數組內的過濾器數組

Question

我有一系列產品，我想過濾所有這些產品。 這里的級別也是一個數組，因為我使用了復選框，因此如果用戶選擇級別1，則應顯示級別1的產品，如果用戶再次選擇級別2，則應顯示級別1和2的產品。 但是，我正在獲得所有產品，而不是所選級別。

這是我的代碼

// sample product data
products = [
{id: "1", sellerLevel: "new", status: "online"},
{id: "1", sellerLevel: "1", status: "offline"},
{id: "1", sellerLevel: "1", status: "offline"},
{id: "1", sellerLevel: "2", status: "online"}
]

function productFilterFromLevel({level}) { // level ["new", "1"]
  return products.filter(function(product) {
    return level.filter(function(lvl) {
      return product.sellerLevel === lvl;
    });
  });
} 

Answer 1

您可以在過濾器中使用Array.includes

const productFilterFromLevel = ({level}, products) => products
  .filter(item => level.includes(item.sellerLevel))

不論是否包含，您都可以像這樣過濾鍵

const productFilterFromLevel = function({level}, products) {
  return products.filter(function(product) {
    return level.indexOf(product.sellerLevel) !== -1
  })
}

如您所見， includes版本更加干凈，但是您始終可以創建自己的包含函數：

const arrayIncludes = function(array, item) {
  return array.indexOf(item) !== -1
}

Answer 2

您有一個對象數組，這是Array.filter的一個很好的用例。

// sample product data
products = [
{id: "1", sellerLevel: "new", status: "online"},
{id: "1", sellerLevel: "1", status: "offline"},
{id: "1", sellerLevel: "1", status: "offline"},
{id: "1", sellerLevel: "2", status: "online"}
]

function filter(products, matches) {
  return products.filter(product => {
    return matches.includes(product.sellerLevel);
  });
}

console.log(filter(products, ['new', '1']));

// [
//   {id: "1", sellerLevel: "new", status: "online"},
//   {id: "1", sellerLevel: "1", status: "offline"},
//   {id: "1", sellerLevel: "1", status: "offline"}
// ]

https://jsfiddle.net/persianturtle/fL8eupoz/3/

Answer 3

這不是簡單匹配那么容易，因為您需要比較sellerLevel ：

function productFilterFromLevel(level) { // level ["new", "1"]
  return products.filter(function(product) {
    for (var i = 0; i < level.length; i++) {
      if (level[i] === 'new') {
        return product.sellerLevel === 'new'
      }
      return +product.sellerLevel <= +level[i]
    }
  });
} 

Answer 4

 // sample product data products = [{ id: "1", sellerLevel: "new", status: "online" }, { id: "1", sellerLevel: "1", status: "offline" }, { id: "1", sellerLevel: "1", status: "offline" }, { id: "1", sellerLevel: "2", status: "online" }] function productFilterFromLevel(level) { return products.filter(function(product) { return level.indexOf(product.sellerLevel) != -1; }); } console.log(productFilterFromLevel(["new", "1"]));