这是什么?

这是一些关于在编程PHP时可能遇到的警告、错误和注意事项的答案,而不知道如何修复它们。这也是一个社区维基,所以每个人都被邀请加入和维护这个列表。

为什么会这样?

Questions like "Headers already sent" or "Calling a member of a non-object" pop up frequently on Stack Overflow. The root cause of those questions is always the same. So the answers to those questions typically repeat them and then show the OP which line to change in their particular case. These answers do not add any value to the site because they only apply to the OP's particular code. Other users having the same error cannot easily read the solution out of it because they are too localized. That is sad because once you understood the root cause, fixing the error is trivial. Hence, this list tries to explain the solution in a general way to apply.

我该怎么做呢?

如果您的问题被标记为此问题的副本,请在下面找到您的错误消息并将修复应用于您的代码。答案通常包含进一步的调查链接,以防仅从一般答案中不清楚。

如果您想投稿,请添加您“最喜欢的”错误消息、警告或通知,每个答案一条,简短描述它的含义(即使它只是突出显示手册页的术语),可能的解决方案或调试方法,以及现有的有价值的问答列表。此外,请随意改进任何现有的答案。

列表

Nothing is seen. The page is empty and white. (also known as White Page/Screen Of Death) Code doesn't run/what looks like parts of my PHP code are output Warning: Cannot modify header information - headers already sent Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given a.k.a. Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource Warning: [function] expects parameter 1 to be resource, boolean given Warning: [function]: failed to open stream: [reason] Warning: open_basedir restriction in effect Warning: Division by zero Warning: Illegal string offset 'XXX' Warning: count(): Parameter must be an array or an object that implements Countable Parse error: syntax error, unexpected '[' Parse error: syntax error, unexpected T_XXX Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE Parse error: syntax error, unexpected T_PAAMAYIM_NEKUDOTAYIM Parse error: syntax error, unexpected 'require_once' (T_REQUIRE_ONCE), expecting function (T_FUNCTION) Parse error: syntax error, unexpected T_VARIABLE Fatal error: Allowed memory size of XXX bytes exhausted (tried to allocate XXX bytes) Fatal error: Maximum execution time of XX seconds exceeded Fatal error: Call to a member function ... on a non-object or null Fatal Error: Call to Undefined function XXX Fatal Error: Cannot redeclare XXX Fatal error: Can't use function return value in write context Fatal error: Declaration of AAA::BBB() must be compatible with that of CCC::BBB()' Return type of AAA::BBB() should either be compatible with CCC::BBB(), or the #[\ReturnTypeWillChange] attribute should be used Fatal error: Using $this when not in object context Fatal error: Object of class Closure could not be converted to string Fatal error: Undefined class constant Fatal error: Uncaught TypeError: Argument #n must be of type x, y given Notice: Array to string conversion (< PHP 8.0) or Warning: Array to string conversion (>= PHP 8.0) Notice: Trying to get property of non-object error Notice: Undefined variable or property "Notice: Undefined Index", or "Warning: Undefined array key" Notice: Undefined offset XXX [Reference] Notice: Uninitialized string offset: XXX Notice: Use of undefined constant XXX - assumed 'XXX' / Error: Undefined constant XXX MySQL: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ... at line ... Strict Standards: Non-static method [<class>::<method>] should not be called statically Warning: function expects parameter X to be boolean/string/integer HTTP Error 500 - Internal server error Deprecated: Arrays and strings offset access syntax with curly braces is deprecated

还看到:

这个符号在PHP中是什么意思?


当前回答

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

其他回答

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

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

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

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?”

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");