api - Google Apps 脚本和外部 API 授权在标头中失败

Question

试图让这个工作。在参数列表之后，我不断收到 Missing ) 。（第 6 行，文件“代码”）解雇。我已经仔细检查了我的括号，但无济于事。我错过了什么吗？

我希望这是一个合理的问题。谢谢你。

function myFunction() {
  var url = "https://company.harvestapp.com/people";
  var headers = {
    "Accept": "application/xml",
    "Content-Type": "application/xml",
    "Authorization": "Basic " + Utilities.base64Encode(dude@dude.com +":"+pw)
     };
  var response = UrlFetchApp.fetch(url,headers);
  var text = response.getResponseCode();
  Logger.log(text);
}

Answer 1

知道了。还要感谢布赖恩。我终于意识到你需要一个“选项”对象来传递方法和标题。

function myFunction() {
    var url = "https://swellpath.harvestapp.com/people/";
    var user = "dude@dude.com";
    var password = "supersecurepw";

    var headers = {
        "Accept": "application/xml",
        "Content-Type": "application/xml",
        "Authorization": "Basic "+ Utilities.base64Encode(user+":"+password)
    };

    var options = {
        "method" : "get",
        "headers" : headers 
    };

    var response = UrlFetchApp.fetch(url,options);

    Logger.log(response);
}

Answer 2

可能会有其他错误，但是您的第一个问题是您需要将“headers”对象添加到“options”对象，然后执行以下操作

    var url = "https://company.harvestapp.com/people";

    var headers = { "Accept": "application/xml",
                   "Content-Type": "application/xml",
                  "Authorization": "Basic " + Utilities.base64Encode(dude@dude.com +":"+pw)
                  };

      var options ={
                    "headers" : headers
                   };
       var response = UrlFetchApp.fetch(url,options);

Answer 3

编辑：

尝试：

   "Authorization": "Basic " + Utilities.base64Encode("dude@dude.com:"+pw)