3

我正在尝试使用电子表格脚本编辑器中的Google Apps 脚本制作 Pastebin.com 粘贴。谁能告诉我我做错了什么？

function postPastebinPost() {
  var options, url, apiKey, payload, response;

  apiKey = <api key goes here>;
  payload = 'Hello World';

  options = {
    'method' : 'post',
    'payload' : payload
  };

  url = 'https://pastebin.com/api/api_post.php'
    + '?api_dev_key=' + apiKey
    + '&api_option=paste'
    + '&api_paste_code=' + encodeURIComponent(payload);

  response = UrlFetchApp.fetch(url, options);
  Logger.log(response);
}

我运行它，我的日志显示Bad API request, invalid api_option. 我已经搜索了解决方案，但我没有找到任何解决方案。

文档：

• Pastebin.com API

• Google Apps 脚本的UrlFetchApp 类

4

2 回答 2

4

参数应在 POST 请求的有效负载中传递。

function postPastebinPost() {

  var apiKey = 'YOUR KEY GOES HERE';
  var text = 'Hello World';

  var payload = {
    api_dev_key: apiKey,
    api_option: 'paste',
    api_paste_code:  text
  };

  var options = {
    method : 'POST',
    payload: payload
  };

  var url = 'https://pastebin.com/api/api_post.php';

  var response = UrlFetchApp.fetch(url, options);
  Logger.log(response.getContentText());
}
于 2018-05-12T13:07:28.720 回答
0

以下是如果用户想要创建一个新的粘贴作为他们自己的 Pastebin 帐户的一部分（而不是 «Paste as a guest»）。这只是对 Amit Agarwal 答案的改编。

function postPastebinPost() {
  var title = 'abc';
  var contents = 'Hello World \n next line of content  \n more text';
  var payload = {
    api_dev_key: 'aa6f3ab...', // https://pastebin.com/api#1
    api_option: 'paste',
    api_paste_name: title,
    api_paste_code: contents,
    api_paste_private: '0', // public paste
    api_user_name: 'diccionario...', // name of your Pastebin account
    api_user_password: 'dk398d...', // password to your Pastebin account
    api_user_key: '39dk3...', // https://pastebin.com/api/api_user_key.html
    };
  var options = {
    method : 'POST',
    payload: payload
    };
  var url = 'https://pastebin.com/api/api_post.php';
  var response = UrlFetchApp.fetch(url, options);
  Logger.log(response.getContentText());
}

整个文档位于https://pastebin.com/api。

于 2019-07-01T18:32:16.283 回答