警告：mysqli_connect(): (28000/1045): 拒绝访问答案

【问题标题】：Warning: mysqli_connect(): (28000/1045): Access denied for警告：mysqli_connect(): (28000/1045): 拒绝访问
【发布时间】：2016-09-27 09:09:21
【问题描述】：

我的网页上出现了两个错误。而且我不太习惯搞数据库，mysql，所以我需要你帮助我！

警告：mysqli_connect(): (28000/1045): Access denied for user 'u467224226_yjygu'@'10.1.2.19' (using password: YES) in /home/u467224226/public_html/galeria/api.php on line 8

警告：mysqli_error() 期望参数 1 为 mysqli，布尔值在 /home/u467224226/public_html/galeria/api.php 第 8 行给出

这是代码。

<?php // connect to the database $user_name = ""; $password = ""; $database = ""; $host_name = "localhost";

$con = mysqli_connect($host_name ,$user_name ,$password,$database) or die("Error " .       mysqli_error($con)); //check connection

echo "Connection opened";

mysql_close($con); ?>

<?php   require_once('include/config.inc.php');     header("Content-type: application/x-javascript");
    
    
            $connect = mysql_connect($CONFIG['dbserver'],$CONFIG['dbuser'],$CONFIG['dbpass'])       or die('Erro ao conectar ao servidor');     $connect_db = mysql_select_db($CONFIG['dbname'], $connect)      or die ('Erro ao conectar ao banco de dados');
            
                
                $resultado = mysql_query("SELECT * FROM `cpg_albums` ORDER BY `cpg_albums`.`aid` DESC LIMIT 0 , 4", $connect)

                    or die('Nenhum album encontrado com esta query');
                                        
                    echo 'document.write(\'';
                    
                    if(mysql_num_rows($resultado) == 0){
                        echo 'Nenhum Ã¡lbum cadastrado';
                    } else { echo '<section id="ultimas_fotos">  ';
                        while($row = mysql_fetch_array($resultado)){
                            echo ' ';
                                $album_id = $row['aid'];
                                $subresult = mysql_query("SELECT * FROM `cpg_pictures` where aid=$album_id order by pid DESC LIMIT 0, 20");
                                
                                if(mysql_num_rows($subresult) == 0){
                                    $album_img = "http://tisdalephotos.com/images/thumbs/thumb_nopic.png";
                                } else {
                                    while($subrow = mysql_fetch_array($subresult)){
                                        $album_img = "http://tisdalephotos.com/albums/".$subrow['filepath'].'thumb_'.$subrow['filename'] .$subrow['datebrowse'];
                                                                                
                                    }
                                }
                                
                                echo '<div id="ultalb"><a href="http://tisdalephotos.com/thumbnails.php?album='.$album_id.' "><img src="'.$album_img.'" alt="" /></a><div id="ultalbt"><a href="http://tisdalephotos.com/thumbnails.php?album='.$album_id.' ">+</a></div></div>';   
                                
                            echo '';
                        }
                    }

echo '</tr></section>';
                    echo '\');';
                            ?>

【问题讨论】：

你有什么想法吗？ SO 不仅仅是一个调试服务。
登录信息（密码）错误？
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标签： mysql database

【解决方案1】：

假设您的密码正确。

您的用户是否拥有正确的权限？以root身份登录然后运行：

select Host, User from mysql.user;

查看为您的用户列出的内容。

如果它丢失了，你可以运行：

CREATE USER 'u467224226_yjygu'@'%%' IDENTIFIED BY 'PASSWORD';
GRANT ALL PRIVILEGES ON * . * TO 'u467224226_yjygu'@'%%';
FLUSH PRIVILEGES;

【讨论】：