不显示 PHP 错误
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Not displaying PHP errors
提问by beetree
I've boiled down the problem and made it clean so that it hopefully will be easier for you to help me.
我已经将问题归结并清理干净，希望您能更轻松地帮助我。
I have a very simple code:
我有一个非常简单的代码：
<?php
echo "Hello world";
?>
This runs perfectly fine.
这运行得很好。
If I run the following code (parse error) I do not get any errors but the text "Hello world" is still displayed:
如果我运行以下代码（解析错误），我不会收到任何错误，但仍会显示文本“Hello world”：
<?php
echo "Hello world";
piwejfoiwjefoijwef
?>
If I place the parse error before the code it does however not display "Hello world":
如果我将解析错误放在代码之前，它不会显示“Hello world”：
<?php
piwejfoiwjefoijwef
echo "Hello world";
?>
When I print phpinfo (in the same file, same directory) I have the following settings: display_errors On display_startup_errors On error_reporting 1
当我打印 phpinfo（在同一个文件，同一个目录中）时，我有以下设置： display_errors On display_startup_errors On error_reporting 1
If I try to also set the error reporting inside the script and run it with the following code I still do not get any errors or warning but the text "Hello world" is displayed:
如果我还尝试在脚本中设置错误报告并使用以下代码运行它，我仍然没有收到任何错误或警告，但会显示文本“Hello world”：
<?php
error_reporting(E_ERROR | E_WARNING | E_PARSE); ini_set('display_errors', '1');
echo "Hello world";
owieufpowiejf
?>
My php.ini file has the following values (and I have restarted Apache):
我的 php.ini 文件具有以下值（并且我已经重新启动了 Apache）：
error_reporting = E_ERROR & ~E_DEPRECATED
display_errors = On
display_startup_errors = On
I am running Apache / PHP / MySQL on the Amazon AMI with on a 64-bit AWS EC2. I am not that knowledgeable with server configurations. The errors started when I transitioned to the Amazon server. Besides error reporting the server and Apache/PHP runs flawlessly.
我正在使用 64 位 AWS EC2 在 Amazon AMI 上运行 Apache/PHP/MySQL。我对服务器配置不太了解。当我转换到亚马逊服务器时，错误就开始了。除了错误报告服务器和 Apache/PHP 运行完美。
Please guide me in what I can do to fix the problem.
请指导我如何解决问题。
Thanks!
谢谢！
回答by Madara's Ghost
That is a notice.
那是一个通知。
error_reporting(E_ALL);
reveals it.
揭示它。
Code I used:
我使用的代码：
<?php
error_reporting(E_ALL); ini_set('display_errors', '1');
echo "Hello world";
owieufpowiejf
?>
Output:
输出：
Hello world
Notice: Use of undefined constant owieufpowiejf - assumed 'owieufpowiejf' in /code/14B4LY on line 5
你好，世界
注意：使用未定义的常量 owieufpowiejf - 在第 5 行的 /code/14B4LY 中假定为 'owieufpowiejf'
That's because it's not a parse error, it thinks of it as a constant and tried to parse it as a string. And placing a normal string is a valid statement.
那是因为它不是解析错误，而是将其视为常量并尝试将其解析为字符串。放置一个普通字符串是一个有效的语句。
回答by Ole
You can try error_reporting(-1)
你可以试试 error_reporting(-1)
-1 is the maximum value error_reportingcan take and always will be.
-1 是error_reporting可以取的最大值，并且永远都是。
回答by Krasimir
Create .htaccess file in your main directory and place:
在您的主目录中创建 .htaccess 文件并放置：
php_flag display_errors on
# 7 stands for E_ERROR | E_WARNING | E_PARSE
php_value error_reporting 7
Exact values of error_reportingconstants could be found in the official documentation http://php.net/manual/en/errorfunc.constants.php
error_reporting可以在官方文档http://php.net/manual/en/errorfunc.constants.php 中找到常量的确切值
Of course you should have mod_rewrite enabled in your server.
当然，您应该在服务器中启用 mod_rewrite。
回答by Jake
In your PHP script try setting error_reporting to E_ALL and see if you get a notice..
在您的 PHP 脚本中，尝试将 error_reporting 设置为 E_ALL 并查看您是否收到通知。
error_reporting(E_ALL)
Check out the documentation: http://www.php.net/manual/en/errorfunc.configuration.php#ini.error-reporting
查看文档：http: //www.php.net/manual/en/errorfunc.configuration.php#ini.error-reporting
回答by Tom Kincaid
I had this problem and fixed it by editing /etc/php.ini to display_errors = On
我遇到了这个问题并通过将 /etc/php.ini 编辑为 display_errors = On 来修复它
回答by mario
Calling error_reporting()in the same script that contains the syntax error is never going to work.
调用error_reporting()包含语法错误的同一个脚本永远不会起作用。
<?php
echo "Hello world";
piwejfoiwjefoijwef
?>
This script in particular does not get you any syntax error, because it does not contain any syntax error. It's just an echostatement and a bare constant in the second line. The trailing semicolon can be omitted right before the ?>
特别是这个脚本不会给你任何语法错误，因为它不包含任何语法错误。它只是echo第二行中的一个语句和一个裸露的常量。尾随分号可以省略?>
You would get a notice, if it hadn't been turned off. Again, you didn't enable E_NOTICEin your other test.
如果它没有被关闭，你会收到通知。同样，您没有E_NOTICE在其他测试中启用。
