[
{
"id": 1,
"subject": "FUNDAMENTALS OF INFORMATION TECHNOLOGY",
"present": "false",
"date": "2018-03-04",
"createdAt": "2018-03-04T14:06:18.000Z",
"updatedAt": "2018-03-04T14:06:18.000Z",
"SemesterId": 1,
"BatchId": 1,
"StudentId": 1,
"Student": {
"id": 1,
"rollnumber": 11111,
"studentname": "LAWANNA HAUGHT",
"dateofbirth": "1997-1-01",
"mobilenumber": "9875449112",
"createdAt": "2018-03-03T14:59:01.000Z",
"updatedAt": "2018-03-03T14:59:01.000Z",
"BatchId": 1
}
},
{
"id": 7,
"subject": "COMPUTER ORGANIZATION",
"present": "true",
"date": "2018-03-04",
"createdAt": "2018-03-04T14:09:12.000Z",
"updatedAt": "2018-03-04T14:09:12.000Z",
"SemesterId": 1,
"BatchId": 1,
"StudentId": 1,
"Student": {
"id": 1,
"rollnumber": 11111,
"studentname": "LAWANNA HAUGHT",
"dateofbirth": "1997-1-01",
"mobilenumber": "9875449112",
"createdAt": "2018-03-03T14:59:01.000Z",
"updatedAt": "2018-03-03T14:59:01.000Z",
"BatchId": 1
}
},
{
"id": 2,
"subject": "FUNDAMENTALS OF INFORMATION TECHNOLOGY",
"present": "true",
"date": "2018-03-04",
"createdAt": "2018-03-04T14:06:18.000Z",
"updatedAt": "2018-03-04T14:06:18.000Z",
"SemesterId": 1,
"BatchId": 1,
"StudentId": 2,
"Student": {
"id": 2,
"rollnumber": 11112,
"studentname": "KAILA HALBROOK",
"dateofbirth": "1997-1-01",
"mobilenumber": "9875449113",
"createdAt": "2018-03-03T14:59:01.000Z",
"updatedAt": "2018-03-03T14:59:01.000Z",
"BatchId": 1
}
},
{
"id": 8,
"subject": "COMPUTER ORGANIZATION",
"present": "false",
"date": "2018-03-04",
"createdAt": "2018-03-04T14:09:12.000Z",
"updatedAt": "2018-03-04T14:09:12.000Z",
"SemesterId": 1,
"BatchId": 1,
"StudentId": 2,
"Student": {
"id": 2,
"rollnumber": 11112,
"studentname": "KAILA HALBROOK",
"dateofbirth": "1997-1-01",
"mobilenumber": "9875449113",
"createdAt": "2018-03-03T14:59:01.000Z",
"updatedAt": "2018-03-03T14:59:01.000Z",
"BatchId": 1
}
}
]
这是一批的出勤记录。 我正在尝试将此数组转换为以下形式。
具有特定StudentId的所有记录必须与一个StudentId和studentname一起使用。
[
{
"StudentId": 1,
"studentname": "LAWANNA HAUGHT",
"rollnumber": 11111,
"attendance": [
{
"subject": "FUNDAMENTALS OF INFORMATION TECHNOLOGY",
"present": 0,
"absent": 1,
"total": 1
},
{
"subject": "COMPUTER ORGANIZATION",
"present": 1,
"absent": 0,
"total": 1
}
]
},
{
"StudentId": 2,
"studentname": "KAILA HALBROOK",
"rollnumber": 11111,
"attendance": [
{
"subject": "FUNDAMENTALS OF INFORMATION TECHNOLOGY",
"present": 1,
"absent": 0,
"total": 1
},
{
"subject": "COMPUTER ORGANIZATION",
"present": 0,
"absent": 1,
"total": 1
}
]
}
]
present是具有相同StudentId和subjectname且present = true
的行数缺少具有相同StudentId和subjectname且present = false
的行数total是具有相同StudentId和subjectname的行数。
这是我在一个StackOverflow答案和使用underscore.js的帮助下编写的代码。 但它找到了所有人的现有，缺席和总价值，而不是学生和主题。
var groups = _.groupBy(list, 'StudentId');
var result = _.map(groups, function(student){
return {
StudentId: student[0].StudentId,
studentname: student[0].Student.studentname,
rollnumber: student[0].Student.rollnumber,
attendances: _.map(student, function(item){
return {
subject:item.subject,
present: _.reduce(item, function(memo, att){
if(att.present=='true')
return memo + 1;
else
return memo;
}, 0),
absent: _.reduce(item, function(memo, att){
if(att.present=='false')
return memo + 1;
else
return memo;
}, 0),
total: _.reduce(item, function(memo, att){
