致命错误:未定义类常量

此错误意味着您试图使用不存在的类常量。与其他“未定义”的通知和警告不同，这是一个致命错误，将立即停止脚本。

首先要检查的是印刷错误。确认在类中定义了常量，并且使用适当的名称空间调用它。还要确认已包含用于解析常量的所有适当文件。

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T_PAAMAYIM_NEKUDOTAYIM

范围解析操作符也被称为“Paamayim Nekudotayim”，源自希伯来语פייםנקדיים”。意思是“双冒号”。

如果您无意中在代码中放入::，则通常会发生此错误。

相关问题:

参考:PHP语法错误;以及如何解决这些问题? PHP中两个冒号是什么意思? 在PHP中::(双冒号)和->(箭头)之间的区别是什么? 意外T_PAAMAYIM_NEKUDOTAYIM，期望T_NS_Separator

文档:

范围解析运算符(::)

警告:除以零

警告信息“除零”是新PHP开发人员最常被问到的问题之一。此错误不会导致异常，因此，一些开发人员偶尔会通过在表达式之前添加错误抑制操作符@来抑制警告。例如:

$value = @(2 / 0);

但是，就像任何警告一样，最好的方法是追踪警告的原因并解决它。警告的原因将来自任何实例，其中你试图除以0，一个变量等于0，或一个变量没有被分配(因为NULL == 0)，因为结果将是'undefined'。

要纠正此警告，您应该重写表达式以检查值是否为0，如果为0，则执行其他操作。如果值为0，要么不除法，要么将值改为1，然后再除法，这样除法的结果相当于只除以额外的变量。

if ( $var1 == 0 ) { // check if var1 equals zero
    $var1 = 1; // var1 equaled zero so change var1 to equal one instead
    $var3 = ($var2 / $var1); // divide var1/var2 ie. 1/1
} else {
    $var3 = ($var2 / $var1); // if var1 does not equal zero, divide
}

相关问题:

警告:除以零 使用PHP和MySQL 在WordPress主题中除以零错误 如何抑制“除零”错误 如何用0求除法?

严格标准:非静态方法[<class>::<method>]不应该被静态调用

当您试图在类上调用静态的非静态方法时发生，并且您在error_reporting()设置中也有E_STRICT标志。

例子:

class HTML {
   public function br() {
      echo '<br>';
   }
}

HTML::br()或$ HTML::br()

实际上，您可以通过不向error_reporting()添加E_STRICT来避免此错误，例如

error_reporting(E_ALL & ~E_STRICT);

因为对于PHP 5.4.0及以上版本，E_STRICT包含在E_ALL [ref]中。但这并不可取。解决方案是将你想要的静态函数定义为实际的静态函数:

public static function br() {
  echo '<br>';
}

或者按常规调用函数:

$html = new HTML();
$html->br();

相关问题:

我如何解决“非静态方法xxx:xxx()不应该在PHP 5.4中被静态调用?”

解析错误:语法错误，意外的T_ENCAPSED_AND_WHITESPACE

在PHP 8.0及以上版本中，该消息改为:

语法错误，意外的字符串内容“”，期望“-”或标识符或变量或数字

当试图在双引号字符串中引用带有引号键的数组值时，当整个复杂变量结构没有包含在{}中时，最常遇到此错误。

错误情况:

这将导致意外的T_ENCAPSED_AND_WHITESPACE:

echo "This is a double-quoted string with a quoted array key in $array['key']";
//---------------------------------------------------------------------^^^^^

可能的修复:

在双引号字符串中，PHP将允许数组关键字字符串不带引号使用，并且不会发出E_NOTICE。所以上面可以写成:

echo "This is a double-quoted string with an un-quoted array key in $array[key]";
//------------------------------------------------------------------------^^^^^

整个复杂数组变量和键可以被括在{}中，在这种情况下，它们应该被引用以避免E_NOTICE。PHP文档对于复杂变量推荐使用这种语法。

echo "This is a double-quoted string with a quoted array key in {$array['key']}";
//--------------------------------------------------------------^^^^^^^^^^^^^^^
// Or a complex array property of an object:
echo "This is a a double-quoted string with a complex {$object->property->array['key']}";

当然，上述任何一种方法的替代方法都是将数组变量连接到，而不是插值它:

echo "This is a double-quoted string with an array variable". $array['key'] . " concatenated inside.";
//----------------------------------------------------------^^^^^^^^^^^^^^^^^^^^^

如需参考，请参阅PHP字符串手册页中的变量解析部分