mysql_connect(): user 'name'@'host'拒绝访问

当您连接到无效或缺少凭据(用户名/密码)的MySQL/MariaDB服务器时，会出现此警告。所以这通常不是代码问题，而是服务器配置问题。

See the manual page on mysql_connect("localhost", "user", "pw") for examples. Check that you actually used a $username and $password. It's uncommon that you gain access using no password - which is what happened when the Warning: said (using password: NO). Only the local test server usually allows to connect with username root, no password, and the test database name. You can test if they're really correct using the command line client: mysql --user="username" --password="password" testdb Username and password are case-sensitive and whitespace is not ignored. If your password contains meta characters like $, escape them, or put the password in single quotes. Most shared hosting providers predeclare mysql accounts in relation to the unix user account (sometimes just prefixes or extra numeric suffixes). See the docs for a pattern or documentation, and CPanel or whatever interface for setting a password. See the MySQL manual on Adding user accounts using the command line. When connected as admin user you can issue a query like: CREATE USER 'username'@'localhost' IDENTIFIED BY 'newpassword'; Or use Adminer or WorkBench or any other graphical tool to create, check or correct account details. If you can't fix your credentials, then asking the internet to "please help" will have no effect. Only you and your hosting provider have permissions and sufficient access to diagnose and fix things. Verify that you could reach the database server, using the host name given by your provider: ping dbserver.hoster.example.net Check this from a SSH console directly on your webserver. Testing from your local development client to your shared hosting server is rarely meaningful. Often you just want the server name to be "localhost", which normally utilizes a local named socket when available. Othertimes you can try "127.0.0.1" as fallback. Should your MySQL/MariaDB server listen on a different port, then use "servername:3306". If that fails, then there's a perhaps a firewall issue. (Off-topic, not a programming question. No remote guess-helping possible.) When using constants like e.g. DB_USER or DB_PASSWORD, check that they're actually defined. If you get a "Warning: Access defined for 'DB_USER'@'host'" and a "Notice: use of undefined constant 'DB_PASS'", then that's your problem. Verify that your e.g. xy/db-config.php was actually included and whatelse. Check for correctly set GRANT permissions. It's not sufficient to have a username+password pair. Each MySQL/MariaDB account can have an attached set of permissions. Those can restrict which databases you are allowed to connect to, from which client/server the connection may originate from, and which queries are permitted. The "Access denied" warning thus may as well show up for mysql_query calls, if you don't have permissions to SELECT from a specific table, or INSERT/UPDATE, and more commonly DELETE anything. You can adapt account permissions when connected per command line client using the admin account with a query like: GRANT ALL ON yourdb.* TO 'username'@'localhost'; If the warning shows up first with Warning: mysql_query(): Access denied for user ''@'localhost' then you may have a php.ini-preconfigured account/password pair. Check that mysql.default_user= and mysql.default_password= have meaningful values. Oftentimes this is a provider-configuration. So contact their support for mismatches. Find the documentation of your shared hosting provider: e.g. HostGator, GoDaddy, 1and1, DigitalOcean, BlueHost, DreamHost, MediaTemple, ixWebhosting, lunarhosting, or just google yours´. Else consult your webhosting provider through their support channels. Note that you may also have depleted the available connection pool. You'll get access denied warnings for too many concurrent connections. (You have to investigate the setup. That's an off-topic server configuration issue, not a programming question.) Your libmysql client version may not be compatible with the database server. Normally MySQL and MariaDB servers can be reached with PHPs compiled in driver. If you have a custom setup, or an outdated PHP version, and a much newer database server, or significantly outdated one - then the version mismatch may prevent connections. (No, you have to investigate yourself. Nobody can guess your setup).

更多的引用:

mysql服务器错误:mysql用户无法访问root用户 mysql_connect():拒绝访问 mysql_select_db()拒绝访问用户" @'localhost'(使用密码:NO) PHPMyAdmin拒绝用户“root”@“localhost”的访问

顺便说一句，你可能不想再使用mysql_*函数了。新来者经常迁移到mysqli，然而这也一样乏味。相反，阅读PDO和准备好的语句。 $db =新PDO("mysql:主机=localhost;dbname=testdb"， "用户名"，"密码");

解析错误:语法错误，意外的“[”

这个错误有两种形式:

变化1

$arr = [1, 2, 3];

