如何将数组中的项与对象中的项匹配？

Question

采用以下数据结构

report = {
   fraud_ids: [2, 3, 4, 15],
}

 fraudTypes: {
      ACH: {
        fraud_category: "fraud",
        fraud_subcategory: "a",
        fraud_type_id: "4",
      },
      'Account takeover': {
        fraud_category: "misc",
        fraud_subcategory: "a",
        fraud_type_id: "2",
      },
      'Advance fee': { 
        fraud_category: "fraud",
        fraud_subcategory: "b",
        fraud_type_id: "8",
      },
      'Against Financial Institution Customer(s)': {
        fraud_category: "cyber",
        fraud_subcategory: "b",
        fraud_type_id: "15",
      },
      'Against Financial Institution(s)': {
        fraud_category: "cyber",
        fraud_subcategory: "a",
        fraud_type_id: "78",
      },
      'Alters or cancels transaction to avoid BSA recordkeeping requirement': {
        fraud_category: "structuring",
        fraud_subcategory: "a",
        fraud_type_id: "3",
      },
   }

我希望将fraudTypes对象与fraud_type_id数组中的正确数字相匹配。当我找到匹配时，我只想返回对象的键。

因此，使用上面的示例，我将返回['Account takeover', 'Alters or cancels transaction to avoid BSA recordkeeping requirement', 'ACH', Against Financial Institution Customer(s)']

我编写了以下逻辑，令我惊讶的是，我得到了一个返回未定义的四项的数组。

const x = report.fraud_ids.map(id => {
  Object.keys(fraudTypes).map(fraudDescription => {
    if (parseInt(fraudTypes[fraudDescription].fraud_type_id, 10) === id) {
      return true;
    }
  });
});

我期望它返回匹配的四个对象，然后我知道我必须编写额外的逻辑来返回键。我做错了什么？

请看codeSandbox。

Answer 1

您没有从第一个回调中返回任何内容，.map也不是正确的内部方法，而是使用.find查找带有匹配ID的键。

​

report={fraud_ids:[2,3,4,15],fraudTypes:{ACH:{fraud_category:"fraud",fraud_subcategory:"a",fraud_type_id:"4"},"Account takeover":{fraud_category:"misc",fraud_subcategory:"a",fraud_type_id:"2"},"Advance fee":{fraud_category:"fraud",fraud_subcategory:"b",fraud_type_id:"8"},"Against Financial Institution Customer(s)":{fraud_category:"cyber",fraud_subcategory:"b",fraud_type_id:"15"},"Against Financial Institution(s)":{fraud_category:"cyber",fraud_subcategory:"a",fraud_type_id:"78"},"Alters or cancels transaction to avoid BSA recordkeeping requirement":{fraud_category:"structuring",fraud_subcategory:"a",fraud_type_id:"3"}}};

const x = report.fraud_ids.map(id => (
  Object.keys(report.fraudTypes).find(fraudDescription =>
    Number(report.fraudTypes[fraudDescription].fraud_type_id) === id)
));
console.log(x);

​

Answer 2

我不知道这是不是你想要的，但到此为止。

const selectedFrauds = report.fraud_ids.map((id, index) => {
  return(
    Object.values(fraudTypes).map((fraud, idx) => {
      return id === +fraud.fraud_type_id ? fraud : "";
    })
  )
})

const cleanSelectedFrauds = selectedFrauds.map((el) => {
  return el.filter((innerEl) => {
    if(innerEl) {
      return innerEl
    }
  })
});


console.log(cleanSelectedFrauds)

这将返回以下数组：

[
  [
    {
      fraud_category: 'misc',
      fraud_subcategory: 'a',
      fraud_type_id: '2'
    }
  ],
  [
    {
      fraud_category: 'structuring',
      fraud_subcategory: 'a',
      fraud_type_id: '3'
    }
  ],
  [
    {
      fraud_category: 'fraud',
      fraud_subcategory: 'a',
      fraud_type_id: '4'
    }
  ],
  [
    {
      fraud_category: 'cyber',
      fraud_subcategory: 'b',
      fraud_type_id: '15'
    }
  ]
]