如何在nodejs上使用对象json数据进行嵌套循环答案

【问题标题】：How to do nested loops with object jSON data on nodejs如何在nodejs上使用对象json数据进行嵌套循环
【发布时间】：2019-01-12 03:46:18
【问题描述】：

我是使用 nodejs 和 mongodb 的新手，现在我使用 nodejs 和 mongodb 构建 restful API。我想用http://jsonapi.org 标准为我的API 使用响应标准。

我需要提前建议如何最好地做到这一点，使我的 API 响应类似于以下 JSON 数据：

HTTP/1.1 200 OK
Content-Type: application/vnd.api+json

{
  "links": {
    "self": "http://example.com/users"
  },
  "data": [{
    "type": "users",
    "id": 5b647bb8998248235a0aab3c,
    "attributes": {
      "username": "luke",
      "email": "luke@mail.com",
      "password": "password",
      "hashpassword":"",
      "oldpassword":"$2a$10$eyt6YV6m2JJrebNxvS0iEuxMCubDXratNJ6/XK797IGvepXBdp9Yq",
      "salt":"2q6eN9U0vWFBsIF1MtB5WrgPiB8pldTS",
      "usertype":"bisnis",
      "userstatus":"not aktif"
    }
  }, {
    "type": "users",
    "id": 5b647bdf998248235a0aab3d,
    "attributes": {
      "username": "ken",
      "email": "ken@mail.com",
      "password": "password",
      "hashpassword":"",
      "oldpassword":"$2a$10$eyt6YV6m2JJrebNxvS0iEuxMCubDXratNJ6/XK797IGvepXBdp9Yq",
      "salt":"2q6eN9U0vWFBsIF1MtB5WrgPiB8pldTS",
      "usertype":"bisnis",
      "userstatus":"not aktif"
    }
  }]
}

如上所述创建输出时，我遇到了嵌套迭代问题。这是我的代码：

const UsersModel = mongoose.model('Users');

//...show list user
exports.listUsers = (req, res) => {
  UsersModel.find({}, (err, result) => {
    if (err) {
      res.send({code: 400, failed: 'Error 400'});
    }
    res.json(result);
  });
};

这是我的结果 JSON：

[
    {
        "type": "users",
        "userstatus": "not aktif",
        "_id": "5b647bb8998248235a0aab3c",
        "username": "luke",
        "email": "luke@mail.com",
        "password": "password",
        "usertype": "bisnis",
        "hashpassword": "$2a$10$eyt6YV6m2JJrebNxvS0iEuxMCubDXratNJ6/XK797IGvepXBdp9Yq",
        "salt": "2q6eN9U0vWFBsIF1MtB5WrgPiB8pldTS",
        "__v": 0
    },
    {
        "type": "users",
        "userstatus": "tidak aktif",
        "_id": "5b647bdf998248235a0aab3d",
        "username": "ken",
        "email": "ken@mail.com",
        "password": "password",
        "usertype": "personal",
        "hashpassword": "$2a$10$hok988mszyIBiXVNjmfifOiPNzXkBRRRynXJS/0qCkvlaBOQs65MO",
        "salt": "IiMvtVYVqTpZFXmYQIM4IlS6PJFVZ3kw",
        "__v": 0
    }
]

这是我解决问题的临时代码。

//...show list user
exports.listUsers = (req, res) => {
    UsersModel.find({}, (err, result) => {
      if (err) {
        res.send({code: 400, failed: 'Error 400'});
      }

      let listData  = [];

      for (let key in result) {
        let data = {};
        let attr = {};

        if (result.hasOwnProperty(key)) {
          data.type = result[key].type;
          data.id = result[key]._id;

          for(let i in result[key]) {

            if(result[key].hasOwnProperty(i)) {
              attr.username = result[key].username;
              attr.email = result[key].email;
              attr.password = result[key].password;
              attr.hashpassword = result[key].hashpassword;
              attr.oldpassword = result[key].oldpassword;
              attr.salt = result[key].salt;
              attr.usertype = result[key].usertype;
              attr.userstatus = result[key].userstatus;
            }
          }

          data.attribute = attr;
          listData.push(data);
        }

      }

      let collections = {
        "meta": {
          "copyright": "Copyright 2018 Kotakku Studio and Lab",
          "authors": [
            "sw. saputra"
          ]
        },
        "link": {
          "self": req.protocol + '://' + req.get('host') + req.originalUrl
        },
        "data": listData
      }

      res.json(collections);
    });
};

如果我的临时代码不正确，请给我建议优雅和最好的方法来解决我的问题。

提前感谢。

【问题讨论】：

您可以使用map 函数并从find 回调中提取所有代码。您可以将 map 函数应用于 result 变量并删除 for，因为 map 函数会遍历整个数组。看地图文档developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

标签： javascript json node.js mongodb express

【解决方案1】：

您需要根据您的要求创建一个合适的 mongoose 架构并将 res 映射到该架构。

//Schema for users
var UsersSchema = mongoose.Schema({
 links: {
   self: String 
 },
  data:[{
    type: String,
    attributes: {
      username: String,
      .
      .
      .
    }
  }]
})

数组中每个文档的 id 将由 MongoDB 自动创建。对于猫鼬模式文档，请参阅http://mongoosejs.com/docs/guide.html 您也可以查看这个简单的 TODO API 以供参考https://github.com/mkujaggi/node-course-todo-api

【讨论】：

【解决方案2】：

在这个条件的陈述中：

if(result[key].hasOwnProperty(i)) {

您应该访问i 的result[key]：

        if(result[key].hasOwnProperty(i)) {
          attr.username = result[key][i].username;
          attr.email = result[key][i].email;
          attr.password = result[key][i].password;
          attr.hashpassword = result[key][i].hashpassword;
          attr.oldpassword = result[key][i].oldpassword;
          attr.salt = result[key][i].salt;
          attr.usertype = result[key][i].usertype;
          attr.userstatus = result[key][i].userstatus;
        }

而且您不需要所有这些.hasOwnProperty 检查，除非您正在修改本机原型，而您可能不应该这样做。特别是对于Array，如果您只是使用for 而不是for-in 正确循环，则没有必要。

exports.listUsers = (req, res) => {
UsersModel.find({}, (err, result) => {
    if (err) {
      res.send({
        code: 400,
        failed: 'Error 400'
      });
    }

    let listData = [];

    let data = {};
    let attr = {};

    data.type = result[key].type;
    data.id = result[key]._id;

    for (let i = 0; i < result[key].length; i++) {

        attr.username = result[key][i].username;
        attr.email = result[key][i].email;
        attr.password = result[key][i].password;
        attr.hashpassword = result[key][i].hashpassword;
        attr.oldpassword = result[key][i].oldpassword;
        attr.salt = result[key][i].salt;
        attr.usertype = result[key][i].usertype;
        attr.userstatus = result[key][i].userstatus;
    }

    data.attribute = attr;
    listData.push(data);
  }

}

let collections = {
  "meta": {
    "copyright": "Copyright 2018 Kotakku Studio and Lab",
    "authors": [
      "sw. saputra"
    ]
  },
  "link": {
    "self": req.protocol + '://' + req.get('host') + req.originalUrl
  },
  "data": listData
}

res.json(collections);
});
};

【讨论】：

哦，我没有注意到数组在 JSON 响应的顶层。应该删除外循环。

【解决方案3】：

我会通过迭代键来做这样的事情（即使我们不知道键名，这也会动态工作）

var test=[
    {
        "type": "users",
        "userstatus": "not aktif",
        "_id": "5b647bb8998248235a0aab3c",
        "username": "luke",
        "email": "luke@mail.com",
        "password": "password",
        "usertype": "bisnis",
        "hashpassword": "$2a$10$eyt6YV6m2JJrebNxvS0iEuxMCubDXratNJ6/XK797IGvepXBdp9Yq",
        "salt": "2q6eN9U0vWFBsIF1MtB5WrgPiB8pldTS",
        "__v": 0
    },
    {
        "type": "users",
        "userstatus": "tidak aktif",
        "_id": "5b647bdf998248235a0aab3d",
        "username": "ken",
        "email": "ken@mail.com",
        "password": "password",
        "usertype": "personal",
        "hashpassword": "$2a$10$hok988mszyIBiXVNjmfifOiPNzXkBRRRynXJS/0qCkvlaBOQs65MO",
        "salt": "IiMvtVYVqTpZFXmYQIM4IlS6PJFVZ3kw",
        "__v": 0
    }
]
var listData=[]
test.forEach((obj)=>{
   var tempObj={attributes:{}}; //initialize tempObj
   Object.keys(obj).forEach((key)=>{
     if(key=="_id" || key=="type"){
      tempObj[key]=obj[key];
     }else{
      tempObj.attributes[key]=obj[key];
      }
   })
   listData.push(tempObj);
})

let collections = {
        "meta": {
          "copyright": "Copyright 2018 Kotakku Studio and Lab",
          "authors": [
            "sw. saputra"
          ]
        },
        "link": {
          "self": 'test url'
        },
        "data": listData
      }
console.log(collections)

【讨论】：