我一直在尝试获取pastebin API,而不是告诉我pastebin链接,只输出原始数据。 PHP代码是这样的:
<?php
$api_dev_key = 'Stackoverflow(fake key)';
$api_paste_code = 'API.'; // your paste text
$api_paste_private = '1'; // 0=public 1=unlisted 2=private
$api_paste_expire_date = 'N';
$api_paste_format = 'php';
$api_paste_code = urlencode($api_paste_code);
$url = 'http://pastebin.com/api/api_post.php';
$ch = curl_init($url);
?>
通常这会将$ api_paste_code上传到pastebin,显示为像pastebin.com/St4ck0v3RFL0W,但我希望它能生成原始数据。
原始数据链接是http://pastebin.com/raw.php?i=,有人可以帮忙吗?
答案 0 :(得分:1)
据我所知,响应包含创建内容时生成的Pastebin URL。像这样的网址:
http://pastebin.com/UIFdu235s
所以你只需要摆脱“http://pastebin.com/”做:
$id = str_replace("http://pastebin.com/", "", $url_received_on_last_step);
然后,将其附加到您提供的原始网址上:
$url_raw = "http://pastebin.com/raw.php?i=".$id;
您将获得原始数据。
答案 1 :(得分:0)
首先请注意,您必须向pastebin.com API发送POST
请求,而不是GET
。所以不要在输入数据上使用urlencode()
!
要从页面网址获取原始粘贴网址,您有几个选项。但最简单的可能是:
$apiResonse = 'http://pastebin.com/ABC123';
$raw = str_replace('m/', 'm/raw.php?i=', $apiResponse);
最后,这是一个完整的例子:
<?php
$data = 'Hello World!';
$apiKey = 'xxxxxxx'; // get it from pastebin.com
$apiHost = 'http://pastebin.com/';
$postData = array(
'api_dev_key' => $apiKey, // your dev key
'api_option' => 'paste', // action to perform
'api_paste_code' => utf8_decode($data), // the paste text
);
$ch = curl_init();
curl_setopt_array($ch, array(
CURLOPT_URL => "{$apiHost}api/api_post.php",
CURLOPT_RETURNTRANSFER => 1,
CURLOPT_POST => 1,
CURLOPT_POSTFIELDS => http_build_query($postData),
));
$result = curl_exec($ch); // on success, some string like 'http://pastebin.com/ABC123'
curl_close($ch);
if ($result) {
$pasteId = str_replace($apiHost, '', $result);
$rawLink = "{$apiHost}raw.php?i={$pasteId}";
echo "Created new paste.\r\n Paste ID:\t{$pasteId}\r\n Page Link:\t{$result}\r\n Raw Link:\t{$rawLink}\r\n";
}
运行上面的代码,输出:
c:\xampp\htdocs>php pastebin.php
Created new paste.
Paste ID: Bb8Ehaa7
Page Link: http://pastebin.com/Bb8Ehaa7
Raw Link: http://pastebin.com/raw.php?i=Bb8Ehaa7