Question

我正在尝试使用电子表格脚本编辑器中的 Google Apps脚本进行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);
}

我运行此操作，我的日志显示为Bad API request, invalid api_option。我已经搜索了解决方案，但我还没找到。

文档：

•Pastebin.com API

•Google Apps脚本UrlFetchApp Class

Answer 1

参数应该在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());
}

Answer 2

如果用户要创建新的粘贴作为其自己的Pastebin帐户的一部分（而不是“以客人身份粘贴”），以下内容是可能的。这只是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。

使用Google Apps脚本和Pastebin.com API发布粘贴

2 个答案: