javascript - 当尝试在 Google Sheet 脚本中使用 e.value 时，我得到 TypeError : Cannot read property "Values" from undefined. (第 7 行，文件 "Code")

Question

这是一个由两部分组成的问题:

1. 我有带有链接表单的 Google 表格。提交表单后，我希望将表单中的响应复制到另一个谷歌表格中，我打算在其中进行更改和重新格式化，然后通过电子邮件发送。下面的脚本是我当前编写的，它在FormSubmit 上设置了一个触发器。但是，我不断收到以下错误:

类型错误:无法从未定义中读取属性“值”。 (第 7 行，文件“代码”)

代码如下:

function formSubmitReply(e)
{    
var t = "1g-wIs6nGxu3mJYA1vKtPCxBLCsvh1upeVGbCokOOTIw"; 
var tName = "AggregationOutput";

//Get information from form and set as variables
  var email = e.Values[2];
  var name = e.Values[3];

// Get template, copy it as a new temp, and save the Doc’s id
var tcopyId = SpreadsheetApp.openById(t).copy(tName+' for '+name).getId();

// Open the temporary document & copy form responses into template copy response sheet
var copyt = SpreadsheetApp.openById (tcopyId);
var copyts = copyt.getSheetByName('Resp');  

// Transfers Data from Form Responses to Temporary file
copyts.getRange('A3').setValue(name);

//Sends copy of template in an email as an excel file
var url = "https://docs.google.com/feeds/download/spreadsheets/Export?key=" + copyt.getId();  
var subject = 'Aggregaton Output for' + name;
var body = url   
MailApp.sendEmail(email, subject, body);

// Deletes temp file
DriveApp.getFileById(tcopyId).setTrashed(true);
}

我的问题的第二部分，即使我可以让代码工作，当在表单中跳过问题时，您会建议什么 - 这不会改变 e.values 中的数组吗？使用最后一行作为问题是我希望人们返回并编辑表单上的响应，然后重新提交，这意味着最后一行并不总是使用的行。

感谢任何和所有帮助。

Answer 1

对于第 1 部分，请尝试以下操作:

function formSubmitReply(e)
{    
var t = "1g-wIs6nGxu3mJYA1vKtPCxBLCsvh1upeVGbCokOOTIw"; 
var tName = "AggregationOutput";

//Get information from form and set as variables
var itemResponses = e.response.getItemResponses();

  var email = itemResponses[2].getResponse();
  var name = itemResponses[3].getResponse();

// Get template, copy it as a new temp, and save the Doc’s id
var tcopyId = SpreadsheetApp.openById(t).copy(tName+' for '+name).getId();

// Open the temporary document & copy form responses into template copy response sheet
var copyt = SpreadsheetApp.openById (tcopyId);
var copyts = copyt.getSheetByName('Resp');  

// Transfers Data from Form Responses to Temporary file
copyts.getRange('A3').setValue(name);

//Sends copy of template in an email as an excel file
var url = "https://docs.google.com/feeds/download/spreadsheets/Export?key=" + copyt.getId();  
var subject = 'Aggregaton Output for' + name;
var body = url   
MailApp.sendEmail(email, subject, body);

// Deletes temp file
DriveApp.getFileById(tcopyId).setTrashed(true);
}