Question

我有一个项目数据，其中包含项目和相关员工的列表。每个员工对象都具有salary属性。我想做的就是找到薪水最高的员工。

代码示例：

var projects = [
  //Object(0)		
  {
    projectName: "Winter",
    projectCode: "O0123",
    employee: [{
        title: "Mr.",
        name: "Tom",
        id: 1005,
        salary: 12345
      },
      {
        title: "Mr.",
        name: "Bunny",
        id: 1009,
        salary: 54321
      },
      {
        title: "Mr.",
        name: "Harris",
        id: 1010,
        salary: 23456
      },
    ]
  },
  //Object(1)
  {
    projectName: "Summer",
    projectCode: "P10406",
    employee: [{
        title: "Mr.",
        name: "Seth",
        id: 1006,
        salary: 1234
      },
      {
        title: "Mr.",
        name: "Sam",
        id: 1011,
        salary: 654321
      },
    ],
  }
]


console.log(projects.length);

let maxSalary = 0;

for (var i = 0; i < projects.length; i++) {
  console.log(projects[i].projectName);

  for (var j = 0; j < projects[i].employee.length; j++) {
    console.log("\t" + projects[i].employee[j].title + projects[i].employee[j].name + "\n" + "\t" + "Salary: " + projects[i].employee[j].salary);

    if (i == 0 && j == 0) {
      maxSalary <= projects[i].employee[j].salary;
    }

    if (projects[i].employee[j].salary > maxSalary) {
      maxSalary = projects[i].employee[j].salary;
    }
  }
}

console.log("Max Salary = " + maxSalary);

请提出任何建议。

Answer 1

只需循环遍历不同的projects，然后循环employees即可获得最大值。

var projects = [{
    projectName: "Winter",
    projectCode: "O0123",
    employee: [
      {title: "Mr.", name: "Tom", id: 1005, salary: 12345},
      {title: "Mr.", name: "Bunny", id: 1009, salary: 54321},
      {title: "Mr.", name: "Harris", id: 1010, salary: 23456}
    ]
  },
  {
    projectName: "Summer",
    projectCode: "P10406",
    employee: [
      {title: "Mr.", name: "Seth", id: 1006, salary: 1234},
      {title: "Mr.", name: "Sam", id: 1011, salary: 654321}
    ]
  }
];

var max = 0;
projects.forEach(p => p.employee.forEach(e => e.salary > max && (max = e.salary)));

console.log(max);

如果您想收到问题中提到的employee，而不是薪水，则可以基本相同，只需退还全部object：

var projects = [{
    projectName: "Winter",
    projectCode: "O0123",
    employee: [
      {title: "Mr.", name: "Tom", id: 1005, salary: 12345},
      {title: "Mr.", name: "Bunny", id: 1009, salary: 54321},
      {title: "Mr.", name: "Harris", id: 1010, salary: 23456}
    ]
  },
  {
    projectName: "Summer",
    projectCode: "P10406",
    employee: [
      {title: "Mr.", name: "Seth", id: 1006, salary: 1234},
      {title: "Mr.", name: "Sam", id: 1011, salary: 654321}
    ]
  }
];

var max = {salary: 0};
projects.forEach(p => p.employee.forEach(e => e.salary > max.salary && (max = e)));

console.log(max);

我认为这些示例将为您提供基本的方法。

Answer 2

在这里，您必须混合使用flatMap，这会将您的数组数组转换为简单的平面数组。

然后您可以使用array reduce领取最高薪水。

const projects = [
    {
        employee: [{
                salary: 12345
            },
            {
                salary: 54321
            },
            {
                salary: 23456
            },
        ]
    },
    {
        employee: [{
                salary: 1234
            },
            {
                salary: 654321
            },
        ]
    }  
];

const salaries = projects.flatMap(project => {
    // here we have array of array, goal is to craft flat array of salary.

    return project.employee.map(employe => {
        // From each employee, we pickup only the salary.
        return employe.salary;
    });

});

const highest = salaries.reduce((accumulator, currentValue) => {
    // If current salary is highest than the previous, we keep track of it.
    if(currentValue > accumulator) {
        accumulator = currentValue;
    }
    return accumulator;
});

Answer 3

您已经有了最高薪水，您需要做的就是尽快获得薪水最高的员工的索引。您的代码应如下所示。

var projects = [
  //Object(0)		
  {
    projectName: "Winter",
    projectCode: "O0123",
    employee: [{
        title: "Mr.",
        name: "Tom",
        id: 1005,
        salary: 12345
      },
      {
        title: "Mr.",
        name: "Bunny",
        id: 1009,
        salary: 54321
      },
      {
        title: "Mr.",
        name: "Harris",
        id: 1010,
        salary: 23456
      },
    ]
  },
  //Object(1)
  {
    projectName: "Summer",
    projectCode: "P10406",
    employee: [{
        title: "Mr.",
        name: "Seth",
        id: 1006,
        salary: 1234
      },
      {
        title: "Mr.",
        name: "Sam",
        id: 1011,
        salary: 654321
      },
    ],
  }
]


console.log(projects.length);

let maxSalary = 0;
let employeeWithMaxSalary = {};

for (var i = 0; i < projects.length; i++) {
  console.log(projects[i].projectName);

  for (var j = 0; j < projects[i].employee.length; j++) {
    console.log("\t" + projects[i].employee[j].title + projects[i].employee[j].name + "\n" + "\t" + "Salary: " + projects[i].employee[j].salary);

    if (i == 0 && j == 0) {
      maxSalary <= projects[i].employee[j].salary;
    }

    if (projects[i].employee[j].salary > maxSalary) {
      maxSalary = projects[i].employee[j].salary;
      employeeWithMaxSalary = projects[i].employee[j];
    }
  }
}

console.log("Max Salary = " + maxSalary);
console.log(employeeWithMaxSalary);

通过使用ES6语法（箭头功能和foreach）来代替嵌套循环，还可以改善代码。

Javascript-查找最大数量

3 个答案: