Pastebin API粘贴数据

Question

我一直在尝试让pastebin API，而不是告诉我pastebin链接，只输出原始数据。PHP代码如下：

<?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中，显示为pastebin.com/St4ck0v3RFL0W，但我希望它生成原始数据。

原始数据链接是http://pastebin.com/raw.php?i=，有人能帮上忙吗？

参考：http://pastebin.com/api

Answer 1

首先，请注意，您必须将POST请求发送到pastebin.com接口，而不是GET。所以不要在输入数据上使用urlencode()！

要从页面url中获取原始粘贴url，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，这里有一个完整的例子：

<?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

Answer 2

据我所知，响应包含创建内容时生成的Pastebin URL。如下所示的Url：

http://pastebin.com/UIFdu235s

所以你需要做的就是去掉"http://pastebin.com/“：

$id = str_replace("http://pastebin.com/", "", $url_received_on_last_step);

然后，将其附加到您提供的原始url中：

$url_raw = "http://pastebin.com/raw.php?i=".$id;

你会得到原始数据。