JavaScript数组5种迭代方法各有什么妙用？

let goods = [{
  name: '时尚女装',
  price: 888.00
}, {
  name: '精品手表',
  price: 666.00
}]
goods.forEach(item => {
    console.log(`${item.name}价格为${item.price}`);
    /**
    * 时尚女装价格为888
    * 精品手表价格为666
    */
})

let childs = [{
  name: '张三',
  age: 1
}, {
  name: '李四',
  age: 2
}, {
  name: '王五',
  age: 2
}]
// 数组中需要每一个项都符合条件才返回true，反之返回false
let isAdults = childs.every(child => child.age >= 3)
console.log(isAdults) // false

let isAdults2 = childs.every(child => child.age >= 1)
console.log(isAdults2) // true

// sex 0：男   1：女
let people1 = [{
  name: '小张',
  sex: 0
}, {
  name: '小刘',
  sex: 0
}, {
  name: '小李',
  sex: 0
}]
// 数组中只要有一个选项符合条件的就返回true，反之返回false
let discount1 = people1.some(p => p.sex === 1)
console.log(discount1) // false 不打折 


let people2 = [{
  name: '小张',
  sex: 0
}, {
  name: '丽丽',
  sex: 1
}, {
  name: '小李',
  sex: 0
}]
// 数组中只要有一个选项符合条件的就返回true，反之返回false
let discount2 = people2.some(p => p.sex === 0)
console.log(discount2) // true 打折

let employees = [{
  name: '张三',
  wages: 7000
}, {
  name: '李四',
  wages: 8000
}, {
  name: '王五',
  wages: 10000
}]
let employeesWages = employees.map(employee=>{
  employee.wages += 10000
  return employee
})
console.log(employeesWages)
console.log(employees) // employees和employeesWages相等，所以使用map时，要注意如不需要修改源数据，就不要修改里面的对象
/**
  [{name: "张三", wages: 17000},
  {name: "李四", wages: 18000},
  {name: "王五", wages: 20000}]
*/
let employeesWages2 = employees.map(employee=>{
  return {
    name: employee.name,
    wages: employee.wages + 10000
  }
})
console.log(employeesWages2)
console.log(employees) // 此时employeesWages2和employees就不同了

let employees = [{
  name: '张三',
  wages: 7000
}, {
  name: '李四',
  wages: 8000
}, {
  name: '王五',
  wages: 10000
}]
let employeesFilter = employees.filter(employee=>{
  return employee.wages >= 8000
})
console.log(employeesFilter)
/**
  [{name: "李四", wages: 8000},
  {name: "王五", wages: 10000}]
*/