按名称打开Goog​​le文档电子表格 [英] Open Google Docs Spreadsheet by name

问题描述

我有一种情况，脚本正在输入数据并将其发送到电子表格。过了一会儿，这张电子表格变得太大了。

现在我们必须手动将项目从主电子表格移动到新电子表格。原因是不是每个人都熟悉代码，并且愿意更改代码中的ID。

我想知道是否有办法按名称打开电子表格。如果没有，是否有更好的方式来实现我们需要的内容（如上所述）？解决方案

更新：DocsList现在已经日落。使用 DriveApp.getFilesByName（

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David提供了一些很好的代码来洗牌周围的数据。如果你真的需要的只是通过名字打开一个电子表格，那么这将做到这一点：

  function getSpreadsheetByName（filename） {
 var files = DocsList.find（filename）; 
 
 for（var i in files）
 {
 if（files [i] .getName（）== filename）
 {
 // open - 无证函数
返回SpreadsheetApp.open（files [i]）; 
 // openById  - 记录但更详细的
 //返回SpreadsheetApp.openById（files [i] .getId（））; 
} 
} 
返回null; 
} 
  

I have a situation where a script is taking input data and sending it to a spreadsheet. After a while, this spreadsheet becomes too big.

Right now we have to manually move the items from the the primary spreadsheet to a new one. The reason is that not everyone is familiar with the code and are willing to change the ID in the code.

I would like to know if there is a way to open the spreadsheet by name. If not, is there a better way of achieving what we need (described above)

解决方案

Update: DocsList has now been sunset. Use DriveApp.getFilesByName(name) instead.

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David has provided some good code for shuffling your data around. If all you really did need was just to open a spreadsheet by name then this will do the trick:

function getSpreadsheetByName(filename) {
  var files = DocsList.find(filename);

  for(var i in files)
  {
    if(files[i].getName() == filename)
    {
      // open - undocumented function
      return SpreadsheetApp.open(files[i]);
      // openById - documented but more verbose
      // return SpreadsheetApp.openById(files[i].getId());
    }
  }
  return null;
}

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