什么也看不见。页面是空白的，白色的。

也被称为白页死亡或白屏幕死亡。当错误报告被关闭并且发生致命错误(通常是语法错误)时，就会发生这种情况。

如果启用了错误日志记录，您将在错误日志中找到具体的错误消息。这通常在一个名为“php_errors.log”的文件中，要么在一个中心位置(例如在许多Linux环境中是/var/log/apache2)，要么在脚本本身的目录中(有时在共享主机环境中使用)。

有时，临时启用错误显示可能更直接。然后，白页将显示错误消息。要小心，因为这些错误对访问网站的每个人都是可见的。

这可以通过在脚本顶部添加以下PHP代码轻松完成:

ini_set('display_errors', 1); error_reporting(~0);

代码将打开错误显示并将报告设置为最高级别。

由于ini_set()是在运行时执行的，它对解析/语法错误没有影响。这些错误将出现在日志中。如果你想在输出中显示它们(例如在浏览器中)，你必须将display_startup_errors指令设置为true。在php.ini或.htaccess或任何其他在运行时之前影响配置的方法中执行此操作。

您可以使用相同的方法设置log_errors和error_log指令来选择您自己的日志文件位置。

查看日志或使用显示，您将得到更好的错误消息和脚本停止的代码行。

相关问题:

PHP的白屏死机 白屏死机! PHP不显示错误消息 PHP在错误时释放500 -在哪里有文档? 如何在PHP中获得有用的错误消息? 关于Stackoverflow的所有PHP“死亡白页”问题

相关的错误:

解析错误:语法错误，意外的T_XXX 致命错误:调用成员函数…在一个非物体上 代码不运行/我的PHP代码的某些部分被输出

致命错误:不能在写上下文中使用函数返回值

这通常发生在直接使用带有empty的函数时。

例子:

if (empty(is_null(null))) {
  echo 'empty';
}

这是因为empty是一个语言结构，而不是一个函数，在5.5之前的PHP版本中，它不能用表达式作为参数调用。在PHP 5.5之前，empty()的参数必须是变量，但在PHP 5.5+中允许任意表达式(例如函数的返回值)。

Empty，尽管它的名字，实际上并不检查一个变量是否为“空”。相反，它检查变量是否存在，或== false。表达式(如示例中的is_null(null))将始终被视为存在，因此这里的empty仅检查它是否等于false。你可以在这里用!替换empty()，例如if (!is_null(null))，或者显式地与false进行比较，例如if (is_null(null) == false)。

相关问题:

致命错误:不能使用函数返回值

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");

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

警告:字符串偏移量为XXX

当您尝试使用方括号语法访问数组元素时，就会发生这种情况，但您是在字符串上执行此操作，而不是在数组上执行此操作，因此该操作显然没有意义。

例子:

$var = "test";
echo $var["a_key"];

如果你认为变量应该是一个数组，看看它来自哪里并修复那里的问题。

MySQL:你的SQL语法有错误;检查手册，对应于您的MySQL服务器版本的正确语法使用近…排队……

这个错误通常是因为您忘记正确转义传递给MySQL查询的数据。

一个不应该做的事情(“坏主意”)的例子:

$query = "UPDATE `posts` SET my_text='{$_POST['text']}' WHERE id={$_GET['id']}";
mysqli_query($db, $query);

这段代码可以包含在一个表单提交的页面中，URL为http://example.com/edit.php?id=10(编辑帖子n°10)

如果提交的文本包含单引号会发生什么?$query将以:

$query = "UPDATE `posts` SET my_text='I'm a PHP newbie' WHERE id=10';

当这个查询被发送到MySQL时，它会抱怨语法错误，因为中间有一个额外的单引号。

为了避免这样的错误，在查询中使用数据之前，必须始终转义数据。

在SQL查询中使用数据之前转义数据也是非常重要的，因为如果您不这样做，您的脚本将对SQL注入开放。SQL注入可能导致记录、表或整个数据库的更改、丢失或修改。这是一个非常严重的安全问题!

文档:

如何防止PHP中的SQL注入? mysql_real_escape_string () mysqli_real_escape_string () 如何从“Bobby表”XKCD漫画的SQL注入工作? 绕过mysql_real_escape_string()的SQL注入