javascript - 按两个属性排序，其中一个属性优先但具有共同值？

Question

问题: 如何按两个属性对 data 数组进行排序:

1. 其中 type 始终位于顶部，
2. 其中计数始终从小到大。

这是我的努力:

var data = [
  {type: 'first', count: '1'},
  {type: 'second', count: '5'},
  {type: 'first', count: '2'},
  {type: 'second', count: '2'},
  {type: 'second', count: '1'},
  {type: 'first', count: '0'},
]

//Expected
var newData = [
  {type: 'first', count: '0'},
  {type: 'first', count: '1'},
  {type: 'first', count: '2'},
  {type: 'second', count: '1'},
  {type: 'second', count: '2'},
  {type: 'second', count: '5'},
]

 //**Pseudo code**//
// Will put the types on top
data.sort((a,b) => a.type === 'first' ? -1:0)

// This will sort the count 
data.sort((a,b) => a.count < b.count ? -1 ? (a.count > b.count ? 1:0)

由于 count 在不同类型之间共享值，我发现很难解决它。如何对这两个属性进行排序，但保持类型始终位于顶部，并始终按从小到大的顺序进行计数？

Answer 1

您可以像这样使用sort()方法。

var data = [
  {type: 'first', count: '1'},
  {type: 'second', count: '5'},
  {type: 'first', count: '2'},
  {type: 'second', count: '2'},
  {type: 'second', count: '1'},
  {type: 'first', count: '0'},
]

var result = data.sort(function(a, b) {
  return ((b.type == 'first' ) - (a.type == 'first')) || (a.count - b.count)
})

console.log(result)