我有这个数组,并希望只获得独特的课程和更新的DateStart。
const classes = [{
Course: {id: 1, Name: 'JS'},
Name: 'JS Morning',DateStart: "2018/09/01"
}, {
Course: {id: 1, Name: 'JS'},
Name: 'JS Afternoon',DateStart: "2018/10/15"
}, {
Course: {id: 1, Name: 'JS'},
Name: 'JS Night',DateStart: "2018/10/01"
}, {
Course: {id: 2, Name: 'Jquery'},
Name: 'JQ Morning',DateStart: "2018/10/01"
}, {
Course: {id: 2, Name: 'Jquery'},
Name: 'JQ Night',DateStart: "2018/09/15"
}];
每个例子,我的结果数组应该是:
const results = [{
DataStart:"2018/09/01",Course{id: 1, Name: "JS"}
},{
DataStart:"2018/09/15",Course{id: 2, Name: "Jquery"}
我知道使用foreach这样做,但我想使用MAP或REDUCE或FILTER。有人能帮助我吗?
谢谢!!!亚历克斯
我们可以按日期先做一个sort()
,在这种情况下更容易比较日期作为数字,取代/
像20180901
,一旦我们有排序数组我们可以做filter()
只保留Course.Name
的第一个记录。
var classes = [{
Course: {id: 1, Name: 'JS'},
Name: 'JS Morning',DateStart: "2018/09/01"
}, {
Course: {id: 1, Name: 'JS'},
Name: 'JS Afternoon',DateStart: "2018/10/15"
}, {
Course: {id: 1, Name: 'JS'},
Name: 'JS Night',DateStart: "2018/10/01"
}, {
Course: {id: 2, Name: 'Jquery'},
Name: 'JQ Morning',DateStart: "2018/10/01"
}, {
Course: {id: 2, Name: 'Jquery'},
Name: 'JQ Night',DateStart: "2018/09/15"
}];
var diff = {}
results = classes.sort((a,b)=>{
return a.DateStart.replace(/\//g, "") - b.DateStart.replace(/\//g, "");
}).filter(a=>{
delete a.Name
if(a.Course.Name in diff){
return false;
}else{
diff[a.Course.Name] = true;
return true;
}
})
console.log(results)
var classes = [{
Course: {id: 1, Name: 'JS'},
Name: 'JS Morning',DateStart: "2018/09/01"
}, {
Course: {id: 1, Name: 'JS'},
Name: 'JS Afternoon',DateStart: "2018/09/01"
}, {
Course: {id: 1, Name: 'JS'},
Name: 'JS Night',DateStart: "2018/09/01"
}, {
Course: {id: 2, Name: 'Jquery'},
Name: 'JQ Morning',DateStart: "2018/09/15"
}, {
Course: {id: 2, Name: 'Jquery'},
Name: 'JQ Night',DateStart: "2018/09/15"
}];
var result = classes.map(x => { return { DateStart: x.DateStart, Course: x.Course } }).filter((x,i,a)=> i == a.findIndex(y => y.DateStart == x.DateStart && y.Course.id == x.Course.id && y.Course.Name == x.Course.Name));
console.log(result);
您可以使用array#reduce
使用id
获取独特的课程。如果你得到一个重复的课程,那么比较DateStart
并存储先前的值。获得对象累加器中的所有唯一课程后,使用Object.values()
获取所有值。
const classes = [{ Course: {id: 1, Name: 'JS'}, Name: 'JS Morning',DateStart: "2018/09/01" }, { Course: {id: 1, Name: 'JS'}, Name: 'JS Afternoon',DateStart: "2018/10/15" }, { Course: {id: 1, Name: 'JS'}, Name: 'JS Night',DateStart: "2018/10/01" }, { Course:{id: 2, Name: 'Jquery'}, Name: 'JQ Morning',DateStart: "2018/10/01" }, { Course: {id: 2, Name: 'Jquery'}, Name: 'JQ Night',DateStart: "2018/09/15" }],
result = Object.values(classes.reduce((r,{Course, DateStart}) => {
if(r[Course.id]) {
if(r[Course.id].DateStart > DateStart)
r[Course.id] = {Course, DateStart};
} else
r[Course.id] = {Course, DateStart};
return r;
},{}));
console.log(result);
您可以使用以下解决方案使用map通过属性从数组中获取唯一值:
classes.map( c => ({ DataStart: c.DateStart, Course: c.Course }))