为什么不PHP接受命令参数$ 1 [英] Why doesn't PHP accept command with argument $1

问题描述

可能重复：结果
  <一href=\"http://stackoverflow.com/questions/12500129/making-variable-from-html-file-work-in-php-file\">Making在PHP文件的HTML文件的工作变

我有一个传递一个参数的PHP文件另一个文件（HTML文件）。它让我与回声检查，它抛回我在HTML文件中写道相同的参数，所以到目前为止，一切都很好。 PHP的文件看起来是这样的。

I have another file (HTML-file) that passes an argument to this PHP-file. It does so I have checked it with "echo" and it throws back the same argument I wrote in the HTML-file, so so far, so good. The PHP-file looks like this.

<?
       $filepath = "/usr/sbin";
       exec("ONE $search -command $filepath ");
       fopen($filepath, "rw");
?>

命令ONE，这是一个bash脚本我写的做一些里grep在文件中有一个参数，那就是$搜索，这在行为传递到这个PHP文件。如果我交流$搜索，以实际的单词搜索内容它就像一个魅力。为什么它不接受$搜索作为参数？它只是似乎忽略它，并抛回一个空白页，但如果我用搜索内容它给回我想要什么。

The command "ONE" which is a bash-script I wrote to do some "greps" in files takes one argument and that is "$search" which in deed is passed over to this PHP-file. And if I exchange "$search" to the actual word "searchword" it works like a charm. Why doesn't it accept "$search" as argument? It just seem to ignore it and throws back a blank page, but if I use "searchword" it gives back just what I want.

推荐答案

您在的fopen 语法错误（可能是抄写错误）。

You have a syntax error in fopen (probably a transcription error).

你的意思是使用 $ _ REQUEST ['搜索'] 而不仅仅是 $搜索？通过传递参数传递给这个PHP文件的HTML文件我假设你通过一个链接或某种异步请求的意思。如果是那样的话，你应该使用 escapeshellarg 。

Do you mean to use $_REQUEST['search'] as opposed to just $search? By "HTML-file that passes an argument to this PHP-file" I assume you mean via a link or asynchronous request of some kind. If that's the case, you should use escapeshellarg.

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