javascript - js mula-mula susun mengikut umur, jika umurnya sama, kemudian susun mengikut atas

Question

var obj = [{
        id : 1,
        age : 20,
            top :5
    },{
        id : 3,
        age : 21,
            top : 6
    },{
        id : 2,
        age : 20,
            top : 8
    }]
  function keysort(property) {
      return function(a, b) {
          var value1 = a[property] == '-' ? 0 : a[property];
          var value2 = b[property] == '-' ? 0 : b[property];
           return value1 - value2;
      }
  }
  var obj1 = obj.sort(keysort('age'));
写一半 不会写了  age相同的情况下  再按照top从高到低排序  想请教下老司机 

Answer 1

Ini sangat panjang lebar...

obj.sort( function(curr,next) {
    return !!( curr.age-next.age )? curr.age-next.age: curr.top-next.top;
} );

Bukankah ini hebat

Answer 2

Guna sahaja apa yang anda bawa

    obj = obj.sort((a, b) =>  { return a.age - b.age || b.top - a.top;} );
    console.log(obj);
    
    

Sebab anda bercakap tentang menyusun atas dari tinggi ke rendah cara penulisan ini ialah semakin besar nombor, semakin dekat dengan bahagian depan
Jika anda mahu nombor yang lebih kecil menjadi lebih kecil, ubah sahaja kedudukan b.atas - a.top to a.top - b.top

Answer 3

Pengalaman dalam talian https://jsfiddle.net/hguyjgs8/1/

//假设top 不大于1000, 大于1000的，适度修改
var obj = [{
  id: 1,
  age: 20,
  top: 5
}, {
  id: 3,
  age: 21,
  top: 6
}, {
  id: 2,
  age: 20,
  top: 8
}]

function pad(num, size) {
  var s = num + "";
  while (s.length < size) s = "0" + s;
  return s;
}

obj.sort((a, b) => pad(a.age, 2) + pad(1000-a.top, 3) > pad(b.age, 2)  + pad(1000-b.top, 3)).forEach((i) => {
    document.writeln(JSON.stringify(i)+'<br>');
});

Answer 4

function keySort (...args) {
    let props = args.map(name => {
        let desc = name[0] === '-'
        if (desc) name = name.substring(1)
        return { desc, name }
    })
    
    return (a, b) => {
        let result = 0
        for (let prop of props) {
            let p = prop.name
            result = prop.desc ? b[p] - a[p] : a[p] - b[p]
            if (result) return result
        }
        return result
    }
}

obj.sort(keySort('age', '-top'))

https://jsfiddle.net/sojxjqpf/

Answer 5

Siapa Dulu

var whoFirst = ['age', 'top']; 

var copy = o => JSON.parse(
    JSON.stringify(o)
); 

var judge = (a, b, whos) => {
    if (whos.length === 0) return 0; 
    
    let key = whos[0]; 
    if (a[key] !== b[key]){
        return a[key] - b[key]; 
    } else {
        return judge(a, b, whos.slice(1)); 
    }
}

Seterusnya

var sorts = arr => {
    let a = copy(arr); 

    a.sort((a, b) => {
        return judge(a, b, whoFirst); 
    }); 
    
    return a; 
}

S

WhoFirst 升序。

var obj = [{
    id : 1,
    age : 20,
        top :5
},{
    id : 3,
    age : 21,
        top : 6
},{
    id : 2,
    age : 20,
        top : 8
},{
    id: 4,
    age: 20, 
    top: 2
},{
    id: 8,
    age: 20, 
    top: 2
},{
    id: 5,
    age: 20, 
    top: 11
},{
    id: 7,
    age: 20, 
    top: 9
},{
    id: 6,
    age: 20, 
    top: 2
},{
    id: 9,
    age: 20, 
    top: 1
}]; 


sorts(obj);