在传递参数时从另一个 php 文件调用 php 文件
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/5389619/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
calling php file fron another php file while passing arguments
提问by su03
i need to call a php inside another php file and pass some arguments also. how can i do this?? i tried
我需要在另一个 php 文件中调用一个 php 并传递一些参数。我怎样才能做到这一点？？我试过
include("http://.../myfile.php?file=$name");
- but gives access denied. i read like v must not set allow_url_open to OFF.
- 但拒绝访问。我读起来像 v 不能将 allow_url_open 设置为 OFF。
if i write like
如果我这样写
$cmd = "/.../myfile.php?file=".$name";
$out =exec($cmd. " 2>&1");
echo $out;
- gives error as /.../myfiles.php?file=hello: no such file or directory.
- 给出错误为 /.../myfiles.php?file=hello: 没有这样的文件或目录。
how can i solve this???
我该如何解决这个问题？？？
回答by Bj?rn
You don't have to pass anything in to your included files, your variables from the calling document will be available by default;
您不必向包含的文件传递任何内容，默认情况下，调用文档中的变量将可用；
File1.php
文件1.php
<?php
$variable = "Woot!";
include "File2.php"; //if in the same folder
File2.php
文件2.php
<?php
echo $variable;
回答by Charlie
the location in your code is incorrect:
您代码中的位置不正确：
$cmd = "/.../myfile.php?file=".$name;
回答by MiPnamic
you can include file over http only if the allow_url_fopenis set to TRUE, and the same parameter allow you to pass variables to the files...
仅当allow_url_fopen设置为 TRUE 时，您才能通过 http 包含文件，并且相同的参数允许您将变量传递给文件...
回答by Stefan Gehrig
You should not include files via HTTP connection - that's almost always a serious security problem.
您不应该通过 HTTP 连接包含文件 - 这几乎总是一个严重的安全问题。
If you must do this, you have to set allow_url_includeandallow_url_fopento ONbut neither is a recommended procedure.
如果您必须这样做，您必须设置allow_url_includeandallow_url_fopentoON但两者都不是推荐的程序。
