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我遇到我的PHP的错误，我不知道它是什么？它说：“访问被禁止！错误禁止访问
您没有权限访问请求的对象，这是无论是读保护无论是否由服务器读取。
如果你认为这是一个服务器错误，请与网站管理员联系。
错误403
twitter_sample.com 的Apache/2.4.3（Win32的）的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 »" /> </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
?>
谢谢你......
有可能是您的网络服务器设置的问题，并没有任何与你的PHP脚本。 – compid
即时通讯在本地主机运行它...它的问题？ –
这可能是文件夹权限问题 - 您使用的是哪个Web服务器，以及您从哪个文件夹提供文件？ – compid