javascript - js first sort by age, if the age is the same, then sort by top

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

This is so long-winded...

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

Isn’t this great?

Answer 2

Just use what you bring

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

Because you are talking about sorting top from high to low. This way of writing means that the larger the number, the higher it is. If you want the smaller number to be smaller, just change the position b.top - a.top to a.top - b.top

Answer 3

Online experience 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

Who First

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)); 
    }
}

Next

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);