用另一个数组对象过滤一个数组对象

Question

我正在尝试过滤数组 localcartArr 的对象以排除 products 数组中包含的任何对象，但它返回相同的 localcartArr 而不排除此对象：

{ "productID": "5f6fd57f8b6f6b0017992443" }

存在于products数组中，它只适用于===运算符，但不适用于不等于运算符!=或!==。

const products = [
  {
    "_id": "60242abc209cbd32d8e85ec8",
    "productID": "5f6fd4f18b6f6b001799243f",
    "quantity": 2
  },
  {
    "_id": "60242b00209cbd32d8e85ec9",
    "productID": "5f6fd57f8b6f6b0017992443",
    "quantity": 1
  }
]
        
let localcartArr = [
  {                
    "productID": "5f6fd57f8b6f6b0017992443",
    "quantity": 2
  },
  {                   
    "productID": "5f6fd12a8b6f6b001799242f",
    "quantity": 1
  },
  {               
    "productID": "5f7a5668a9baa50017d495e8",
    "quantity": 1
  }
]

let filterd = localcartArr.filter(local => {
  return products.some(product => {
    return local.productID !==  product.productID    
  });
});

Answer 1

为了反转“一些/等于”，您需要返回“不是每个/不等于”。

这是De Morgan's laws在布尔逻辑中的一个原理。

const products = [{
  "_id": "60242abc209cbd32d8e85ec8",
  "productID": "5f6fd4f18b6f6b001799243f",
  "quantity": 2
}, {
  "_id": "60242b00209cbd32d8e85ec9",
  "productID": "5f6fd57f8b6f6b0017992443",
  "quantity": 1
}];

const localcartArr = [
  { "productID": "5f6fd57f8b6f6b0017992443", "quantity": 2 },
  { "productID": "5f6fd12a8b6f6b001799242f", "quantity": 1 },
  { "productID": "5f7a5668a9baa50017d495e8", "quantity": 1 }
];

let filtered1 = localcartArr.filter(local =>
  products.some(product =>
    local.productID === product.productID));
    
let filtered2 = localcartArr.filter(local =>
  !products.every(product =>
    local.productID !== product.productID));

console.log(filtered1);
console.log(filtered2);

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这是相同的程序，但经过了简化：

const products = [{
  "_id": "60242abc209cbd32d8e85ec8",
  "productID": "5f6fd4f18b6f6b001799243f",
  "quantity": 2
}, {
  "_id": "60242b00209cbd32d8e85ec9",
  "productID": "5f6fd57f8b6f6b0017992443",
  "quantity": 1
}];

const cart = [
  { "productID": "5f6fd57f8b6f6b0017992443", "quantity": 2 },
  { "productID": "5f6fd12a8b6f6b001799242f", "quantity": 1 },
  { "productID": "5f7a5668a9baa50017d495e8", "quantity": 1 }
];

const filterEqual = (a, b, c) => a.filter(x => b.some(y => (c(x) === c(y))));
const filterNotEqual = (a, b, c) => a.filter(x => !b.every(y => (c(x) !== c(y))));

const filtered1 = filterEqual(cart, products, ({productID}) => productID);
const filtered2 = filterNotEqual(cart, products, ({productID}) => productID);

console.log(filtered1);
console.log(filtered2);

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Answer 2

使用您要匹配的属性进行过滤并排除不匹配的属性。

一些例子：

const products = [{
    "_id": "60242abc209cbd32d8e85ec8",
    "productID": "5f6fd4f18b6f6b001799243f",
    "quantity": 2
  },
  {
    "_id": "60242b00209cbd32d8e85ec9",
    "productID": "5f6fd57f8b6f6b0017992443",
    "quantity": 1
  }
];

let localcartArr = [{
    "productID": "5f6fd57f8b6f6b0017992443",
    "quantity": 2
  },
  {
    "productID": "5f6fd12a8b6f6b001799242f",
    "quantity": 1
  },
  {
    "productID": "5f7a5668a9baa50017d495e8",
    "quantity": 1
  }
];
// find matches
const myArrayFiltered = localcartArr.filter((el) => {
  return products.some((f) => {
    return f.productID === el.productID;
  });
});

console.log(myArrayFiltered);

//filter out matches
const notIn = localcartArr.filter((el) => {
  return myArrayFiltered.some((f) => {
    return f.productID !== el.productID;
  });
});

console.log(notIn);

// to put it together (for clarity)
// not that this is not super efficient if you have a large pair of sets
const notInX = localcartArr.filter((el) => {
  return  localcartArr.filter((el) => {
  return products.some((f) => {
    return f.productID === el.productID;
  });
}).some((f) => {
    return f.productID !== el.productID;
  });
});
console.log(notInX);