Question

我收到了这个错误：

警告：mysql_fetch_assoc（）要求参数1为资源，布尔值在第70行的C：\ xampp \ htdocs \ core \ functions \ users.php中给出。

注册成功，登录成功;问题是它没有向我显示用户数据并给我这个错误。我是PHP的新手。

这是我的代码users.php：

  function user_data($user_id) {
        $data = array();
        $user_id = (int)$user_id;
        $func_num_args = func_num_args();
        $func_get_args = func_get_args();
    if ($func_num_args > 1) {
        unset($func_get_args[0]);
    $fields = '`' . implode ('`, `', $func_get_args) . '`';

    $data = mysql_fetch_assoc(mysql_query("SELECT $fields FROM `users` WHERE `user_id` = $user_id"));

    return $data;
        }
}
    function login($username, $password){
        $user_id = user_id_from_username($username);

        $username = sanitize($username);
        $password = md5($password);

        return (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username`='$username' AND `password`='$password'"), 0)==1) ? $user_id : false;
}
function logged_in() {
    return (isset($_SESSION['user_id'])) ? true : false;
}
function user_exists($username) {


       $username = sanitize($username);
      return (mysql_result( mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username` ='$username' "), 0) == 1) ? true : false;
      }
    function user_active($username) {
        $username = sanitize($username);
        return(mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username' AND `active` =1"), 0) == 1) ? true : false;
    }

我的register.php

 <?php
     include 'core/init.php'; 
    include 'overall/headerr.php'; 
    if (empty($_POST) === false) { 
    $required_fields = array('username', 'password', 'password_again', 'first_name');
    foreach($_POST as $key =>$value) {
    if(empty($value) && in_array($key, $required_fields) === true) {
        $errors[] = 'Fields marked with an asterisk required';
        break 1;
    }
    }
    if (empty($errors) === true){
        if (user_exists($_POST['username']) === true) {
            $errors[] = 'Sorry, the username \'' . $_POST['username'] . '\' is already taken.'; 
        }
    if (preg_match("/\\s/", $_POST['username'])== true) {
        $errors[]='Your username must not contain any spaces.';
    }
        if (strlen($_POST['password']) < 6 ) {
            $errors[] = 'Your password must be at least 6 characters';
        }
        if ($_POST['password'] !== $_POST['password_again']) {
                    $errors[] = 'Your password do not match';

        }

        }
        }

    ?>

    <h1>Register</h1>
    <?php 
    if (isset($_GET['success']) === true && empty ($_GET['success']) === false) {
        echo 'You\'ve been registered successfully!';
    }else{
    if (empty($_POST) === false && empty ($errors) === true) {
    $register_data = array(
    'username' => $_POST['username'],
    'password' => $_POST['password'],
    'first_name' => $_POST['first_name'],
    'contact' => $_POST['contact'],
    'information' => $_POST['information']

Answer 1

首先，您应该切换到msqli，因为msql在PHP 5.5.0中已弃用，并且正在PHP 7.0.0中删除

这是一个重复的问题，answer可以帮助您

<?PHP
$sql = 'some query...';
$result = mysql_query($q);

if (! $result){
    throw new My_Db_Exception('Database error: ' . mysql_error());
}

while($row = mysql_fetch_assoc($result)){
    //handle rows.
}

注意检查（！$ result） - 如果你的$ result是一个布尔值，它肯定是假的，这意味着有一个数据库错误，这意味着你的查询可能不好。 / p>

以下是来自msqli_fetch_assoc()

的php.net的一些额外资源

http://php.net/manual/en/mysqli-result.fetch-assoc.php

以下是msql_fetch_assoc()被弃用的信息

http://php.net/manual/en/function.mysql-fetch-assoc.php

警告：mysql_fetch_assoc（）期望参数1是资源，给定布尔值

1 个答案: