Google Apps 脚本和外部 API 授权在标头中失败

Question

试图让它发挥作用。我在参数列表后不断收到 Missing ) 。（第 6 行，文件“Code”）驳回。我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。

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。也感谢布莱恩。我终于意识到您需要一个“选项”对象来传递方法和标头。

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);
}