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

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

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

代码不运行/我的PHP代码的某些部分被输出

如果你的PHP代码没有任何结果，或者你在网页中看到你的PHP源代码输出的部分文字，你可以很确定你的PHP实际上没有被执行。如果在浏览器中使用“查看源代码”，则可能会看到完整的PHP源代码文件。因为PHP代码嵌入在<?php ?>标记时，浏览器将尝试将它们解释为HTML标记，结果可能看起来有些混乱。

要真正运行PHP脚本，您需要:

执行脚本的web服务器 将文件扩展名设置为.php，否则web服务器不会将其解释为* 通过web服务器访问你的。php文件

*除非你重新配置，否则一切都可以配置。

最后一点尤其重要。只需双击该文件，就可以在浏览器中使用如下地址打开它:

file://C:/path/to/my/file.php

这完全绕过了您可能正在运行的任何web服务器，并且文件没有得到解释。你需要在你的web服务器上访问文件的URL，可能是这样的:

http://localhost/my/file.php

您可能还想检查是否使用了短的打开标记<?而不是<?php和您的php配置已关闭短打开标记。

还可以看到PHP代码没有被执行，而是代码显示在页面上

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");

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。