php - Error Access forbidden -
i encounter error on php , dont know is... says "access forbidden!
you don't have permission access requested object. either read-protected or not readable server.
if think server error, please contact webmaster.
error 403
twitter_sample.com apache/2.4.3 (win32) openssl/1.0.1c php/5.4.7"
my code is
<?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 pic twitpic, , 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 posted :</span> <textarea cols="35" rows="2" name="message"></textarea></p> <p><span style="vertical-align:text-top;height:40px;font-weight:bold;">choose image upload: </span><input type="file" name="media" /></p> <p style="width:250px;text-align:right;margin-top:50px;"><input type="submit" value="upload »" /> </p> </form> <sup>script powered <a href="http://www.digimantra.com/">www.digimantra.com</a></sup> </body> can skip posting update twitter passing third argument false or skipping it. if want upload image programmatically, without user input or form can using following code. make sure image path correctly mention, else throw error. <?php $file='file_to_be_uploaded.gif'; $postfields = array(); $postfields['username'] = 'twitter_username'; $postfields['password'] = 'twitter_password'; $postfields['message'] = 'message posted' ; $postfields['media'] = "@$file"; //be sure prefix @, else wont upload $t=new twitpic($postfields,true,true); $t->post(); ?> and
<?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 $return=''; /* * @param1 array of data uploaded * @param2 if passed true display result in xml format, default false * @param3 if passed true update status twitter,default false */ function __construct($data,$return=false,$tweet=false) { $this->post_data=$data; if(empty($this->post_data) || !is_array($this->post_data)) //validates 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 curl request based on nature of request 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, can add more { switch($code) { case 0: echo 'think, forgot pass data'; break; default: echo 'something broke !!'; break; } exit; } } //class ends here ?> thank you...
i don't know if issue, in localhost must use:
$postfields['username'] =addslashes( $_post['username'])
rather than
$postfields['username'] = $_post['username']
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