blocks|key|5881729|text|下面是使用mysqli连接数据库的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5881730|<?php
$servername+=+"localhost";
$username+=+"username";
$password+=+"password";
$dbname+=+"dbname";
$db+=+new+mysqli($servername,+$username,+$password,+$dbname);
?>|code-block|syntax|javascript|5881731|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is how you should connect your database using mysqli :
<pre><code>&lt;?php
$servername = &quot;localhost&quot;;
$username = &quot;username&quot;;
$password = &quot;password&quot;;
$dbname = &quot;dbname&quot;;
$db = new mysqli($servername, $username, $password, $dbname);
?&gt;
</code></pre>

blocks|key|1794282|text|您的密码可能无效。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1794283|编辑:在我看来，您实际上没有将参数传递给连接函数，您是否包含/导入了变量？|1794284|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

You may have an invalid password.

Edit: To me it looks like you're failing to actually pass parameters to the connection function, are you including / importing the variables?

blocks|key|1797416|text|你的DB_HOST就是你的mysql服务器ip+(大部分时间是本地主机)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1797417|您的DB_USER是Mysql的用户(大多数时间是根用户)|1797418|您的DB_PASS是Mysql用户的密码(大多数时间为无)|1797419|您的DB_NAME是您在Mysql服务器中的数据库名称。|1797420|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

your DB_HOST is your mysql server ip (most time is localhost)

your DB_USER is your User of Mysql (most time is root)

your DB_PASS is your Password of Mysql User (most time is none)

your DB_NAME is your name of your database in Mysql server

blocks|key|1794306|text|您在尝试连接到数据库时传递了无效的凭据(用户名和/或密码)。这里不能回答要传递什么，因为只有您(数据库管理员)知道现有的用户帐户。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1794307|要将用户帐户添加到MySQL数据库，请参阅MySQL+:：MySQL+5.7参考手册:：6.3.2添加用户帐户%5E。这需要以root用户身份连接，并提供root密码(如果已设置)。|1794308|第二个是第一个错误的结果，使用了错误的错误函数。对于mysqli_connect()，你必须调用PHP:+mysqli::$connect_error+-+Manual%5E而不是mysqli_error()错误:隐藏复制代码$connection+=+mysqli_connect+($host，$user，$password)或die('Not+connected+:+Ah+sh*t‘。mysqli_connect_error())；|1794309|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

You have passed invalid credentials (user name and/or password) when trying to connect to the database. What to pass can't be answered here because only you (the database administrator) would know about the existing user accounts.

To add an user account to a MySQL database see MySQL :: MySQL 5.7 Reference Manual :: 6.3.2 Adding User Accounts[^]. This requires connecting as root and providing the root password if one has been set.

The second is a result of the first and of using the wrong error function. For mysqli_connect(), you have to call PHP: mysqli::$connect_error - Manual[^] instead of mysqli_error() upon errors:
Hide Copy Code
$connection = mysqli_connect ($host, $user, $password) or die('Not connected : Ah sh*t ' . mysqli_connect_error());

blocks|key|1797434|text|这个问题很棘手。但是，我如何绕过它是很简单的。在你mysqli_connect(_,_,_,_)中按"localhost"，"user"，"password"，"database_name"的顺序排列它们；|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1797435|entityMap^0|P|N|1E|B|1Q|6|1X|A|28|F|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|G|$]]

