이와 같은 JavaScript 클래스가 있다고 가정 해 봅시다.
var DepartmentFactory = function(data) {
this.id = data.Id;
this.name = data.DepartmentName;
this.active = data.Active;
}
그런 다음 해당 클래스의 인스턴스를 여러 개 만들어서 배열에 저장한다고 가정 해 봅시다.
var objArray = [];
objArray.push(DepartmentFactory({Id: 1, DepartmentName: 'Marketing', Active: true}));
objArray.push(DepartmentFactory({Id: 2, DepartmentName: 'Sales', Active: true}));
objArray.push(DepartmentFactory({Id: 3, DepartmentName: 'Development', Active: true}));
objArray.push(DepartmentFactory({Id: 4, DepartmentName: 'Accounting', Active: true}));
이제는에 의해 생성 된 객체 배열을 갖게됩니다 DepartmentFactory
. array.sort()
이 객체 배열 DepartmentName
을 각 객체 의 속성 별로 정렬 하는 방법을 사용하려면 어떻게해야 합니까?
이 array.sort()
방법은 문자열 배열을 정렬 할 때 잘 작동합니다.
var myarray=["Bob", "Bully", "Amy"];
myarray.sort(); //Array now becomes ["Amy", "Bob", "Bully"]
그러나 객체 목록으로 어떻게 작동합니까?
답변
다음과 같이해야합니다.
objArray.sort(function(a, b) {
var textA = a.DepartmentName.toUpperCase();
var textB = b.DepartmentName.toUpperCase();
return (textA < textB) ? -1 : (textA > textB) ? 1 : 0;
});
참고 : 대 / 소문자를 변경하면 대 / 소문자를 구분하지 않습니다.
답변
유니 코드를 지원하려면
objArray.sort(function(a, b) {
return a.DepartmentName.localeCompare(b.DepartmentName);
});
답변
var DepartmentFactory = function(data) {
this.id = data.Id;
this.name = data.DepartmentName;
this.active = data.Active;
}
// use `new DepartmentFactory` as given below. `new` is imporatant
var objArray = [];
objArray.push(new DepartmentFactory({Id: 1, DepartmentName: 'Marketing', Active: true}));
objArray.push(new DepartmentFactory({Id: 2, DepartmentName: 'Sales', Active: true}));
objArray.push(new DepartmentFactory({Id: 3, DepartmentName: 'Development', Active: true}));
objArray.push(new DepartmentFactory({Id: 4, DepartmentName: 'Accounting', Active: true}));
function sortOn(property){
return function(a, b){
if(a[property] < b[property]){
return -1;
}else if(a[property] > b[property]){
return 1;
}else{
return 0;
}
}
}
//objArray.sort(sortOn("id")); // because `this.id = data.Id;`
objArray.sort(sortOn("name")); // because `this.name = data.DepartmentName;`
console.log(objArray);
데모 : http://jsfiddle.net/diode/hdgeH/
답변
// Sorts an array of objects "in place". (Meaning that the original array will be modified and nothing gets returned.)
function sortOn (arr, prop) {
arr.sort (
function (a, b) {
if (a[prop] < b[prop]){
return -1;
} else if (a[prop] > b[prop]){
return 1;
} else {
return 0;
}
}
);
}
//Usage example:
var cars = [
{make:"AMC", model:"Pacer", year:1978},
{make:"Koenigsegg", model:"CCGT", year:2011},
{make:"Pagani", model:"Zonda", year:2006},
];
// ------- make -------
sortOn(cars, "make");
console.log(cars);
/* OUTPUT:
AMC : Pacer : 1978
Koenigsegg : CCGT : 2011
Pagani : Zonda : 2006
*/
// ------- model -------
sortOn(cars, "model");
console.log(cars);
/* OUTPUT:
Koenigsegg : CCGT : 2011
AMC : Pacer : 1978
Pagani : Zonda : 2006
*/
// ------- year -------
sortOn(cars, "year");
console.log(cars);
/* OUTPUT:
AMC : Pacer : 1978
Pagani : Zonda : 2006
Koenigsegg : CCGT : 2011
*/
답변
데모
var DepartmentFactory = function(data) {
this.id = data.Id;
this.name = data.DepartmentName;
this.active = data.Active;
}
var objArray = [];
objArray.push(new DepartmentFactory({Id: 1, DepartmentName: 'Marketing', Active: true}));
objArray.push(new DepartmentFactory({Id: 2, DepartmentName: 'Sales', Active: true}));
objArray.push(new DepartmentFactory({Id: 3, DepartmentName: 'Development', Active: true}));
objArray.push(new DepartmentFactory({Id: 4, DepartmentName: 'Accounting', Active: true}));
console.log(objArray.sort(function(a, b) { return a.name > b.name}));
답변
objArray.sort((a, b) => a.DepartmentName.localeCompare(b.DepartmentName))
답변
이렇게 해
objArrayy.sort(function(a, b){
var nameA=a.name.toLowerCase(), nameB=b.name.toLowerCase()
if (nameA < nameB) //sort string ascending
return -1
if (nameA > nameB)
return 1
return 0 //default return value (no sorting)
});
console.log(objArray)