return memo + 1;
}, 0)
}
})
})
}
}
);
答案 0 :(得分：1)
好的，这有点奇怪了。如果您正在编写后端编码，那么在查询端处理此问题可能是有意义的。话虽这么说，你可以使用reduce to＆＃34; collapse＆＃34;将数组转换为基于StudentId的对象。在此过程中，您可以构建记录的外观形状。我并不认为这是使用reduce的标准方法，因为您正在转换数组。话虽如此，我无法想到使用地图的简单方法。这个操作的结果给了我们一个对象，我们可以使用Object.values（）将它返回到一个数组。我不了解现在，缺席和总数的用例。
这是JSFiddle。
function isolateRecords(data) {
return Object.values(data.reduce((accum, current) => {
let attenance = {
subject: current.subject,
present: current.present === "true" ? 1 : 0,
absent: current.present === "false" ? 1 : 0,
total: 0
};
attenance.total = attenance.present + attenance.absent;
if (accum[current.StudentId]) {
accum[current.StudentId].attendance.push(attenance);
} else {
accum[current.StudentId] = {
StudentId: current.StudentId,
studentname: current.Student.studentname,
rollnumber: current.Student.rollnumber,
attendance: [attenance]
}
}
return accum;
}, {}));
}
isolateRecords(myOriginalArrayOfRecords);
答案 1 :(得分：1)
您可以通过寻找strunden以及稍后寻找出席来使用嵌套方法。
Eqvar data = [{ id: 1, subject: "FUNDAMENTALS OF INFORMATION TECHNOLOGY", present: "false", date: "2018-03-04", createdAt: "2018-03-04T14:06:18.000Z", updatedAt: "2018-03-04T14:06:18.000Z", SemesterId: 1, BatchId: 1, StudentId: 1, Student: { id: 1, rollnumber: 11111, studentname: "LAWANNA HAUGHT", dateofbirth: "1997-1-01", mobilenumber: "9875449112", createdAt: "2018-03-03T14:59:01.000Z", updatedAt: "2018-03-03T14:59:01.000Z", BatchId: 1 } }, { id: 7, subject: "COMPUTER ORGANIZATION", present: "true", date: "2018-03-04", createdAt: "2018-03-04T14:09:12.000Z", updatedAt: "2018-03-04T14:09:12.000Z", SemesterId: 1, BatchId: 1, StudentId: 1, Student: { id: 1, rollnumber: 11111, studentname: "LAWANNA HAUGHT", dateofbirth: "1997-1-01", mobilenumber: "9875449112", createdAt: "2018-03-03T14:59:01.000Z", updatedAt: "2018-03-03T14:59:01.000Z", BatchId: 1 } }, { id: 2, subject: "FUNDAMENTALS OF INFORMATION TECHNOLOGY", present: "true", date: "2018-03-04", createdAt: "2018-03-04T14:06:18.000Z", updatedAt: "2018-03-04T14:06:18.000Z", SemesterId: 1, BatchId: 1, StudentId: 2, Student: { id: 2, rollnumber: 11112, studentname: "KAILA HALBROOK", dateofbirth: "1997-1-01", mobilenumber: "9875449113", createdAt: "2018-03-03T14:59:01.000Z", updatedAt: "2018-03-03T14:59:01.000Z", BatchId: 1 } }, { id: 8, subject: "COMPUTER ORGANIZATION", present: "false", date: "2018-03-04", createdAt: "2018-03-04T14:09:12.000Z", updatedAt: "2018-03-04T14:09:12.000Z", SemesterId: 1, BatchId: 1, StudentId: 2, Student: { id: 2, rollnumber: 11112, studentname: "KAILA HALBROOK", dateofbirth: "1997-1-01", mobilenumber: "9875449113", createdAt: "2018-03-03T14:59:01.000Z", updatedAt: "2018-03-03T14:59:01.000Z", BatchId: 1 } }],
grouped = [];
data.forEach(({ subject, present, StudentId, Student: { rollnumber, studentname } }) => {
var student = grouped.find(g => StudentId === g.StudentId),
attendance;
if (!student) {
student = { StudentId, studentname, rollnumber, attendance: [] };
grouped.push(student);
}
attendance = student.attendance.find(a => a.subject === subject);
if (!attendance) {
attendance = { subject, present: 0, absent: 0, total: 0 };
student.attendance.push(attendance);
}
attendance[{ true: 'present', false: 'absent' }[present]]++;
attendance.total++;
});
console.log(grouped);