是否有一种惯用的Javascript方法，可以避免为失败的`Array.find（）`提供默认值？

Question

我有用于动态创建对象的工作代码：

// dummy sources
const departments = [
    { code: 'DA', title: 'Department A' },
    { code: 'DB', title: 'Department B' },
    { code: 'DC', title: 'Department C' },
    { code: 'DD', title: 'Department D' },
]

const reviewers = ['Department A','Pub Lunches','Department C','Department D']

const validationObject = {}

// actual code
reviewers.forEach( (reviewer, index) => {
    const dept = departments.find( dept => dept.title === reviewer) || { code: reviewer.replace(/[ ]/g, '_').replace(/[/W]/g, '') }
    validationObject[dept.code.toLowerCase()+'_reviewer'] = {blah:'blah'}
})

console.log(validationObject)

[可以，但是创建虚拟对象只是为了克服未定义结果的code元素时（当部门中没有匹配的元素时）克服错误的样子。

我是否缺少一种惯用的方法来做到这一点？

Answer 1

在ES6中应该是这样的

const dept = { code: reviewer.replace(/[ ]/g, '_').replace(/[/W]/g, ''), ...departments.find( dept => dept.title === reviewer) }

Answer 2

也许在分配前使用dept运算符检查if的值：

reviewers.forEach( (reviewer, index) => {
    const dept = departments.find( dept => dept.title === reviewer);
    if (dept)
        validationObject[dept.code.toLowerCase()+'_reviewer'] = {blah:'blah'}
})

或一种解决方案，但使用了额外的内存：

const unique = new Map(departments.map(s => [s.title, s.code]));
reviewers.forEach( (reviewer, index) => {
    unique.get(reviewer) ? validationObject[unique.get(reviewer)] = {blah:'blah'} : '';
})

console.log(validationObject)

示例：

// dummy sources
const departments = [
    { code: 'DA', title: 'Department A' },
    { code: 'DB', title: 'Department B' },
    { code: 'DC', title: 'Department C' },
    { code: 'DD', title: 'Department D' },
]

const reviewers = ['Department A','Pub Lunches','Department C','Department D']

const validationObject = {}

const unique = new Map(departments.map(s => [s.title, s.code]));
reviewers.forEach( (reviewer, index) => {
    unique.get(reviewer) ? validationObject[unique.get(reviewer).toLowerCase()+'_reviewer'] = {blah:'blah'} : '';
})

console.log(validationObject)

Answer 3

真诚地，我猜是打字稿。 https://www.typescriptlang.org/