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

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");

其他回答

"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

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

警告:count():参数必须是数组或实现Countable的对象

不言而喻的;传递给count()函数的参数必须是可数的,通常是数组。

可能的问题是传递了一个标量值,如字符串或整数,或者对象没有实现Countable接口。在有问题的变量上使用var_dump()可以显示是否是这种情况。

警告:open_basedir限制生效

此警告可以与与文件和目录访问相关的各种函数一起出现。它警告一个配置问题。

当它出现时,这意味着已禁止访问某些文件。

警告本身不会破坏任何东西,但如果文件访问被阻止,脚本通常无法正常工作。

修复方法通常是更改PHP配置,相关设置称为open_basedir。

有时使用了错误的文件或目录名称,解决方法是使用正确的名称。

相关问题:

Open_basedir限制生效。文件(/)不在允许路径内: 所有PHP的“警告:open_basedir限制在有效”的查询在Stackoverflow

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?