Question

有一系列对象

ar = [
    {name: 'a', count: 1},
    {name: 'c', count: 2},
    {name: 'f', count: 0},
    {name: 'e', count: 0},
    {name: 'b', count: 3},
    {name: 'a', count: 0},
];

我正在尝试通过以下方式对它进行排序

计数> 0
按字母顺序命名

所以结果数组将是

ar = [
    {name: 'a', count: 1},
    {name: 'b', count: 3},
    {name: 'c', count: 2},
    {name: 'a', count: 0},
    {name: 'e', count: 0},
    {name: 'f', count: 0},
];

我尝试了几种多重排序方法，结果总是很接近，但是计数为0的部分没有正确排序。

Answer 1

您可以减去count > 0返回的布尔值。这将优先排序count > 0的项目。如果两个项目具有相同的优先级，请使用localeCompare

根据name属性对它们进行排序

const arr = [
  { name: 'a', count: 1 },
  { name: 'c', count: 2 },
  { name: 'f', count: 0 },
  { name: 'e', count: 0 },
  { name: 'b', count: 3 },
  { name: 'a', count: 0 },
];

arr.sort((a, b) => (b.count > 0) - (a.count > 0) || a.name.localeCompare(b.name))

console.log(arr)

Answer 2

好的。为什么不使用简单明了的方式？我的意思是将一个数组分为2个数组：一个数组，其中所有项的count> 0，第二个数组，其余的项。然后按字母顺序对两者进行排序。然后将concat阵列退回。

    const ar = [
        {name: 'a', count: 1},
        {name: 'c', count: 2},
        {name: 'f', count: 0},
        {name: 'e', count: 0},
        {name: 'b', count: 3},
        {name: 'a', count: 0},
    ];
    
    const pCount = [];
    const nCount = [];
    ar.forEach((item) => {
        if (item.count > 0) {
            pCount.push(item);
        } else {
            nCount.push(item);
        }
    });
    
    pCount.sort((a, b) => a.name.localeCompare(b.name));
    nCount.sort((a, b) => a.name.localeCompare(b.name));
    
    const result = [...pCount, ...nCount]; // concat 
    console.log(result);

以特定方式对对象进行多排序数组

2 个答案: