2011-05-14 61 views
2

我试图让一个php api请求简单地粘贴到pastebin,我在http://pastebin.com/api找到一个例子,它是相当海峡,所以我不认为会有任何麻烦。但这个例子似乎并没有奏效。我不断收到响应PasteBin API错误?不允许简单的发布

Bad API request, invalid api_option 

但是你可以清楚地看到它设置它创建的字符串中api_option=paste ...

,并在文件中,它说

Creating A New Paste, [Required Parameters] 
Include all the following POST parameters when you request the URL: 

1. api_dev_key - which is your unique API Developers Key. 
2. api_option - set as 'paste', this will indicate you want to create a new paste. 
3. api_paste_code - this is the text that will be written inside your paste. 

Leaving any of these parameters out will result in an error. 

所以... 。我认为它看起来不错,除了它提供的例子。

任何人有任何想法这里发生了什么?

<?php 


$api_dev_key   = '1234'; // your api_developer_key 
$api_paste_code   = 'some random text to test'; // your paste text 
$api_paste_private  = '0'; // 0=public 1=private 
$api_paste_name   = 'savelogtest'; // name or title of your paste 
$api_paste_expire_date = '10M'; 
$api_paste_format  = 'php'; 
$api_user_key   = ''; // if invalid key or no key is used, the paste will be create as a guest 
$api_paste_name   = urlencode($api_paste_name); 
$api_paste_code   = urlencode($api_paste_code); 


$url    = 'http://pastebin.com/api/api_post.php'; 
$ch     = curl_init($url); 

curl_setopt($ch, CURLOPT_POST, true); 
curl_setopt($ch, CURLOPT_POSTFIELDS, 'api_option=paste&api_user_key='.$api_user_key.'&api_paste_private='.$api_paste_private.'&api_paste_name='.$api_paste_name.'&api_paste_expire_date='.$api_paste_expire_date.'&api_paste_format='.$api_paste_format.'&api_dev_key='.$api_dev_key.'&api_paste_code='.$api_paste_code.''); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); 
curl_setopt($ch, CURLOPT_VERBOSE, 1); 
curl_setopt($ch, CURLOPT_NOBODY, 0); 

$response   = curl_exec($ch); 
echo $response; 


?> 

回答

1

API示例正常工作。我只是跑你的密码(当然改变了$ api_dev_key,和它的工作第一次输出:http://pastebin.com/eyn9tWNS

尝试,并在你的脚本的顶部添加此:

error_reporting(E_ALL); 
    ini_set("display_errors", "on"); 

它应该给你一些更好错误报告。