Javascript：用第一个属性对对象数组进行排序，当相等时与另一个排序

Question

我有一个数组，如下所示：

let myarray = 
[{id:1 , name: "mark" , birthday: "1990-08-18"},
{id:2 , name: "fred" , birthday: "1990-08-17"},
{id:3 , name: "franck" , birthday: "1990-08-16"},
{id:4 , name: "mike" , birthday: "1990-08-15"},
{id:5 , name: "jean" , birthday: "1990-08-17"}]

我正在按“生日”对这些对象进行排序

myarray.sort((a, b) => new Date(b.birthday).getTime() - new Date(a.birthday).getTime());

但有时，两个对象的生日值相同（生日相同），在那种情况下我希望排序将这两个对象（不是所有对象）与更高的进行比较“id"（id高的优先）

建议 ?

Answer 1

需要在sort函数中单独添加一个case如下：

let myarray = [
{id:1 , name: "mark" , birthday: "1990-08-18"},
{id:2 , name: "fred" , birthday: "1990-08-17"},
{id:3 , name: "franck" , birthday: "1990-08-16"},
{id:4 , name: "mike" , birthday: "1990-08-15"},
{id:5 , name: "jean" , birthday: "1990-08-17"}
]

myarray.sort((a, b) => {
     let dif = new Date(b.birthday).getTime() - new Date(a.birthday).getTime();
     if(dif != 0) return dif;
     else return b.id - a.id;
});
console.log(myarray);

Javascript：用第一个属性对对象数组进行排序，当相等时与另一个排序

Javascript: sort an array of objects with a first attribute , and when equal sort with another

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