Question

给出以下对象数组：

var data = [
  {fruit: "apples", stock: false, season: true},
  {fruit: "peaches", stock: true, season: false},
  {fruit: "oranges", stock: false, season: false},
  {fruit: "pears", stock: false, season: true},
]

和这两个数组：

var fruits = ["apples", "peaches"]
var inv = ["stock"]

如何从data过滤对象，以便保留对象：

如果它们在fruit数组中有fruits；和
它们具有inv数组中的属性，该属性设置为true

因此，在上面的示例中，只有桃子可以存活：

  var result = [
    {fruit: "peaches", stock: true, season: false}
  ]

Answer 1

我认为您可以只将过滤器与some()和includes()配合使用：

var data = [ { fruit: "apples", stock: false, season: true }, { fruit: "peaches", stock: true, season: false }, { fruit: "oranges", stock: false, season: false }, { fruit: "pears", stock: false, season: true }],
    fruits = ["apples", "peaches"],
    inv = ["stock"],
    result = data.filter(o => fruits.includes(o.fruit) && inv.every(k => o[k]));
   
console.log(result);

Answer 2

您可以使用Array#includes在Fruits数组中进行筛选，并通过Array#every迭代map来检查属性。

inv

Answer 3

您可以在此处使用filter和every的组合。

var data = [
  {fruit: "apples", stock: false, season: true},
  {fruit: "peaches", stock: true, season: false},
  {fruit: "oranges", stock: false, season: false},
  {fruit: "pears", stock: false, season: true},
]

var fruits = ["apples", "peaches"]
var inv = ["stock"]

var result= data.filter(a=> fruits.some(b=> b == a.fruit) && inv.every(k => a[k]));
console.log(result)

Answer 4

var output = data.filter(val => {
    if (fruits.includes(val.fruit)) {
        return inv.filter(prop => {
            if (val[prop] == true)
                return val
        }).length > 0
    }
})
console.log(output)

如果您只想过滤通过，我认为这可以。

使用值数组过滤对象数组？

4 个答案: