如何从JSON列表中获取特定数据

Question

我有这个JSON

{
  "doctors": [
    {
      "id": 8,
      "schedules": [
        {
          "id": 8,
          "totime": "11:17",
          "dayId": 2,
          "location": "Somajiguda",
          "fromtime": "10:17",
          "hospitalId": 5,
          "day": "Tuesday",
          "hospital": "Yashoda"
        }
      ],
      "username": "d1",
      "degree": "DA(Anaesthesia)",
      "email": "1@2.com",
      "imagePath": "",
      "department": "Bio-Chemistry",
      "name": "d1",
      "userid": 51,
      "gender": "Male",
      "mobile": "1234567900"
    },
    {
      "id": 10,
      "schedules": [
        {
          "id": 10,
          "totime": "12:35",
          "dayId": 2,
          "location": "Somajiguda",
          "fromtime": "11:35",
          "hospitalId": 5,
          "day": "Tuesday",
          "hospital": "Yashoda"
        }
      ],
      "username": "d3",
      "degree": "BDS",
      "email": "d3@d3.com",
      "imagePath": "",
      "department": "Bio-Chemistry",
      "name": "d3",
      "userid": 56,
      "gender": "Male",
      "mobile": "1234567890"
    },
    {
      "id": 1,
      "schedules": [
        {
          "id": 1,
          "totime": "12:55",
          "dayId": 1,
          "location": "Somajiguda",
          "fromtime": "11:55",
          "hospitalId": 5,
          "day": "Monday",
          "hospital": "Yashoda"
        }
      ],
      "username": "doctor",
      "degree": "BDS",
      "email": "",
      "imagePath": null,
      "department": "Critical Care",
      "name": "doctor",
      "userid": 4,
      "gender": "Male",
      "mobile": "1234567890"
    },
    {
      "id": 7,
      "schedules": [
        {
          "id": 7,
          "totime": "11:17",
          "dayId": 2,
          "location": "Somajiguda",
          "fromtime": "11:17",
          "hospitalId": 5,
          "day": "Tuesday",
          "hospital": "Yashoda"
        }
      ],
      "username": "donald",
      "degree": "DA(Anaesthesia)",
      "email": "donald@doctor.com",
      "imagePath": "",
      "department": "Bio-Chemistry",
      "name": "donald",
      "userid": 47,
      "gender": "Male",
      "mobile": "1234567989"
    },
    {
      "id": 6,
      "schedules": [
        {
          "id": 6,
          "totime": "11:15",
          "dayId": 1,
          "location": "Somajiguda",
          "fromtime": "11:15",
          "hospitalId": 5,
          "day": "Monday",
          "hospital": "Yashoda"
        }
      ],
      "username": "john",
      "degree": "BDS",
      "email": "john@john.com",
      "imagePath": null,
      "department": "Anesthesiology",
      "name": "john",
      "userid": 46,
      "gender": "Male",
      "mobile": "1234567890"
    },
    {
      "id": 5,
      "schedules": [
        {
          "id": 5,
          "totime": "13:11",
          "dayId": 2,
          "location": "Somajiguda",
          "fromtime": "12:11",
          "hospitalId": 5,
          "day": "Tuesday",
          "hospital": "Yashoda"
        }
      ],
      "username": "sknayak",
      "degree": "BDS",
      "email": "sknayak@sknayak.com",
      "imagePath": "",
      "department": "Anesthesiology",
      "name": "sknayak",
      "userid": 38,
      "gender": "Male",
      "mobile": "1234567890"
    },
    {
      "id": 2,
      "schedules": [
        {
          "id": 2,
          "totime": "16:26",
          "dayId": 6,
          "location": "Somajiguda",
          "fromtime": "15:26",
          "hospitalId": 5,
          "day": "Saturday",
          "hospital": "Yashoda"
        }
      ],
      "username": "drsukant",
      "degree": "BDS",
      "email": "",
      "imagePath": null,
      "department": "Anesthesiology",
      "name": "sukant",
      "userid": 9,
      "gender": "Male",
      "mobile": "1234567890"
    }
  ]
}

在这个JSON中有一个唯一的字段ID。我通过一个像这样的ajax获取这个json

var test=$.ajax({  
    type: "GET", 
    url: projectUrl+"getDoctors",  
    dataType:"json",
    jsonp: true,
    async:false 
}).responseText;
console.log(test);

正如您在JSON中看到的那样，有一个字段id。例如id = 8 username是d1，id = 10，username是d3。我将id存储在会话中。例如，如果id是8，那么我只想要ID为8的那些详细信息（用户名d1，电子邮件1@2.com .....）。

那么如何将JSON过滤为特定值。

Answer 1

您可以为特定项目创建一个计算式 ：

self.doctors = ko.observableArray();
self.d3Doctor = ko.computed(function() {
    return ko.utils.arrayFirst(self.doctors(), function(obj) {
        return obj.id === 8;
    });
});

现在，您只需要担心填充doctors observableArray即可：

$.getJSON(projectUrl+"getDoctors", function(response) {
    ko.utils.arrayPushAll(yourViewModel.doctors, response.doctors);
});

这允许以下内容：

<div data-bind="if: d3Doctor()">
    <h3>Doctor with ID 8</h3>
    <p>Name: <span data-bind="text: d3Doctor().name"></span></p>
    <p>E-mail: <span data-bind="text: d3Doctor().email"></span></p>
    <p>Address: <span data-bind="text: d3Doctor().address"></span></p>
    ...
</div>

Answer 2

您可以使用each来查找它，请尝试以下操作：

$(document).ready(function() {
    $.ajax({
        type: "GET",
        url: projectUrl+"getDoctors",
        dataType: "json",
        jsonp: true,
        async: false
    }).done(function(data) {
        $.each(data.doctors, function(i, v) {
            if (v.id == '8') {
                console.log('found', v.username, v.email);
                return false;
            }
        });
    });
});

Answer 3

可能是这个吗？：

function findById(data, id){
     for(var i=0;i<data.length;i++){
          if(data[i].id === id) return { username:data[i].username, email:data[i].email};
     }
     // Or, you can use $.grep like this:
     // var foundDoctors = $.grep(data,function(e){return e.id === id;})
     // return foundDoctors.length > 0 && foundDoctors[0] || null;

     // Or, you can use filter() method from Array (IE 9+)
}

findById(jsondata["doctors"], 8);

Answer 4

是。 我可以回答。

以我的方式，我个人更喜欢将所有数据转换为数组并进行迭代或操作。

如本问题所述 ，我们已经将数据推送为以下格式：

doctors.push({id:currPat.id,name:currPat.username});

因此，现在我可以对doctors数组使用数组filter功能：

var result = doctors.filter(function(currentObject) {
    return currentObject.id === 8;
});

console.log(result); // {id: 8, name:d1}

或者，您也可以在JQuery中使用.map() ，只需使该函数检查data.id并返回所需的对即可。

而且，如果您要争取nano性能。 我想我的方法比.map()更好