Question

我在php上遇到错误，我不知道它是什么......它说“禁止访问！”

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错误403

twitter_sample.com Apache / 2.4.3（Win32）OpenSSL / 1.0.1c PHP / 5.4.7“

我的代码是

<?php

if($_POST)
{

  $file=$_FILES['media'];
  $postfields = array();

  $postfields['username'] = $_POST['username'];

  $postfields['password'] = $_POST['password'];
  $postfields['message'] = $_POST['message'];
  $postfields['media'] = "@$file[tmp_name]";

  $t=new twitpic($postfields,true,true);
  $t->post();
  exit;
}

?>

<style type="text/javascript">
  *{font-family:verdana;}
  span{font-size:12px;color:#393939;}
  h3{font-size:14px;color:#5AAAF7;}
</style>
<body>

  <h3>Upload your pic to twitpic, and post status on twitter</h3>
  <form method="post"  enctype="multipart/form-data" action="<?= $_SERVER[PHP_SELF] ?>"   >
    <p><span style="height:40px;font-weight:bold;margin-right:56px;">Twitter Username :</span><input type="text" name="username" /></p>
    <p><span style="height:40px;font-weight:bold;margin-right:61px;">Twitter Password:</span><input type="password" name="password" /></p>

    <p><span style="vertical-align:text-top;height:40px;font-weight:bold;margin-right:28px;">Message to be posted :</span> <textarea cols="35" rows="2" name="message"></textarea></p>
    <p><span style="vertical-align:text-top;height:40px;font-weight:bold;">Choose an image to upload: </span><input type="file" name="media" /></p>
    <p style="width:250px;text-align:right;margin-top:50px;"><input type="submit" value="Upload&nbsp;&raquo;" /> </p>
  </form>
  <sup>Script powered by <a href="http://www.digimantra.com/">www.digimantra.com</a></sup>
</body>

You can skip posting update to twitter by passing the third argument as false or just by skipping it. If you want to upload image programmatically, without the user input or the form then you can do it using the following code. Make sure the image path is correctly mention, else it will throw an error.

<?php
$file='file_to_be_uploaded.gif';
$postfields = array();

$postfields['username'] = 'twitter_username';

$postfields['password'] = 'twitter_password';
$postfields['message'] = 'Message to be posted' ;
$postfields['media'] = "@$file"; //Be sure to prefix @, else it wont upload

$t=new twitpic($postfields,true,true);
$t->post();
?>

和

<?php

class twitpic
{
  /* 
   * variable declarations
   */
  var $post_url='http://twitpic.com/api/upload';
  var $post_tweet_url='http://twitpic.com/api/uploadAndPost';
  var $url='';
  var $post_data='';
  var $result='';
  var $tweet='';
  var $return='';

/*
* @param1 is the array of data which is to be uploaded
* @param2 if passed true will display result in the XML format, default is false
* @param3 if passed true will update status twitter,default is false
*/

  function __construct($data,$return=false,$tweet=false)
  {
    $this->post_data=$data;
    if(empty($this->post_data) || !is_array($this->post_data)) //validates the data
      $this->throw_error(0);
    $this->display=$return;
    $this->tweet=$tweet;

  }

  function post()
  {
    $this->url=($this->tweet)?$this->post_tweet_url:$this->post_url; //assigns URL for curl request based on the nature of request by user
    $this->makeCurl();
  }
  private function makeCurl()
  {
    $curl = curl_init();
    curl_setopt($curl, CURLOPT_CONNECTTIMEOUT, 2);
    curl_setopt($curl, CURLOPT_HEADER, false);
    curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1);
    curl_setopt($curl, CURLOPT_BINARYTRANSFER, 1);
    curl_setopt($curl, CURLOPT_URL, $this->url);
    curl_setopt($curl, CURLOPT_POST, 3);
    curl_setopt($curl, CURLOPT_POSTFIELDS, $this->post_data);
    $this->result = curl_exec($curl);
    curl_close($curl);
    if($this->display)
    {
      header ("content-type: text/xml");
      echo $this->result ;
    }

  }
  private function throw_error($code) //handles few errors, you can add more

  {
    switch($code)
    {
      case 0:
        echo 'Think, you forgot to pass the data';
        break;
      default:
        echo 'Something just broke !!';
        break;
    }
    exit;
  }
} //class ends here

?>

谢谢......

Answer 1

我不知道这是否是问题，但在localhost中你必须使用：

$ postfields ['username'] = addslashes（$ _POST ['username']）

而不是

$ postfields ['username'] = $ _POST ['username']

错误访问被禁止

1 个答案: