Question

我面临一个小问题的问题。我想使用JavaScript for循环以下面的格式创建一个JSON数组。

var mainJson = {
        "people": [{
                "name": "Ravi",
                "role": "team member",
                "appointments": [{
                        "info": "room 1",
                        "type": "Type01"
                        "title": "Meet John Miller"
                    }, {
                        "info": "room 2",
                        "type": "Type02"
                        "title": "Meet Mitchell"
                    }]
            ]
        }
    };

要创建以上JSON，我有2个jsons，如下所示..

var subJson1 = [{
            "RowId": "00272727",
            "Rowlabel": "Ravi",
            "role": "team member"  
}]

var subJson2 = [{
      "info": "room 1",
      "RowId": "00272727",
      "type": "Type01",
      "title": "Meet John Miller"
      }, {
      "info": "room 2",
      "type": "Type02",
      "RowId": "00272727",
      "title": "Meet Mitchell"
      }]

此处同时包含 SubJson1 和 subJson2 ＆＃34; RowID ＆＃34;是常见的参数。基于 RowId ，我们必须创建 mainJson 。在人员中的 mainJson 数组对象将超过1。我只是只显示一个项目以供理解。

我试图通过for循环来实现这一点。但我面临着解决这个问题的问题。有人可以帮助我使用 subJson1 和 subJson2

创建 mainJson

提前谢谢。

Answer 1

子JSON值是否总是以相同的格式出现？如果是这样，您可以跳过for循环的麻烦，只需通过函数传递值。

编辑：请参阅下面的编辑答案。

const subJson1 = [
    {
        "RowId": "00272727",
        "Rowlabel": "Ravi",
        "role": "team member"  
    }
]

const subJson2 = [
    {
    "info": "room 1",
    "RowId": "00272727",
    "type": "Type01",
    "title": "Meet John Miller"
    }, 
    {
    "info": "room 2",
    "type": "Type02",
    "RowId": "00272727",
    "title": "Meet Mitchell"
    }
]

const consolidateJSON = (subJson1, subJson2) => {
    return {
        people: [
            {
                name: subJson1[0].Rowlabel,
                role: subJson1[0].role,
                appointments: subJson2.map( item => {
                    const {info, type, title} = item;
                    return {
                        info,
                        type,
                        title
                    }
                } )
            }
        ]
    }
}

Answer 2

以下是使用ES6 select [ZRDocs].* from [ZRDocs] left join ([ZRDocItems] IN "D:\2222.mdb") AS abc on [ZRDocs].[ID] = [abc].[DocID];按RowId映射约会人员的方法：

const subJson1 = [{
  "RowId": "00272727",
  "Rowlabel": "Ravi",
  "role": "team member"  
}]

const subJson2 = [{
  "info": "room 1",
  "RowId": "00272727",
  "type": "Type01",
  "title": "Meet John Miller"
  }, {
  "info": "room 2",
  "type": "Type02",
  "RowId": "00272727",
  "title": "Meet Mitchell"
}]

/**
 * Get people, mapped with their appointments
 * 
 * @people {Object[]}
 * @appointments {Object[]}
*/
const getPeopleWithAppointments = (people, appointments) => ({
  people: people.map(person => ({
    ...person,
    appointments: appointments.filter(({ RowId }) => RowId === person.RowId)
  }))
})

console.log(getPeopleWithAppointments(subJson1, subJson2))

Answer 3

试试这个：

＆＃13;

var subJson1 = [{
  "RowId": "00272727",
  "Rowlabel": "Ravi",
  "role": "team member"  
}];

var subJson2 = [{
  "info": "room 1",
  "RowId": "00272727",
  "type": "Type01",
  "title": "Meet John Miller"
}, {
  "info": "room 2",
  "type": "Type02",
  "RowId": "00272727",
  "title": "Meet Mitchell"
}];

var mainJson = {"people" : []};

mainJson.people.push({"appointments" : subJson2.map(item => {
    delete item.RowId;
    return item;
  })
});

mainJson.people[0].name = subJson1[0].Rowlabel;
mainJson.people[0].role = subJson1[0].role;

console.log(mainJson);

＆＃13;

Answer 4

如果您只是尝试合并两个对象，那么有许多实用程序会执行此操作，我会遵循这些实用程序。因此this等问题也提供了几种方法。

但是，你似乎并不是在寻找这个。在您的情况下，您似乎想要创建一个具有新结构的新对象，并将其他两个对象插入其中。为此，您不会使用循环，而只需执行以下操作：

var subJson1 = {
    "RowId": "00272727",
    "Rowlabel": "Ravi",
    "role": "team member"  
}

var subJson2 = [
    {
        "info": "room 1",
        "type": "Type01"
        "title": "Meet John Miller"
    },
    {
        "info": "room 2",
        "type": "Type02"
        "title": "Meet Mitchell"
    }
];

var mainJson = {
    "people": [{
        "name": subJson1.Rowlabel,
        "role": subJson1.role,
        "appointments": subJson2
    }]
};

请记住;你不需要在Javascript（.js文件）中放置你的密钥，只能在JSON（.json文件）中。

javascript如何通过组合两个JSON来创建一个JSON

4 个答案: