></li>
                    <li>Phone Number: <input type="text" name="phone_number"></li>
                    <li>Street Number: <input type="text" name="address_no"> Street Line 1: <input type="text" name="address_street1"></li>
                    <li>Street Line 2 (Apt or Unit Number): <input type="text" name="address_street2"></li>
                    <li>City: <input type="text" name="address_city"> State: <input type="text" name="address_state"> Zip: <input type="text" name="address_zip"> </li>
                </fieldset>
                <input type="submit" name="update" value="Update Customer">
                <input type="submit" name="delete" value="Delete Customer">
        </div>

Answer 1

检查$stmt->bind_param行中变量的顺序。

Answer 2

试试这个：我已经使代码更整洁，更容易理解，并使用bind_param()方法重新排序了客户ID。

$stmt = $mysqli->prepare("
    UPDATE customer 
    SET fName=?, 
        lName=?, 
        email=?, 
        phone_number=?, 
        address_no=?, 
        address_street1=?, 
        address_street2=?, 
        address_city=?, 
        address_state=?, 
        address_zip=? 
     WHERE customer_id = ?
");

if(!$stmt){
    echo "Prepare failed: "  . $stmt->errno . " " . $stmt->error;
}

$paramBinding = $stmt->bind_param("sssiissssii",
                      $_POST['fName'], 
                      $_POST['lName'], 
                      $_POST['email'], 
                      $_POST['phone_number'], 
                      $_POST['address_no'],
                      $_POST['address_street1'], 
                      $_POST['address_street2'], 
                      $_POST['address_city'], 
                      $_POST['address_state'], 
                      $_POST['address_zip'],
                      $_POST['customer_id']
);

if(!$paramBinding){
    echo "Bind failed: "  . $stmt->errno . " " . $stmt->error;
}

修改然后循环选项：

foreach($stmt->fetch() AS $row){ echo '<option value="'. $row['customer_id'] . '"> ' . $row['customer_id'] . '</option>\n'; }

我应该如何解决这个mysqli准备声明？

2 个答案: