如何使用JavaScript中另一个函数的变量

Question

我无法从我的函数UserInfo访问变量，所有变量均未定义。 如何访问变量并将其显示在函数中， seeInfoUser

let UserName;
let UserAge;
let UserBirthPlace;
let UserDream;  

let UserInfo = function(){
  let UserName = prompt("What is your name:");
  let UserAge = prompt("How old are you: ");
  let UserBirthPlace = prompt("Where were you born: ")
  let UserDream = prompt("What is your Greatest Dream: ");
}

let seeInfoUser = function (){              
  let UserInformation = ` ${UserName} is  ${UserAge} and he was born in ${UserBirthPlace} and his greatest dream is ${UserDream}`
  return UserInformation
}

let result = seeInfoUser(UserInfo());
console.log(result)

Answer 1

您正在UserInfo中重新声明变量，这将导致它们隐藏已在更高范围中声明的变量。 只需删除函数内部变量分配上的let关键字，即可使用已声明的变量，而不是重新声明较小的作用域变量。

 // These variables will be available in the current scope and descendent scopes let UserName; let UserAge; let UserBirthPlace; let UserDream; let UserInfo = function(){ // ...So, don't re-declare the variables - just use them! UserName = prompt("What is your name:"); UserAge = prompt("How old are you: "); UserBirthPlace = prompt("Where were you born: ") UserDream = prompt("What is your Greatest Dream: "); } let seeInfoUser = function (){ // You really don't need to declare a variable if all you are going to do is return its value return ` ${UserName} is ${UserAge} and he was born in ${UserBirthPlace} and his greatest dream is ${UserDream}`; } let result = seeInfoUser(UserInfo()); console.log(result) 

Answer 2

代码的问题在于变量的范围。 在javascript中，用let声明的所有变量都具有块作用域。 并且您在UserInfo函数中重新声明了它们，因此您应该只使用已经声明的变量。

 let UserName; let UserAge; let UserBirthPlace; let UserDream; let UserInfo = function() { UserName = prompt("What is your name:"); UserAge = prompt("How old are you: "); UserBirthPlace = prompt("Where were you born: ") UserDream = prompt("What is your Greatest Dream: "); } let seeInfoUser = function() { let UserInformation = ` ${UserName} is ${UserAge} and he was born in ${UserBirthPlace} and his greatest dream is ${UserDream}` return UserInformation } let result = seeInfoUser(UserInfo()); console.log(result)