PHP/MySQL : After inserting the value into a table the value of the variable gets lost and cannot print it out at the end of the file?

Question

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Answer 1

我尝试整理您的代码并删除多余的代码。

//Create User
$email = $_POST['email']; // you have to verify if this is an email or html etc.
$conectar = new mysqli(HOST, USER, PASS, DATABASE);

$query = "INSERT INTO usuarios 
        (userEmail) 
      VALUES 
        (?)";
$stmt = $conectar->prepare($query);
$stmt->bind_param('s',$email);
$stmt->execute();
$userID = $stmt->insert_id;

$stmt->close();//close statement

//assign a role to the newly created user

$query = "INSERT INTO rolesUsuarios 
        (userID, nombreRol) 
      VALUES 
        (?, 'registered')";
$stmt = $conectar->prepare($query);
$stmt->bind_param('i',$userID);
$stmt->execute();
$stmt->close();
$conectar->close();
echo $userID; //Here the content of $userID

首先，您不必为每个语句创建一个新的数据库连接。 第二:为了安全起见，请准备好您的声明。

如果$userID 为空，生成一个error_log($userID);在你 $userID 得到它的值后，如果它是空的，可能有其他错误。