Question

我正在尝试创建一个登录/注册系统，我不断收到错误：警告：mysql_query（）期望参数2是资源，在第8行的/home/roscapex/public_html/core/functions/users.php中给出整数

警告：mysql_result（）需要至少2个参数，第8行/home/roscapex/public_html/core/functions/users.php中给出的参数为1

我的代码是：

<?php
function user_exists($username) {

$username = sanitize($username);

$query = mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username'");

return (mysql_result(mysql_query($query, 0) == 1) ? true : false);

}
?>

注意这可能是重复但我尝试了所有其他相关问题。

Answer 1

您的代码中存在多个错误。让我注释一下：

<?php
function user_exists($username) {

$username = sanitize($username);

$query = mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username'"); // $query now contains the result of the query (a Resource)

// Here you try to run the resource *again* trough mysql_query, with with 2 parameters (which results in the warning). I am not quite sure what you intended.
return (mysql_result(mysql_query($query, 0) == 1) ? true : false);

}
?>

我修好了。

<?php
function user_exists($username) {
  $username = sanitize($username);  // This hopefully santizes the username. Better use mysql_real_escape_string($username)

  $query = "SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username'";

  $result = mysql_query($query) or die(mysql_error());  // Check if query was successful
  $row = mysql_fetch_array($result); // fetch the first row
  return ($row[0] > 1);  // If there is one or more users with this name, return true.

}
// omit the closing ?>, because it causes trouble, when there are any chars behind it (headers alreay sent error etc.)

警告：mysql_result（）期望参数1是资源，在第6行的/home/roscapex/public_html/core/functions/users.php中给出布尔值

1 个答案: