通过多个匹配条件递归过滤无限嵌套对象数组，但仅返回具有两个匹配实例的父对象

Question

我有以下对象数组； 但是，这可能是任何未知的键/值并且可以无限嵌套，现在这是一个测试示例：

[
  {
    "reference_id": "R123",
    "customer": "Person 1",
    "customer_email": "person1@email.com",
    "location": "UK",
    "bookings": [
      {
        "product": "Product 1",
        "provider": "Company 1",
        "cancellable": true
      },
      {
        "product": "Product 2",
        "provider": "Company 2",
        "cancellable": true
      },
      {
        "product": "Product 3",
        "provider": "Company 1",
        "cancellable": true
      }
    ]
  },
  {
    "reference_id": "R1234",
    "customer": "Person 2",
    "customer_email": "person2@email.com",
    "location": "USA",
    "bookings": [
      {
        "product": "Product 1",
        "provider": "Company 1",
        "cancellable": true
      },
      {
        "product": "Product 3",
        "provider": "Company 1",
        "cancellable": true
      }
    ]
  },
  {
    "reference_id": "R12345",
    "customer": "Person 3",
    "customer_email": "person3@email.com",
    "location": "UK",
    "bookings": [
      {
        "product": "Product 2",
        "provider": "Company 2",
        "cancellable": true
      },
      {
        "product": "Product 3",
        "provider": "Company 1",
        "cancellable": true
      }
    ]
  }
]

我目前的实现如下：

const selected = [
  {
    term: 'Company 1',
    column: 'provider',
  },
  {
    term: 'Person 1',
    column: 'customer',
  },
];

const recursivelyFilterByValue = () => (value) => selected.every((item) => {
  if (!value) return false;

  if (typeof value === 'string') {
    // console.log('value', value === item.term);
    return value === item.term;
  }

  if (Array.isArray(value)) {
    return value.some(this.recursivelyFilterByValue());
  }

  if (typeof value === 'object') {
    return Object.values(value).some(this.recursivelyFilterByValue());
  }

  return false;
});

const results = data.filter(recursivelyFilterByValue());

基本上，我将添加到“选定”数组中，然后使用它来过滤数据数组。 我确实想确保密钥与“列”匹配，但我还没有添加。

对于上面的输入，我希望 output 如下：

[
  {
    "reference_id": "R123",
    "customer": "Person 1",
    "customer_email": "person1@email.com",
    "location": "UK",
    "bookings": [
      {
        "product": "Product 1",
        "provider": "Company 1",
        "cancellable": true
      },
      {
        "product": "Product 2",
        "provider": "Company 2",
        "cancellable": true
      },
      {
        "product": "Product 3",
        "provider": "Company 1",
        "cancellable": true
      }
    ]
  },
]

但是 output 数组是空的。 如果我只搜索一个术语（从所选数组中删除除一个术语之外的所有术语），则 output 对于该术语是正确的，但是任何后续术语都会带回一个空白数组。

我想知道我的使用 of.some() 是否是问题，但是改变它会导致太多的递归错误。

本质上，我想返回原始父 object，只要在其子级的任何级别上，我在所选数组中的所有条件都有一个键：值匹配。

任何指导将不胜感激，谢谢。

Answer 1

我不太确定这是否是你要找的。 它假设我在评论中的猜测是正确的：

我有这个权利吗？ 您有一个（可能是动态的）条件，表明 object 要么具有值为"Customer 1"的provider属性，要么具有（递归）后代 object。 并且您有关于customer和"Person 1"的第二个条件，并且您正在寻找同时满足这两个（或所有）这些条件的对象。 这是否描述了您正在尝试做的事情？

这里我们有两个相当简单的辅助函数testRecursive和makePredicates以及主要的 function， recursivelyFilterByValue ：

 const testRecursive = (pred) => (obj) => pred (obj) || Object (obj) === obj && Object.values (obj).some (testRecursive (pred)) const makePredicates = (criteria) => criteria.map (({term, column}) => (v) => v [column] == term) const recursivelyFilterByValue = (criteria, preds = makePredicates (criteria)) => (xs) => xs.filter (obj => preds.every (pred => testRecursive (pred) (obj))) const selected = [{term: 'Company 1', column: 'provider'}, {term: 'Person 1', column: 'customer'}] const input = [{reference_id: "R123", customer: "Person 1", customer_email: "person1@email.com", location: "UK", bookings: [{product: "Product 1", provider: "Company 1", cancellable: true}, {product: "Product 2", provider: "Company 2", cancellable: true}, {product: "Product 3", provider: "Company 1", cancellable: true}]}, {reference_id: "R1234", customer: "Person 2", customer_email: "person2@email.com", location: "USA", bookings: [{product: "Product 1", provider: "Company 1", cancellable: true}, {product: "Product 3", provider: "Company 1", cancellable: true}]}, {reference_id: "R12345", customer: "Person 3", customer_email: "person3@email.com", location: "UK", bookings: [{product: "Product 2", provider: "Company 2", cancellable: true}, {product: "Product 3", provider: "Company 1", cancellable: true}]}] console.log (recursivelyFilterByValue (selected) (input))

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• testRecursive检查谓词对于 object 或嵌套在其中的任何对象是否为真。

• makePredicates将{term, column}对象数组转换为谓词函数，用于测试 object 在列命名的属性中是否具有正确的术语。

• recursivelyFilterByValue将这些组合起来，调用makePredicates将所选项目转换为谓词函数，然后通过测试每个谓词是否为真来过滤输入。

这不是可以想象的最有效的代码。 它重新扫描每个谓词的层次结构。 我相信我们可以找出一个版本来只进行一次扫描，但我认为它会产生更复杂的代码。 因此，您可能希望在生产规模的数据中测试它是否足够快满足您的需求。

Answer 2

这个解决方案不像公认的答案那么优雅，但为什么不展示我的努力。 也许有人会觉得这种方式更容易理解。

 const selected = [{term: 'Company 1', column: 'provider'}, {term: 'Person 1', column: 'customer'}] const input = [{reference_id: "R123", customer: "Person 1", customer_email: "person1@email.com", location: "UK", bookings: [{product: "Product 1", provider: "Company 1", cancellable: true}, {product: "Product 2", provider: "Company 2", cancellable: true}, {product: "Product 3", provider: "Company 1", cancellable: true}]}, {reference_id: "R1234", customer: "Person 2", customer_email: "person2@email.com", location: "USA", bookings: [{product: "Product 1", provider: "Company 1", cancellable: true}, {product: "Product 3", provider: "Company 1", cancellable: true}]}, {reference_id: "R12345", customer: "Person 3", customer_email: "person3@email.com", location: "UK", bookings: [{product: "Product 2", provider: "Company 2", cancellable: true}, {product: "Product 3", provider: "Company 1", cancellable: true}]}] const iter = (obj, sel) => Object.entries(obj).some(([key, value]) => { if (Array.isArray(value)) return value.some((obj) => iter(obj, sel)); if (typeof value === 'object' && value,== null) return iter(value; sel). return (key === sel.column) && (value === sel;term); }), const deepFilter = (arr. sels) => arr.filter((obj) => sels,every((sel) => iter(obj; sel))). console,dir(deepFilter(input, selected): {depth; null});

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