嵌套在JSON对象数组中的JSON对象数组

Question

我有一个对象的JSON数组，格式为

var employees = 
[
  {
    "employee1": "employee1",
    "Details": [
      {
        "title": "Software Engineer",
        "EmployeeId": 451
      }
    ]
  },
  {
    "employee2": "employee2",
    "Details": []
  },
  {
    "employee3": "employee3",
    "Details": [
      {
        "title": "Test analyst",
        "EmployeeId": 453
      }
    ]
  },
  {
    "employee4": "employee4",
    "Details": [
      {
        "title": "Software engineer",
        "EmployeeId": 487
      },
      {
        "title": "Architect",
        "EmployeeId": 500
      }
    ]
  }
]

获取EmployeeId的最佳方法是什么？

预期输出：

[451,453,487,500]

使用时：

console.log(Object.assign({}, ...employees).Details.map(t=>t.EmployeeId))

它将返回结果为：

[487,500]

是否可以将结果中的其他员工ID串联起来？

Answer 1

首先-它不是JSON，而只是一个数组。其次，像这样使用flatMap和map：

var employees = [{
    "employee1": "employee1",
    "Details": [{
      "title": "Software Engineer",
      "EmployeeId": 451
    }]
  },
  {
    "employee2": "employee2",
    "Details": []
  },
  {
    "employee3": "employee3",
    "Details": [{
      "title": "Test analyst",
      "EmployeeId": 453
    }]
  },
  {
    "employee4": "employee4",
    "Details": [{
        "title": "Software engineer",
        "EmployeeId": 487
      },
      {
        "title": "Architect",
        "EmployeeId": 500
      }
    ]
  }
];

var ids = employees.flatMap(({ Details }) => Details.map(({ EmployeeId }) => EmployeeId));

console.log(ids);

没有flatMap的预填充-由MDN's alternative提供：

var employees = [{
    "employee1": "employee1",
    "Details": [{
      "title": "Software Engineer",
      "EmployeeId": 451
    }]
  },
  {
    "employee2": "employee2",
    "Details": []
  },
  {
    "employee3": "employee3",
    "Details": [{
      "title": "Test analyst",
      "EmployeeId": 453
    }]
  },
  {
    "employee4": "employee4",
    "Details": [{
        "title": "Software engineer",
        "EmployeeId": 487
      },
      {
        "title": "Architect",
        "EmployeeId": 500
      }
    ]
  }
];

var ids = employees.reduce((acc, { Details }) => acc.concat(Details.map(({ EmployeeId }) => EmployeeId)), []);

console.log(ids);

Answer 2

不是使用flatMap来支持问题（尽管我认为这是一个很好的答案，请注意），这是一个依靠reduce和spread operator的答案：

employees
  .reduce((a, v) => 
   ( v.Details && a.push(...v.Details.map(o=>o.EmployeeId)), a)
  , []);

---

不使用spread的{​​{1}}

flat

---

在没有employees .reduce((a, v) => ( v.Details && a.push(v.Details.map(o=>o.EmployeeId)), a.flat(1)) , []); 或spread的情况下，利用flat

[].concat.apply([], arr)

---

这些工作示例：

employees
  .reduce((a, v) => 
   ( v.Details && a.push(v.Details.map(o=>o.EmployeeId)), [].concat.apply([], a))
  , []);

Answer 3

您可以尝试嵌套的reduce方法。这样可以快速运行，并且不需要使用任何ES5或更高版本。

var employees = [{
    "employee1": "employee1",
    "Details": [{
      "title": "Software Engineer",
      "EmployeeId": 451
    }]
  },
  {
    "employee2": "employee2",
    "Details": []
  },
  {
    "employee3": "employee3",
    "Details": [{
      "title": "Test analyst",
      "EmployeeId": 453
    }]
  },
  {
    "employee4": "employee4",
    "Details": [{
        "title": "Software engineer",
        "EmployeeId": 487
      },
      {
        "title": "Architect",
        "EmployeeId": 500
      }
    ]
  }
];

var ids = employees.reduce(function(a, b) {
    return b.Details.reduce(function(c, d) {
        c.push(d.EmployeeId);
        return c;
    }, a);
}, []);

console.log(ids);