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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 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= ,有人可以帮忙吗?

参考: http: //pastebin.com/api

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2 回答 2

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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/”的做法:

$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:

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

你会得到原始数据。

于 2012-10-24T00:44:39.877 回答
0

首先,请注意您必须向POSTpastebin.com API 发送请求,而不是GET. 所以不要urlencode()在你的输入数据上使用!

要从页面 url 获取原始粘贴 url,您有多种选择。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
于 2013-04-26T20:54:42.480 回答