Firebase数据库规范化

问题描述

所以我做了一个简单的应用程序，一个三轮车巡逻应用程序，你可以报告鲁莽的三轮车司机（我们这里没有一个问题在我们的城市）登录和填写表格。报表包含：

   -  created_at 
  -  description 
  -  lat 
  -  lng 
  -  plateNumber 
  

到目前为止，这是我的json的结构：

有没有更干净的方法来做到这一点？到目前为止，它的工作，虽然例如，我想通过 plateNumber 的所有报告，我认为这将是棘手。

PS：我知道这不是一个关系数据库，虽然我是开放的建议

规范化数据库...当我遇到这个问题时，我停止编写代码并开始考虑我的应用程序/ web。 ...

• 我想显示所有报表的列表，按板块排序的另一个列表并由其他人订购


• 或者在所有报告列表中按板块？

排序

如果您的答案是第一个，那么您需要生成节点作为索引，如下所示：

 报告{
 report_id {
 created_at：
描述：
 lat：
 lang：
 locati on：
 plate_number：
} 
} 
 reports_idx_plate {
 plate_number1 {
 report_id1：true
 report_id2： true
 report_id3：true
 report_id4：true
} 
} 
  

如果您的答案是第二个，那么您的最佳选择就是这样：

 报告{
 plate_number {
 report_id {
 created_at：
 description：
 lat：
 lang： 
 location：
 plate_number：
} 
} 
} 
  

在简历中，如果您需要从子节点获取列表，则需要规范化节点并生成节点作为索引。
让我知道，如果我已经帮助你，并且编程的很好！

So I made a simple app, a Tricycle Patrol app which you can report reckless tricycle drivers (no 1 problem here in our city) by logging in and filling up forms. The report form contains:

- created_at
- description
- lat
- lng
- plateNumber

And so far, this is the structure of my json:

Is there a cleaner way to do this? So far it works though for example I want to get all reports by plateNumber I think it would be tricky.

Any suggestions would be much appreciated.

PS: I know it's not a relational database though I'm open for suggestions

解决方案

Now as you say you need normalize database ... When I have this problem, I stop writing code and start thinking about my application / web ...

• I want to show a list of all reports, another list sorted by plate and ordered by others?
• Or in all lists of reports are sorted by plate?

If your answer is the first, then you need to generate nodes as indexes, like these:

reports{
    report_id{
        created_at:" "
        description:" "
        lat:" "
        lang:" "
        location:" "
        plate_number:" "
    }
}
reports_idx_plate{
    plate_number1{
        report_id1:"true"
        report_id2:"true"
        report_id3:"true"
        report_id4:"true"
    }
}

If your answer is the second one, then your best option is this one:

reports{
    plate_number{
         report_id{
            created_at:" "
            description:" "
            lat:" "
            lang:" "
            location:" "
            plate_number:" "
        }
    }
}

In resume if you need to get list from child node, you need to normalize the nodes and generate the nodes as index.Let me know if I have helped you and good programming!

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