使用参数调用位于另一台服务器中的PHP文件，并读取另一台服务器PHP中的变量值 [英] Call PHP file located in another server with parameter and read variable values in another server PHP

问题描述

在B服务器.php文件中包含A服务器.php文件，并从A服务器Php文件获取变量值

include A server .php file in B server .php file and get variable values from A server Php File

我正在尝试调用位于与RPG(Report Program Generator，AS400)连接的另一台服务器中的PHP文件.我想从我的Web服务器上调用该PHP文件，并希望可以访问变量和函数.

I'm trying to call PHP file which is located in another server which is connected with RPG(Report Program Generator, AS400). I would like to call that PHP file from my web server and would like to have access to variable and functions.

我尝试了包含"但不起作用

I tried Include but not working

我想用参数调用RPG端的.PHP文件.

I would like to call .PHP file which in RPG side with parameter.

A.php

 <?php

  include("http://10.1.1.12/a/file.php?email=[e-mail]&name=abc");

  echo $a; //values from file.php
  ?>

file.php

<?php
 $email = $_REQUEST['email'];
 $name = $_REQUEST['name'];
 $a = "testing";
 echo $a;
 ?>

推荐答案

出于安全原因，默认情况下在php.ini中禁用了使用 include 来获取另一台服务器的响应的方法，很可能您不会这样做.不能使用它.改用 file_get_contents .

Getting the response of another server using include is disabled by default in the php.ini for security reasons, most likely you won’t be able to use it. Use file_get_contents instead.

在您的 file.php 中，您可以使用数据进行json响应并回显它:

In your file.php you can make a json response using your data and echo it:

<?php
$email = $_REQUEST['email'];
$name = $_REQUEST['name'];
$a = "testing";

header('Content-type: application/json');
echo json_encode(
    'email' => $email,
    'name' => $name,
    'a' => $a
);
?>

在 A.php 中，您需要解析json字符串以获取数据:

And in the A.php you need to parse the json string to get your data:

<?php
$data = json_decode(file_get_contents('http://10.1.1.12/a/file.php?email=[e-mail]&name=abc'));
echo $data['email'], ' ', $data['name'], ' ', $data['a'];
?>

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