您不需要在查询中传递连接。
解决方案:
$sql = "SELECT * FROM users WHERE email='$email'";
$response = $conn->query($sql);
while($res = $response->fetch_array()){
$name=$res['nameofuser']; //just an example
}
echo $name;
真正的解决方案(准备stmt):
$sql = "SELECT * FROM users WHERE email=?";
$response = $conn->prepare($sql);
$response->bind_param('s',$email);
if(!$response->execute()){
echo "Error query: " . $response->error . ".";
}
$result=$response->get_result();
while($res = $result->fetch_array()){
$name=$res['nameofuser']; //just an example
}
echo $name;
'提示'添加到真正的解决方案检查查询是否完成。