PHP MySQLi输出问题

我的PHP脚本有一个问题，我想从数据库中输出“name”中的数据，但它不起作用

代码如下：

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "myDB";

$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$stmt = $conn->prepare("SELECT * FROM products WHERE id = ?");
$stmt->bind_param("s", $_GET['id']);

$stmt->execute();

$result = mysqli_fetch_assoc($stmt);

echo $result['name'];

错误是：

警告：mysqli_fetch_assoc（）希望参数1是mysqli_result，对象在第95行的C:\xampp\htdocs\p\cart\index.php中给出

你知道我做错了什么吗？如果能帮上忙，我们将不胜感激

感谢您的建议

使用

bind_result（）

.After

$stmt->execute（）

$stmt-execute（）

返回一个

mysql\u stmt

对象

mysqli\u fetch\u assoc（）

需要一个

mysql\u结果

对象作为参数，而不是

mysql\u stmt

对象

因此，要在准备好并执行查询后使用mysqli_fetch_assoc（），您必须使用

$stmt->get_result（）

将

mysql\u stmt

对象转换为

mysql\u结果

对象，如下所示

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "myDB";

$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$stmt = $conn->prepare("SELECT * FROM products WHERE id = ?");
$stmt->bind_param("s", $_GET['id']);

$stmt->execute();

$result = mysqli_fetch_assoc($stmt);    
$row = $result->fetch_assoc();

echo $result['name'];

您还可以使用

$stmt->bind_result（$name）

但要做到这一点，我们需要知道查询返回列的顺序。或者您需要使您的查询更加具体

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "myDB";

$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$stmt = $conn->prepare("SELECT name FROM products WHERE id = ?");
$stmt->bind_param("s", $_GET['id']);

$stmt->execute();

$stmt->bind_result($name);
$stmt->fetch();

echo 'Name is ' . $name;

---

您应该显示错误或查看服务器错误日志。您可以将mysqli设置为抛出异常，也可以在每次mysqli调用后检查错误。我的错误是警告：mysqli_fetch_assoc（）希望参数1是mysqli_result，对象在第95行的C:\xampp\htdocs\p\cart\index.php中给出

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "myDB";

$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$stmt = $conn->prepare("SELECT name FROM products WHERE id = ?");
$stmt->bind_param("s", $_GET['id']);

$stmt->execute();

$stmt->bind_result($name);
$stmt->fetch();

echo 'Name is ' . $name;