This issue is tricky. But, how i went around it was straightforward. 
in your <code>mysqli_connect(_,_,_,_)</code> arrange them in this order of <code>"localhost"</code>, <code>"user"</code>, <code>"password"</code>, <code>"database_name"</code>;

blocks|key|1794344|text|将您的密码改回为空|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1794345|转到mysqladmin+root+(C:\xampp\mysql\bin)并键入mysqladmin+--user=root+--password=oldpassword+password"“|1794346|请求新密码时不要键入任何字符，只需按enter键并输入即可|1794347|viola刷新您的phpmyadmin页面，|1794348|通过命令窗口更改密码导致的问题，只需通过phpmyadmin页面更改密码即可避免此类错误，http://localhost/phpmyadmin|offset|length|1794349|entityMap|0|LINK|mutability|MUTABLE|url|http://localhost/phpmyadmin^0|0|0|0|0|19|R|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|V|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@$J|Y|K|Z|1|10]]|A|$]]|$1|L|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|S]]]]

change back your password to null 

go to mysqladmin root (C:\xampp\mysql\bin) and type 
mysqladmin --user=root --password=oldpassword password""

dont type any character when request new password, just push enter and enter

viola refresh your phpmyadmin page, 

the problem caused by changing password via command window, just change the password via phpmyadmin page to avoid such error, <a href="http://localhost/phpmyadmin" rel="nofollow noreferrer">http://localhost/phpmyadmin</a>

blocks|key|2353975|text|我在xampp+(32位)上安装OpenCart时也遇到了同样的问题。我正在完成我认为被请求的信息，而实际上，所有我被要求的都是数据库的名称。将所有字段保留为oriinal，只保留数据库的名称。保留密码为空。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2353976|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

I have had the same problem trying to install OpenCart on xampp (32bit). I was completing the information that I thought was being requested when in fact all I was being asked for was the name of the database. Leave all fields as oriinal just the name of your DB.
Leave password blank.

blocks|key|2353988|text|确保已完成此映射|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2353989|路径\xampp\phpMyAdmin\congig，Inc.|2353990|/*+Authentication+type+and+info+*/
$cfg['Servers'][$i]['auth_type']+=+'config';
$cfg['Servers'][$i]['user']+=+'root';
$cfg['Servers'][$i]['password']+=+YOUR_DEFAULT_PASSWORD;+|code-block|syntax|javascript|2353991|而你的连接文件|2353992|+<?php

$servername+=+"localhost";
$username+=+"root";
$password+=+YOUR_DEFAULT_PASSWORD;
$db+=+"ecommerece";


$con+=+mysqli_connect($servername,+$username,+$password,$db);|2353993|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

make sure this mapping is done
IN path\xampp\phpMyAdmin\congig,inc
<pre><code>/* Authentication type and info */
$cfg['Servers'][$i]['auth_type'] = 'config';
$cfg['Servers'][$i]['user'] = 'root';
$cfg['Servers'][$i]['password'] = YOUR_DEFAULT_PASSWORD; 
</code></pre>
and you connection file
<pre><code> &lt;?php

$servername = &quot;localhost&quot;;
$username = &quot;root&quot;;
$password = YOUR_DEFAULT_PASSWORD;
$db = &quot;ecommerece&quot;;


$con = mysqli_connect($servername, $username, $password,$db);
</code></pre>

my localhost is saying error like <code>Warning: mysqli::__construct(): (HY000/1045): Access denied for user 'root'@'localhost' (using password: YES) in C:\xampp\htdocs\registration\includes\db.php on line 3</code> on the browser I have all things done here's the image of the error in browser.

<a href="https://i.stack.imgur.com/4KDLH.jpg" rel="nofollow noreferrer">1</a>: <a href="https://i.stack.imgur.com/4KDLH.jpg" rel="nofollow noreferrer">here is the image which is facing the error</a>

and also if I am doing something wrong in my code editor then here's the image of the db.php <a href="https://i.stack.imgur.com/9oWqC.jpg" rel="nofollow noreferrer">db.php files code</a>

(HY000/1045) Access denied for user 'root'@'localhost'(using password: YES)

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我的本地主机说错误，就像浏览器上的Warning: mysqli::__construct(): (HY000/1045): Access denied for user 'root'@'localhost' (using password: YES) in C:\xampp\htdocs\registration\i...

问(HY000/1045)拒绝访问用户‘root’@‘localhost’(使用密码: YES)
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问(HY000/1045)拒绝访问用户‘root’@‘localhost’(使用密码: YES)EN