通过对PHP脚本的间接调用访问$ _REQUEST

Question

I want to access $_REQUEST within a PHP script that generates a JavaScript file. 我想在生成JavaScript文件的PHP脚本中访问$ _REQUEST。 Here is my barebones index.php page: 这是我的准系统index.php页面：

<html>
<head>
  <script src="phpGeneratedScript.js.php"></script>
</head>
<body>
  <?php echo json_encode($_REQUEST); ?>
</body>
</html>

And here is my barebones .js.php file: 这是我的准系统.js.php文件：

<?php echo "alert('".json_encode($_REQUEST)."')"; ?>

When I visit index.php?key=value , I see a query displayed in the page, but not in the alert. 当我访问index.php?key=value ，我看到一个查询显示在页面中，但没有显示在警报中。 Is there a way to access $_REQUEST (or $_GET or $_POST) from within a PHP script that is not directly include d? 是否可以从不直接include d的PHP脚本中访问$ _REQUEST（或$ _GET或$ _POST）？

EDIT: To partially answer my own question - this adds the query string to the call to the php file that returns the JavaScript file: 编辑：要部分回答我自己的问题-这会将查询字符串添加到对返回JavaScript文件的php文件的调用中：

<html>
<head>
  <?php
  $query = parse_url($_SERVER['REQUEST_URI'], PHP_URL_QUERY);
  echo
  "<script src='phpGeneratedScript.js.php?$query'></script>"
  ?>
</head>
<body>
  <?php echo json_encode($_REQUEST); ?>
</body>
</html.

However, this does not handle an POST data which may be sent with the original request. 但是，这不处理可能与原始请求一起发送的POST数据。

Following drew010's suggestions, I tried this: 按照drew010的建议，我尝试了以下方法：

html

<html>
<head>
  <?php
  $query = http_build_query($_REQUEST);
  echo
  "<script src='phpGeneratedScript.js.php?$query'></script>"
  ?>
</head>
<body>
  <?php
    if (!$_POST) { ?>
      <form action="test.php" method="post">
        <!--input type="text" name="name" value="value"--><br>
        <input type='submit' name="submit" value='Submit'>
      </form>
 <? } else {
      echo json_encode($_REQUEST);
      echo "<br />";
      echo json_encode($_POST);
    }
  ?>
</body>

js.php

<?php
echo "console.log('".json_encode($_REQUEST)."');";
echo "console.log('".json_encode($_POST)."')";
?>

Here is the output in the browser window, after I click on the Submit button: 单击“提交”按钮后，这是浏览器窗口中的输出：

{"submit":"Submit","PHPSESSID":"b6eb4d1d73fc75976a16031003f58fa0"}
{"submit":"Submit"}

Both $_REQUEST and $_POST are correctly populated. $ _REQUEST和$ _POST都已正确填充。

Here is the output in the console: 这是控制台中的输出：

{"submit":"Submit","PHPSESSID":"b6eb4d1d73fc75976a16031003f58fa0"}
[]

The $_POST data has been added to the $_REQUEST data, but it is not visible in $_POST. $ _POST数据已添加到$ _REQUEST数据中，但是在$ _POST中不可见。 Is this what you would expect, or is there a way to adjust my scripts to show $_POST data correctly? 这是您所期望的，还是可以调整脚本以正确显示$ _POST数据？

Answer 1

Instead of: 代替：

$query = parse_url($_SERVER['REQUEST_URI'], PHP_URL_QUERY);
echo "<script src='phpGeneratedScript.js.php?$query'></script>";

Try: 尝试：

$query = http_build_query($_REQUEST);

parse_url() with PHP_URL_QUERY will only return GET data from the query string. 使用PHP_URL_QUERY parse_url()将仅从查询字符串返回GET数据。

http_build_query() will create a URL-encoded string from both GET and POST data. http_build_query()将根据GET和POST数据创建一个URL编码的字符串。

通过对PHP脚本的间接调用访问$ _REQUEST

问题描述

1 个解决方案

解决方案1
1 已采纳 2014-02-27 02:23:33

通过对PHP脚本的间接调用访问$ _REQUEST

问题描述

1 个解决方案

解决方案1 1 已采纳 2014-02-27 02:23:33

解决方案1
1 已采纳 2014-02-27 02:23:33