Question

我正在尝试为每个用户生成用户ID和关联的6位令牌。对于活动用户，值必须是唯一的。为了确保唯一性，我以递归方式使用相应的函数。

在页面加载时生成值。除了（大多数情况下）当我快速重新加载页面时，我的脚本应该工作它会产生错误。除非我删除mysqli_close($dbLink)语句。问题在于我关闭链接还是事实上函数是递归使用的，组合还是可能是完全不同的东西？

错误：

Warning: mysqli_prepare(): Couldn't fetch mysqli in C:\xampp\htdocs\project\functions.php on line 68

Warning: mysqli_stmt_bind_param() expects parameter 1 to be mysqli_stmt, null given in C:\xampp\htdocs\project\functions.php on line 69

Warning: mysqli_stmt_execute() expects parameter 1 to be mysqli_stmt, null given in     C:\xampp\htdocs\project\functions.php on line 70

Warning: mysqli_close(): Couldn't fetch mysqli in C:\xampp\htdocs\project\functions.php on line 71

代码：

$user_id = generateUserId();
$token = generateToken($user_id);

function generateUserId(){
    global $dbLink;

    $user_id = openssl_random_pseudo_bytes(128);
    $user_id = hash('md5', $user_id);

    $query = "SELECT * FROM users WHERE user_id = '{$user_id}' AND user_active = 1";
    $result_set = mysqli_query($dbLink, $query);
    if (mysqli_num_rows($result_set) != 0) {
        generateUserId();
    }

    return $user_id;
}

function generateToken($user_id){
    global $dbLink;

     //generate Token with different numbers
    $numbers = array(0, 1, 2, 3, 4, 5, 6, 7, 8, 9);
    $token = array();

    while (count($token) < 6) { 
        $number = array_rand($numbers);
        if (!in_array($number, $token)) {
            array_push($token, $number);
        }
    }

    $token = implode($token);

    //check if token exists
    $query = "SELECT * FROM users WHERE user_token = '{$token}' AND user_active = 1";
    $result_set = mysqli_query($dbLink, $query);
    if (mysqli_num_rows($result_set) != 0) {
        generateToken($user_id);
    }

    //save token and user id in database

    $query = "INSERT INTO users ";
    $query.= " (user_id, user_token)";
    $query.= " VALUES (?, ?)";
    $stmt = mysqli_prepare($dbLink, $query);
    mysqli_stmt_bind_param($stmt, 'ss', $user_id, $token);
    mysqli_stmt_execute($stmt);
    mysqli_close($dbLink);

    return $token;
}

Answer 1

问题在于关闭链接，你应该在调用函数后关闭它，而不是在里面。

顺便说一下，如果要在函数内部关闭它，请考虑在调用递归函数后不执行任何类型的sql语句或函数。

我会做的是提取此代码：

//save token and user id in database

$query = "INSERT INTO users ";
$query.= " (user_id, user_token)";
$query.= " VALUES (?, ?)";
$stmt = mysqli_prepare($dbLink, $query);
mysqli_stmt_bind_param($stmt, 'ss', $user_id, $token);
mysqli_stmt_execute($stmt);

在您的功能之外，这样您在关闭它之后就不会使用该链接。

另外......如果你的函数进行递归...让我说10次，插入该值10次是没有意义的，不是吗？

更简单的是，只需添加其他内容即可将其插入基础案例中：

function generateToken($user_id){
global $dbLink;

 //generate Token with different numbers
$numbers = array(0, 1, 2, 3, 4, 5, 6, 7, 8, 9);
$token = array();

while (count($token) < 6) { 
    $number = array_rand($numbers);
    if (!in_array($number, $token)) {
        array_push($token, $number);
    }
}

$token = implode($token);

//check if token exists
$query = "SELECT * FROM users WHERE user_token = '{$token}' AND user_active = 1";
$result_set = mysqli_query($dbLink, $query);
if (mysqli_num_rows($result_set) != 0) {
    generateToken($user_id);
}else{

  //save token and user id in database

  $query = "INSERT INTO users ";
  $query.= " (user_id, user_token)";
  $query.= " VALUES (?, ?)";
  $stmt = mysqli_prepare($dbLink, $query);
  mysqli_stmt_bind_param($stmt, 'ss', $user_id, $token);
  mysqli_stmt_execute($stmt);
  mysqli_close($dbLink);
}


return $token;

}

递归函数和mysqli关闭错误

1 个答案: