根据Google表格电子表格中的动态数据更新Google表单字段 [英] Update Google Forms field based on dynamic data from a Google Sheets spreadsheet

问题描述

在 Google表单中，我们有一个下拉选择列表.

In a Google Form we have a pull down selection list.

项目列表可能会在每分钟之间变化.数据的主副本位于google docs电子表格的一列中.

The list of items may change from minute to minute. The master copy of the data is in a column in a google docs spreadsheet.

当用户访问表单时，我们希望他们看到下拉列表的最新副本，就像电子表格中的最新副本一样.

When a user accesses the form, we want them to see the latest copy of the pull down list as is current in the spreadsheet.

我没有看到使用触发器执行此操作的明显方法.例如，onOpen仅触发表单管理员，而不触发用户.有什么方法可以让我自动保持此表单的最新状态?

I don't see an obvious way to do this with triggers. onOpen, for example, only triggers for administrators of the form, not users. Is there a way I can keep this form up-to-date automatically?

这是我第一次使用Google文档，但是我在Excel脚本编写方面经验丰富.

This is my first time in google docs, but I'm very experienced in excel scripting.

推荐答案

工作流程与您所描述的有所不同.由于Google Apps脚本无法与访问表单的最终用户进行交互，因此无法对此做出反应.但是，您可以对电子表格中的更改做出反应，并更新表单选项以为下一个最终用户做准备.

The work flow needs to be a bit different than what you've described. Since there is no ability for Google Apps Script to interact with an end user accessing a Form, there's no way to react to that. However, you can react to changes in the spreadsheet, and update the form options in preparation for the next end user.

• 根据电子表格中选项的更新方式，您可以在此处使用可安装的触发器来对编辑或更改做出反应.

• Depending on how the options get updated in the spreadsheet, you can use an installable trigger there, to react to an edit or a change.

如果该表单与包含选项的电子表格相关联，则可以使用 FormApp.openById() 或.openByUrl().

If the form is associated with the spreadsheet containing the choices, you can get a handle on the form using Spreadsheet.getFormUrl(), then pass the URL to FormApp.openByUrl(). Otherwise you will need to provide the ID or URL explicitly to FormApp.openById() or .openByUrl(), respectively.

该电子表格触发器功能可以使用 ListItem.setChoiceValues() .

That spreadsheet trigger function can modify the available selections for the relevant question by using ListItem.setChoices() or ListItem.setChoiceValues().

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