PHP shell 脚本的问题：“无法打开输入文件”

Problems with a PHP shell script: "Could not open input file"

提问by Mike L.

Ok, I am tryingto create an email logger, that uses a PHP shell script. I have set up CPanel to pipe emails to my script. I am sure this is all configured properly. However I am having problems with the script, well any script for that matter when running it from the shell.

好的，我正在尝试创建一个使用 PHP shell 脚本的电子邮件记录器。我已经设置了 CPanel 来通过管道将电子邮件发送到我的脚本。我确信这一切都配置正确。但是，我遇到了脚本问题，以及从 shell 运行它时与此相关的任何脚本。

here is an example.

这是一个例子。

#!/usr/local/bin/php –q
<?php

/* Read the message from STDIN */
$fd = fopen("php://stdin", "r");
$email = ""; // This will be the variable holding the data.
while (!feof($fd)) {
$email .= fread($fd, 1024);
}
fclose($fd);
/* Saves the data into a file */
$fdw = fopen("mail.txt", "w+");
fwrite($fdw, $email);
fclose($fdw);
/* Script End */
?>

Real simple, right? Read from STDIN and write to a file...I thought something was wrong, not able to read STDIN for some reason. Hosting provider allows it, allow_url_open and allow_url_include are both on.

真的很简单吧？从 STDIN 读取并写入文件...我以为出了什么问题，由于某种原因无法读取 STDIN。主机提供商允许它，allow_url_open 和 allow_url_include 都打开。

When executing the script via SSH I get the following error: Could not open input file: aq

通过 SSH 执行脚本时，出现以下错误：无法打开输入文件：aq

So once again I thought that was the script telling me, that is could not read from STDIN

所以我再一次认为这是脚本告诉我的，那就是无法从 STDIN 读取

So I tried just a simple script.

所以我只尝试了一个简单的脚本。

#!/usr/local/bin/php –q
<?php
echo 'Hello World';
?>

Same thing: Could not open input file: aq

同样的事情：无法打开输入文件：aq

So it appears that the PHP program is telling me it is unable to open the script? The script is located in $HOME/mail/forward (CHMOD 755) and the script itself is CHMOD 755, as well the file mail.txt is CHMOD 755 I am really stumped on this.

所以看起来PHP程序告诉我它无法打开脚本？该脚本位于 $HOME/mail/forward (CHMOD 755) 中，脚本本身是 CHMOD 755，而文件 mail.txt 也是 CHMOD 755 我真的被这个难住了。

采纳答案by Daniel Trebbien

Have you tried:

你有没有尝试过：

#!/usr/local/bin/php

I.e. without the -qpart? That's what the error message "Could not open input file: -q" means. The first argument to phpif it doesn't look like an option is the name of the PHP file to execute, and -qis CGI only.

即没有-q部分？这就是错误消息“无法打开输入文件：-q”的含义。php如果它看起来不像选项的第一个参数是要执行的 PHP 文件的名称，并且-q仅是 CGI。

EDIT:A couple of (non-related) tips:

编辑：几个（不相关的）提示：

1. You don't need to terminate the last block of PHP with ?>. In fact, it is often better not to.
2. When executed on the command line, PHP defines the global constant STDINto fopen("php://stdin", "r"). You can use that instead of opening "php://stdin"a second time: $fd = STDIN;

1. 你不需要用?>. 事实上，通常最好不要这样做。
2. 在命令行上执行时，PHP 将全局常量定义STDIN为fopen("php://stdin", "r"). 您可以使用它而不是"php://stdin"第二次打开：$fd = STDIN;

回答by JustinP

I just experienced this issue and it was because I was trying to run a script from the wrong directory.. doh! It happens to the best of us.

我刚刚遇到了这个问题，这是因为我试图从错误的目录中运行脚本..doh！它发生在我们中最好的人身上。

回答by programer

When you use php CLI argument -q doesn't exist.

当您使用 php CLI 参数时 -q 不存在。

I had the same problem when I wrote script in the Windows (eclipse) and I tried run them on Linux. Every line in file from Windows is ended by \r\n. I had to delete \r in first line that contained parser path:

当我在 Windows (eclipse) 中编写脚本并尝试在 Linux 上运行它们时，我遇到了同样的问题。来自 Windows 的文件中的每一行都以 \r\n 结束。我不得不删除包含解析器路径的第一行中的 \r ：

When \r was deleted from first line (mcedit shown \r as ^M) script ran correctly.

当从第一行删除 \r 时（mcedit 显示 \r 为 ^M）脚本正确运行。

回答by Joseph Lust

Windows Character Encoding Issue

Windows 字符编码问题

I was having the same issue. I was editing files in PDT Eclipse on Windows and WinSCPing them over. I just copied and pasted the contents into a nano window, saved, and now they worked. Definitely some Windows character encoding issue, and not a matter of Shebangs or interpreter flags.

我遇到了同样的问题。我在 Windows 上的 PDT Eclipse 中编辑文件并 WinSCPing 它们。我只是将内容复制并粘贴到纳米窗口中，保存，现在它们可以工作了。肯定是一些 Windows 字符编码问题，而不是 Shebangs 或解释器标志的问题。

回答by Ram Shengale

I landed up on this page when searching for a solution for “Could not open input file”error. Here's my 2 cents for this error.

在搜索“无法打开输入文件”错误的解决方案时，我登陆了此页面。这是我对这个错误的 2 美分。

I faced this same error while because I was using parameters in my php file path like this:

我遇到了同样的错误，因为我在我的 php 文件路径中使用了这样的参数：

/usr/bin/php -q /home/**/public_html/cron/job.php?id=1234

But I found out that this is not the proper way to do it. The proper way of sending parameters is like this:

但我发现这不是正确的方法。发送参数的正确方式是这样的：

/usr/bin/php -q /home/**/public_html/cron/job.php id=1234

Just replace the "?"with a space " ".

只需用"?"空格替换" "。

回答by jsherk

For me the problem was I had to use /usr/bin/php-cgicommand instead of just /usr/bin/php

对我来说，问题是我必须使用/usr/bin/php-cgi命令而不仅仅是/usr/bin/php

php-cgi is the command run when accessed thru web browser.

php-cgi 是通过 Web 浏览器访问时运行的命令。

php is the CLI command line command.

php 是 CLI 命令行命令。

Not sure why php cli is not working, but running with php-cgi instead fixed the problem for me.

不知道为什么 php cli 不起作用，但是使用 php-cgi 运行反而为我解决了这个问题。

回答by steampowered

Due to windows encoding issue for me

由于我的 Windows 编码问题

I experienced this "Could not open input file" error. Then I obtained the file using wgetfrom another linux system, and the error did not occur.

我遇到了“无法打开输入文件”错误。然后我wget从另一个linux系统中获取了该文件，并没有出现错误。

The error ws only occurring for me when the file transited through windows.

该错误仅在文件通过 Windows 传输时发生。