使用映射将数据保存在 Firebase 中答案

【问题标题】：Saving data in firebase with mapping使用映射将数据保存在 Firebase 中
【发布时间】：2019-12-22 07:18:57
【问题描述】：

我只是将数组传递给 firebase 来存储数据，但这不是从数组中查询数据的有效方法。所以，我打算以映射格式存储数据。但不知道该怎么做。

{
   "ClassDetails": 
   [
     {
       "Code": "CIS695",
       "Instructor": "Dr. Antony",
       "Location": "BB444",
       "Name": "Capstone project"
     }
   ],
   "PersonalDetials": [
     {
       "Email": "rshah2@murraystate.edu",
       "FName": "Ravi",
       "LName": "Shah",
       "Major": "MSIS"
     }
   ]
 }

想改成这种格式

{
   "ClassDetails": 
     "CIS695"{
       "Code": "CIS695",
       "Instructor": "Dr. Antony",
       "Location": "BB444",
       "Name": "Capstone project"
       }
       "CIS690"{
       "Code": "CIS690",
       "Instructor": "Dr. Smith",
       "Location": "BB444",
       "Name": "Project Management"
       }
   "PersonalDetials": 
     {
       "Email": "rshah2@murraystate.edu",
       "FName": "Ravi",
       "LName": "Shah",
       "Major": "MSIS"
     }
 }

尝试了所有方法，但找不到合适的解决方案。Data structure in firebase

【问题讨论】：

标签： angular typescript firebase ionic-framework google-cloud-firestore

【解决方案1】：

考虑到data的结构不会改变，试试这个：

const data = {
  "ClassDetails": [{
    "Code": "CIS695",
    "Instructor": "Dr. Antony",
    "Location": "BB444",
    "Name": "Capstone project"
  }],
  "PersonalDetials": [{
    "Email": "rshah2@murraystate.edu",
    "FName": "Ravi",
    "LName": "Shah",
    "Major": "MSIS"
  }]
};

const mappedData = { ...data
};
mappedData.ClassDetails = {};
for (const item in data.ClassDetails) {
  const itemToAdd = { ...data.ClassDetails[item]
  };
  mappedData.ClassDetails[itemToAdd.Code] = itemToAdd;
}
mappedData.PersonalDetials = mappedData.PersonalDetials[0];

console.log(mappedData);

【讨论】：

【解决方案2】：

试试这样：

  output = {};
  constructor() {
    let classDetails = {};
    this.input.ClassDetails.forEach(item => {
      classDetails[item.Code] = item;
    });
    this.output["ClassDetails"] = classDetails;
    this.output["PersonalDetials"] = this.input.PersonalDetials[0];
    console.log(this.output);
  }

Working Demo

【讨论】：