警告:数组到字符串的转换

注意:数组到字符串的转换

(PHP 7.4之前的一个通知，PHP 8.0之后的一个警告)

如果你试图将数组视为字符串，就会发生这种情况:

$arr = array('foo', 'bar');

echo $arr;  // Notice: Array to string conversion
$str = 'Something, ' . $arr;  // Notice: Array to string conversion

数组不能简单地与字符串回显或连接，因为结果没有很好地定义。PHP将使用字符串“Array”来代替数组，并触发通知，指出这可能不是预期的结果，您应该检查这里的代码。你可能想要这样的东西:

echo $arr[0];  // displays foo
$str = 'Something ' . join(', ', $arr); //displays Something, foo, bar

或者循环数组:

foreach($arr as $key => $value) {
    echo "array $key = $value";
    // displays first: array 0 = foo
    // displays next:  array 1 = bar
}

如果这个通知出现在您意想不到的地方，这意味着您认为是字符串的变量实际上是一个数组。这意味着你的代码中有一个错误，使这个变量成为一个数组，而不是你期望的字符串。

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:调用未定义的函数XXX

在尝试调用尚未定义的函数时发生。常见的原因包括缺少扩展和包含，有条件的函数声明，函数声明中的函数或简单的拼写错误。

例1 -条件函数声明

$someCondition = false;
if ($someCondition === true) {
    function fn() {
        return 1;
    }
}
echo fn(); // triggers error

在这种情况下，fn()将永远不会被声明，因为$someCondition不为真。

例2 -函数声明中的函数

function createFn() 
{
    function fn() {
        return 1;
    }
}
echo fn(); // triggers error

在这种情况下，只有在调用createFn()时才声明fn。注意，后续调用createFn()将触发一个关于重新声明现有函数的错误。

在PHP内置函数中也可以看到这种情况。尝试在官方手册中搜索该函数，并检查它属于哪个“扩展”(PHP模块)，以及哪些版本的PHP支持它。

如果缺少扩展，请安装该扩展并在php.ini中启用它。请参阅PHP手册中的安装说明，以了解函数在扩展中的显示。你也可以使用你的包管理器(例如Debian或Ubuntu中的apt, Red Hat或CentOS中的yum)，或共享托管环境中的控制面板来启用或安装扩展。

如果该函数是在您正在使用的PHP的新版本中引入的，那么您可以在手册或评论部分中找到替代实现的链接。如果它已从PHP中删除，请查找有关原因的信息，因为可能不再需要它了。

如果缺少include，请确保在调用函数之前包含声明该函数的文件。

如果出现错别字，请改正错别字。

相关问题:

https://stackoverflow.com/search?q=Fatal+error%3A+Call+to+undefined+function

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

HTTP错误500 -内部服务器错误

HTTP状态代码500和典型的Apache或浏览器警告是一个非常广泛的消息。这不是实际的错误。要弄清楚是web服务器配置错误(.htaccess)还是PHP致命错误，你必须查看error.log。

你通常可以在下面找到webservers日志:

/var/log/apache2 on Linux servers, often used for local and virtual hosts. /var/www/_user12345_/logs or similar on shared hosting plans. Usually there's a logs/ directory alongside each htdocs/ folder. C:\xampp\apache\logs\error.log for WAMP/XAMPP distributions of Apache+PHP. Alternatively just use a file search feature to locate anything called "error.log". Or look into your Apache httpd.conf and its ErrorLog directive. /var/log/nginx/nginx_error.log for NGINX. C:\inetpub\logs\LogFiles for IIS. Luckily this is uncommon still, but journalctl -r -u apache2.service could also hold parts of the log on Linux setups.

它是一个文本文件。搜索与错误时间最接近的条目，并使用错误消息的重要部分(从“PHP error:…”到“in line…”)进行进一步的google搜索。

[Mon 22:10] [:error] [pid 12345] [client 127.0.0.1] FastCGI: server "/fcgi/p73" stderr: PHP message:PHP error: Unfiltered inputvariable $_JSON['pokestop_lng'] in filyfile.php on line 845

对于FPM设置，你经常会看到致命的PHP错误。而旧的mod_php(共享主机)配置经常混合警告和通知(通常也值得检查)。

如果没有配置为使用系统或Apache日志机制，您可能还需要查看PHP的error.log。一般来说，保留默认值并启用error_display + error_reporting来显示具体的错误会更简单。HTTP 500全捕获页面不过是PHP死机白屏幕的一个变种。

参见:

500内部服务器错误的php文件，而不是html 我怎么能让PHP显示错误，而不是给我500内部服务器错误 如何在IIS7上显示和记录PHP错误? 如何修复WordPress内部服务器的500个错误

解析错误:语法错误，意外的“[”

这个错误有两种形式:

变化1

$arr = [1, 2, 3];

这个数组初始化式语法只在PHP 5.4中引入;它将在此之前的版本上引发解析器错误。如果可能的话，升级你的安装或使用旧的语法:

$arr = array(1, 2, 3);

请参见本手册中的示例。

变化2

$suffix = explode(',', 'foo,bar')[1];

PHP 5.4中还引入了数组解引用函数结果。如果无法升级，你需要使用一个(临时)变量:

$parts = explode(',', 'foo,bar');
$suffix = $parts[1];

请参见本手册中的示例。