Javascript “如何修复”；初始值设定项不提供此绑定元素的值，并且绑定元素没有默认值。”；打字稿？

我正在将用JavaScript编写的Apollo GraphQLAPI项目迁移到TypeScript。我在查找用户代码块时出错，说

var-idArg:any
初始值设定项没有为此绑定元素提供值，并且绑定元素没有默认值。ts（2525）

目前我在不知道实际解决方案的情况下添加了

any

，警告消失了

  async findOne({ id: idArg }: any = {}) {
    const user = await this.knex('users')
      .where('id', idArg)
      .first();

    if (!user) return;
    return user;
  }

不过我还是想知道实际的解决办法。我是否应该添加

编号

类型而不是

任何

？但当我这样做时，错误是

类型“{}”不能分配给类型“number”.ts（2322）

---

请提供帮助。

根据您想要完成的任务，有很多方法可以解决此问题

// The compiler checks the object { id: '1' } and it knows it has an id property
var { id } = { id: '1' }

/* The compiler is confused. It check the object {} and it knows it doesn't have 
a property id1, so it is telling you it doesn't know where to get the value 
for id1
*/
var { id1 } = {}

/* In this case the compiler knows the object doesn't have the property id2 but
since you provided a default value it uses it 'default value'.
 */
var { id2 = 'default value' } = {}

/* In your case there are a couple of solutions: */

// 1) Provide the value in the initializer
function findOne({ id: idArg } = { id: 'value here' }) {
    console.log(id)
}
findOne()

// 2) Provide a default value
function findOne1({ id: idArg = 'value here 1' } = {}) {}

// 3) Provide initializer and type definition
function findOne2({ id: idArg}: { id?: number } = {}) {}

// 3) Do not provide initializer
function findOne3({ id: idArg}: { id: number }) {}

---

我有一个小问题，如果我用默认值

const{rToken=undefined，theData=undefined}={…data}修复这个问题

tslint会抱怨

没有不必要的初始值设定项

。没有值的变量是

未定义的

，因此

rToken=undefined

是不必要的。我已经知道了这一点，但我不知道TSC为什么会在没有它的情况下抛出错误。

// The compiler checks the object { id: '1' } and it knows it has an id property
var { id } = { id: '1' }

/* The compiler is confused. It check the object {} and it knows it doesn't have 
a property id1, so it is telling you it doesn't know where to get the value 
for id1
*/
var { id1 } = {}

/* In this case the compiler knows the object doesn't have the property id2 but
since you provided a default value it uses it 'default value'.
 */
var { id2 = 'default value' } = {}

/* In your case there are a couple of solutions: */

// 1) Provide the value in the initializer
function findOne({ id: idArg } = { id: 'value here' }) {
    console.log(id)
}
findOne()

// 2) Provide a default value
function findOne1({ id: idArg = 'value here 1' } = {}) {}

// 3) Provide initializer and type definition
function findOne2({ id: idArg}: { id?: number } = {}) {}

// 3) Do not provide initializer
function findOne3({ id: idArg}: { id: number }) {}