这个数组初始化式语法只在PHP 5.4中引入;它将在此之前的版本上引发解析器错误。如果可能的话，升级你的安装或使用旧的语法:

$arr = array(1, 2, 3);

请参见本手册中的示例。

变化2

$suffix = explode(',', 'foo,bar')[1];

PHP 5.4中还引入了数组解引用函数结果。如果无法升级，你需要使用一个(临时)变量:

$parts = explode(',', 'foo,bar');
$suffix = $parts[1];

请参见本手册中的示例。

致命错误:允许的内存大小XXX字节已耗尽(已尝试分配XXX字节)

没有足够的内存来运行脚本。PHP已达到内存限制并停止执行。这个错误是致命的，脚本停止。内存限制的值可以在php.ini文件中配置，也可以使用ini_set('memory_limit'， '128 M');在脚本中(将覆盖php.ini中定义的值)。内存限制的目的是防止一个PHP脚本占用所有可用内存，导致整个web服务器瘫痪。

The first thing to do is to minimise the amount of memory your script needs. For instance, if you're reading a large file into a variable or are fetching many records from a database and are storing them all in an array, that may use a lot of memory. Change your code to instead read the file line by line or fetch database records one at a time without storing them all in memory. This does require a bit of a conceptual awareness of what's going on behind the scenes and when data is stored in memory vs. elsewhere.

如果在脚本没有执行内存密集型工作时发生此错误，则需要检查代码以查看是否存在内存泄漏。memory_get_usage函数是您的朋友。

相关问题:

所有“Fatal error: Allowed memory size of XXX bytes exhausted”问题

致命错误:[TraitA]和[TraitB]在[ClassC]组合中定义相同的属性([$x])

当类试图使用多个trait时发生，其中两个或多个trait定义了同名的属性，且属性具有不同的初始值。

例子:

<?php
trait TraitA
{
    public $x = 'a';
}
trait TraitB
{
    public $x = 'b';
}
class ClassC
{
    use TraitA, TraitB;
}

问题:虽然可以解决相互竞争的方法之间的冲突，但目前没有语法可以解决两个相互竞争的属性之间的冲突。此时唯一的解决方案是重构;也就是说，避免属性名之间产生致命错误的冲突。

相关问题:

PHP特征:如何解决属性名称冲突? 特征-属性与父类冲突

注意:试图获取非对象错误的属性

在没有对象时试图访问对象的属性时发生。

非对象通知的一个典型例子是

$users = json_decode('[{"name": "hakre"}]');
echo $users->name; # Notice: Trying to get property of non-object

在本例中，$users是一个数组(因此不是一个对象)，它没有任何属性。

这类似于访问数组中不存在的索引或键(参见注意:未定义索引)。

这个例子简化了很多。大多数情况下，这样的通知表示一个未检查的返回值，例如，如果一个对象不存在，或者只是一个意外的非对象值(例如，在Xpath结果中，具有意外格式的JSON结构，具有意外格式的XML等)，库将返回NULL，但代码不会检查这样的条件。

由于这些非对象通常会被进一步处理，因此在对非对象调用对象方法时经常会发生致命错误(参见:致命错误:调用成员函数…在非对象上)停止脚本。

通过检查错误条件和/或变量是否符合期望，可以很容易地防止这种情况。下面是一个带有DOMXPath示例的通知:

$result  = $xpath->query("//*[@id='detail-sections']/div[1]");
$divText = $result->item(0)->nodeValue; # Notice: Trying to get property of non-object

问题是访问第一项的nodeValue属性(字段)，而没有检查它是否存在于$result集合中。相反，通过将变量分配给代码所操作的对象，可以使代码更加显式:

$result  = $xpath->query("//*[@id='detail-sections']/div[1]");
$div     = $result->item(0);
$divText = "-/-";
if (is_object($div)) {
    $divText = $div->nodeValue;
}
echo $divText;

相关的错误:

注意:未定义索引 致命错误:调用成员函数…在一个非物体上

警告:function()期望参数X为布尔值(或整数，字符串等)

如果传递给函数的参数类型错误(PHP不能自动转换)，就会抛出警告。此警告标识出哪个参数有问题，以及预期的数据类型。解决方案:将指定的参数更改为正确的数据类型。

例如下面的代码:

echo substr(["foo"], 23);

输出结果:

PHP警告:substr()期望参数1是字符串，数组给定