将JSON响应转换为Google表格（Google Apps脚本）

Question

我一直试图通过以下代码在Google Apps脚本中将JSON响应插入到Google工作表中，但由于某种原因，我在尝试运行时遇到错误。

请查看截图及以下代码。

    function myFunction() {

  var key_67 = 'YYYYYYYYYYYYYYYYYY';
  var ss_67 = SpreadsheetApp.openById(key_67);
  var sheet_67 = ss_67.getActiveSheet();
  sheet_67.getRange('A1:AZ10000').clearContent();
 var url = 'https://creator.zoho.com/api/json/arfater/view/Leads_Report?authtoken=XXXXXXXXXXXXXXXXXXXX&scope=creatorapi&zc_ownername=ipekuet';
 var response = UrlFetchApp.fetch(url);
var json = response.getContentText();
var data = JSON.parse(json);
Logger.log(data);



  var stats=[]; //create empty array to hold data points



  //The following lines push the parsed json into empty stats array

    stats.push(data.Yearly_Sales); //temp
    stats.push(data.Email); //dewPoint
    stats.push(data.Phone); //visibility

  //append the stats array to the active sheet 
  sheet_67.appendRow(stats)

}

Answer 1

所以基于邮递员应用程序的JSON响应是

var zohoipekuetview65 = {"Leads":[{"Yearly_Sales":"$ 1,000.00","Email":"test@zoho.com","Phone":"123-032-03323","P‌otentially":50,"Stat‌e":"NY","ZipCode":"1‌​0036","Street":"1515 Broadway","Country":"USA","ID":"2198633000000063029","City":‌​"New York","Name":"Arfater Rahman"}]};

当我按原样使用该响应时：

function JsonResponse(){
 var json ='var zohoipekuetview65 = {"Leads":[{"Yearly_Sales":"$ 1,000.00","Email":"test@zoho.com","Phone":"123-032-03323","P‌​otentially":50,"Stat‌​e":"NY","ZipCode":"1‌​0036","Street":"1515 Broadway","Country":"USA","ID":"2198633000000063029","City":‌​"New York","Name":"Arfater Rahman"}]} '
 var data = JSON.parse(json);
Logger.log(data); 
}

我得到了和你一样的错误：

  

SyntaxError：意外的令牌：v

这让我相信您对API的回复有这个术语var zohoipekuetview65（不确定为什么？也许是一个错误）

以下代码将响应字符串拆分为仅提供JSON响应

function trialParse(){
var json ='var zohoipekuetview65 = {"Leads":[{"Yearly_Sales":"$ 1,000.00","Email":"test@zoho.com","Phone":"123-032-03323","P‌​otentially":50,"Stat‌​e":"NY","ZipCode":"1‌​0036","Street":"1515 Broadway","Country":"USA","ID":"2198633000000063029","City":‌​"New York","Name":"Arfater Rahman"}]} '
Logger.log(JsonResponse(json))
}
function JsonResponse(response){
  Logger.log(response)
  var json = response.split("=")[1]
  var data = JSON.parse(json);
  Logger.log(data);
  return data  
}

只需使用var data = JsonResponse(json)

在代码中调用上述函数即可

最后注意：如Jordan Rhea所述，您可以使用Logger.log(json)输出对日志的响应。要查看日志，请转到视图＆gt;日志，它会显示您从Api收到的回复。