致命错误:当不在对象上下文时使用$this

$是PHP中的一个特殊变量，不能被赋值。如果在不存在该对象的上下文中访问该对象，则给出此致命错误。

此错误可能发生:

If a non-static method is called statically. Example: class Foo { protected $var; public function __construct($var) { $this->var = $var; } public static function bar () { // ^^^^^^ echo $this->var; // ^^^^^ } } Foo::bar(); How to fix: review your code again, $this can only be used in an object context, and should never be used in a static method. Also, a static method should not access the non-static property. Use self::$static_property to access the static property. If code from a class method has been copied over into a normal function or just the global scope and keeping the $this special variable. How to fix: Review the code and replace $this with a different substitution variable.

相关问题:

致命错误:当不在对象上下文中时使用$this 致命错误:当不在对象上下文中时使用$this 所有“使用$this时不在对象上下文”的问题在Stackoverflow

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。

致命错误:调用成员函数…在一个非物体上

发生在类似于xyz->method()的代码中，其中xyz不是对象，因此不能调用该方法。

这是一个致命错误，将停止脚本(向前兼容性注意:它将从PHP 7开始成为一个可捕获的错误)。

大多数情况下，这表明代码缺少对错误条件的检查。在调用一个对象的方法之前，验证它确实是一个对象。

一个典型的例子是

// ... some code using PDO
$statement = $pdo->prepare('invalid query', ...);
$statement->execute(...);

在上面的例子中，查询不能被准备，并且prepare()会将false赋值给$statement。尝试调用execute()方法将导致致命错误，因为false是一个“非对象”，因为该值是一个布尔值。

弄清楚为什么函数返回一个布尔值而不是一个对象。例如，检查$pdo对象中最近发生的错误。关于如何调试的细节将取决于如何处理特定函数/对象/类的错误。

如果->准备失败，那么你的$pdo数据库句柄对象没有被传递到当前作用域。找到它被定义的地方。然后将其作为参数传递，将其作为属性存储，或通过全局作用域共享。

另一个问题可能是有条件地创建一个对象，然后尝试在该条件块之外调用一个方法。例如

if ($someCondition) {
    $myObj = new MyObj();
}
// ...
$myObj->someMethod();

如果试图在条件块之外执行方法，则可能无法定义对象。

相关问题:

调用非对象上的成员函数 列出所有PHP“致命错误:调用成员函数…”关于Stackoverflow的问题

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");

"注意:未定义索引"或"警告:未定义数组键"

当您试图通过数组中不存在的键访问数组时发生。

一个典型的Undefined Index通知的例子是(demo)

$data = array('foo' => '42', 'bar');
echo $data['spinach'];
echo $data[1];

数组中菠菜和1都不存在，会触发E_NOTICE。在PHP 8.0中，这是一个E_WARNING。

解决方案是确保在访问索引之前存在索引或偏移量。这可能意味着您需要修复程序中的一个错误，以确保这些索引在您期望它们存在时确实存在。或者这可能意味着你需要使用array_key_exists或isset来测试索引是否存在:

$data = array('foo' => '42', 'bar');
if (array_key_exists('spinach', $data)) {
    echo $data['spinach'];
}
else {
    echo 'No key spinach in the array';
}

如果你有这样的代码:

<?php echo $_POST['message']; ?>
<form method="post" action="">
    <input type="text" name="message">
    ...

那么$_POST['message']将不会在这个页面第一次加载时设置，你将得到上述错误。只有在提交表单并再次运行此代码时，数组索引才会存在。你通常用以下方法检查:

if ($_POST)  ..  // if the $_POST array is not empty
// or
if ($_SERVER['REQUEST_METHOD'] == 'POST') ..  // page was requested with POST

相关问题:

参考文献:“Notice: Undefined variable”和“Notice: Undefined index” 所有关于Stackoverflow的PHP“注意:未定义的索引”问题 http://php.net/arrays

致命错误:超过XX秒的最大执行时间

每个PHP页面请求或脚本调用都度量PHP代码执行了多长时间。如果达到配置的限制，脚本将使用此消息中止。

注意，时间通常不包括发生在PHP“外部”的事情，例如等待数据库结果的时间，或使用shell_exec执行的外部程序等。例外情况是当PHP在Windows上运行时，测量的时间是“时钟时间”，并且确实包括这些外部等待时间。

常见的原因

无限循环。写错的while, do…while或for循环可能永远不会完成，这意味着PHP将永远持续运行。甚至foreach循环也可以是无限的，例如循环遍历一个Iterator对象或生成器函数。 庞大的数据集。即使您的循环不是无限的，如果它正在做很多工作，如果它正在处理大量的结果，它可能需要很长时间才能完成。

改变限制

如果你知道你有一个缓慢的过程，你可以配置时间限制:

在php.ini中，使用max_execution_time设置。 在脚本运行时，使用set_time_limit函数。注意，调用这个函数会将测量的时间重置为零，因此执行set_time_limit(10);意思是“再给10秒钟，不管脚本已经花了多长时间”。

这两种机制都应该给出一个秒数，或者特殊值0，表示“不限制”。设置“无限制”对于你真正想在后台永远运行的命令行脚本是最有用的;对于网页，最好设置一些有限值，即使它非常大，以防止代码中的错误导致整个系统失去响应。