"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

注意:未定义变量

在尝试使用以前未定义的变量时发生。

一个典型的例子是

foreach ($items as $item) {
    // do something with item
    $counter++;
}

如果您之前没有定义$counter，上面的代码将触发通知。

正确的方法是在使用变量之前设置它

$counter = 0;
foreach ($items as $item) {
    // do something with item
    $counter++;
}

类似地，变量在其作用域之外是不可访问的，例如在使用匿名函数时。

$prefix = "Blueberry";
$food = ["cake", "cheese", "pie"];
$prefixedFood = array_map(function ($food) {
  // Prefix is undefined
  return "${prefix} ${food}";
}, $food);

这应该使用use来传递

$prefix = "Blueberry";
$food = ["cake", "cheese", "pie"];
$prefixedFood = array_map(function ($food) use ($prefix) {
  return "${prefix} ${food}";
}, $food);

注意:未定义的属性

此错误的含义大致相同，但指的是对象的属性。重用上面的示例，这段代码将触发错误，因为没有设置counter属性。

$obj = new stdclass;
$obj->property = 2342;
foreach ($items as $item) {
    // do something with item
    $obj->counter++;
}

相关问题:

所有关于Stackoverflow的PHP“注意:未定义变量”问题 “注意:未定义的变量”，“注意:未定义的索引”，“注意:未定义的偏移量”使用PHP 参考:什么是变量作用域，哪些变量可以从哪里访问，什么是“未定义变量”错误?

警告:[function]期望参数1是给定的布尔值

(一个更通用的警告:mysql_fetch_array()期望参数1是resource，给定布尔值)

资源是PHP中的一种类型(就像字符串、整数或对象一样)。资源是一个不透明的blob，本身没有固有的有意义的值。资源特定于一组PHP函数或扩展并由其定义。例如，Mysql扩展定义了两种资源类型:

MySQL模块中使用了两种资源类型。第一个是数据库连接的链接标识符，第二个是保存查询结果的资源。

cURL扩展定义了另外两种资源类型:

．.． 一个卷曲手柄和一个卷曲多手柄。

当var_dump时，这些值看起来像这样:

$resource = curl_init();
var_dump($resource);

resource(1) of type (curl)

这就是大多数资源的全部，某种类型((curl))的数字标识符((1))。

您携带这些资源并将它们传递给不同的函数，这些资源对这些函数有一定的意义。通常，这些函数在后台分配特定的数据，而资源只是它们用来在内部跟踪这些数据的引用。

“…期望参数1为resource，给定布尔值”的错误通常是一个未检查的操作的结果，该操作应该创建一个资源，但却返回false。例如，fopen函数有这样的描述:

返回值 成功时返回文件指针资源，错误时返回FALSE。

因此在这段代码中，$fp要么是类型为(stream)的资源(x)，要么为false:

$fp = fopen(...);

如果你没有检查fopen操作是否成功或失败，因此$fp是一个有效的资源还是假的，并将$fp传递给另一个期望资源的函数，你可能会得到上述错误:

$fp   = fopen(...);
$data = fread($fp, 1024);

Warning: fread() expects parameter 1 to be resource, boolean given

你总是需要错误检查函数的返回值，这些函数试图分配资源，可能会失败:

$fp = fopen(...);

if (!$fp) {
    trigger_error('Failed to allocate resource');
    exit;
}

$data = fread($fp, 1024);

相关的错误:

警告:mysql_fetch_array()期望参数1是resource, boolean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。

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:闭包类的对象不能转换为字符串

这意味着您没有执行匿名函数(或闭包)。

在这个例子中，使用箭头函数会抛出一个错误:

echo $fn = fn($x = 42) => $x;

或者对任何匿名函数

echo function($x = 42) { return $x; };

要解决这个错误，您需要执行闭包。

echo $fn = (fn($x = 42) => $x)(30); // take notice of the 2 sets of brackets

echo (function($x = 42) { return $x; })(30);

这种语法叫做IIFE，是在PHP 7中添加的。