Question

我有一个记录错误的基本控制器，我希望在日志消息中包含 ALL 请求数据，所有标头和请求正文。我怎么能在CodeIgniter中做到这一点？

Answer 1

相当于$this->input的CodeIgniter file_get_contents('php://input')包含来自foo var_dump($this->input)的输出的所有Body请求（如果有）：

object(CI_Input)[9]
  public 'ip_address' => boolean false
  public 'user_agent' => boolean false
  public '_allow_get_array' => boolean true
  public '_standardize_newlines' => boolean true
  public '_enable_xss' => boolean false
  public '_enable_csrf' => boolean false
  protected 'headers' => 
    array (size=0)
      empty
  public 'security' => &
    object(CI_Security)[8]
      protected '_xss_hash' => string '' (length=0)
      protected '_csrf_hash' => string '' (length=0)
      protected '_csrf_expire' => int 7200
      protected '_csrf_token_name' => string 'ci_csrf_token' (length=13)
      protected '_csrf_cookie_name' => string 'ci_csrf_token' (length=13)
      protected '_never_allowed_str' => 
        array (size=10)
          'document.cookie' => string '[removed]' (length=9)
          'document.write' => string '[removed]' (length=9)
          '.parentNode' => string '[removed]' (length=9)
          '.innerHTML' => string '[removed]' (length=9)
          'window.location' => string '[removed]' (length=9)
          '-moz-bindin.......//and many other data.

并且对于标题laso，你在PHP 5.4中使用了headers_list()的新命令。此外，许多其他数据可以包含在 CI seeion 中，例如：ip_address, user_agent ,last_activity

Answer 2

对于标题详细信息

$this->load->library('user_agent');
$reqUserIp = $this->input->ip_address();
$userAgent = $this->agent->agent_string();
$reqHeaders = $this->input->request_headers();

Answer 3

您可以使用帖子控制器hook。

在 application / config / config.php 。

然后在 application / config / hooks.php ;

中

$hook['post_controller'] = array(
                        'class'    => 'PostController',
                        'function' => 'log_it',
                        'filename' => 'PostController.php',
                        'filepath' => 'hooks',
                        'params'   => array()
                        );

在 application / hooks / PostController.php中创建文件

<?php

class PostController {

private $ci;
private $filepath;

public function __construct() {
    $this->ci =& get_instance();
    $log_path = ($this->ci->config->item('log_path') != '') ? $this->ci->config->item('log_path') : APPPATH.'logs/';
    $this->filepath = $log_path.'action_log';
}

public function log_it($log_message=NULL) {
    if ($log_message == NULL) {
        $log_message = date('d/m/Y H:i:s')."::";
        $log_message .= $this->ci->session->userdata('username')."::";
        $log_message .= $this->ci->input->ip_address()."::";
        $log_message .= $this->get_client_ip()."::";
        $log_message .= $this->ci->router->class."::";
        $log_message .= $this->ci->router->method."::";
        $log_message .= uri_string()."::";
        if (isset($_POST)) {
            foreach ($_POST as $key => $value) {
                $log_message .= "[$key] => $value;";
            }
        }
        $log_message .= "\r\n";
    }


    $fp = fopen($this->filepath, FOPEN_WRITE_CREATE);
    fwrite($fp, $log_message);
    fclose($fp);
}

private function get_client_ip() {
    if (!empty($_SERVER['HTTP_CLIENT_IP'])) {   //check ip from share internet
        $ip = $_SERVER['HTTP_CLIENT_IP'];
    } elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])) {   //to check ip is pass from proxy
        $ip = $_SERVER['HTTP_X_FORWARDED_FOR'];
    } else {
        $ip = $_SERVER['REMOTE_ADDR'];
    }
    return $ip;
  }
 }

?>

如何在CodeIgniter中获取请求路径？

3 个答案:

对于标题详细信息