如果在循环中不满足某个条件，则返回的循环函数

Question

我需要递归调用 UsernameFn（cc 后面的行），直到调用它的条件循环变得不真实。如果我将循环完全放在 UsernameFn 函数之外，控制台会显示 Username 未定义，即使 var Username 是全局声明的。我该怎么办？

[它说要添加更多细节。我没有更多的细节。它说要添加更多细节。我没有更多的细节。它说要添加更多细节。我没有更多的细节。它说要添加更多细节。我没有更多的细节。 ]

// dataset for username 

const alpha = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'];
const num = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '0'];
var counter = 0;
var Username;

// determine random length of the username
var UNlength = Math.random() * 11;
UNlength = UNlength.toFixed(0);

function UsernameFn() {
  // generate random nums as pickers
  var PickerN = Math.random() * 10;
  PickerN = PickerN.toFixed(0);
  var PickerA = Math.random() * 26;
  PickerA = PickerA.toFixed(0);
  var PickerG = Math.random() * 10;
  PickerG = PickerG.toFixed(0);

  // Pickers for the first character in "Username"
  var PickerUN = Math.random() * 10;
  PickerUN = PickerUN.toFixed(0);
  var PickerUA = Math.random() * 26;
  PickerUA = PickerUA.toFixed(0);
  var PickerUG = Math.random() * 10;
  PickerUG = PickerUG.toFixed(0);

  // choose randomly btw num and alpha. If Math.random 
  // is an even num, option = alpha. If odd num, option = num
  if (PickerG % 2 == 0) {
    var Chosen = num[PickerN];
  } else {
    var Chosen = alpha[PickerA];
  }

  if (PickerUG % 2 == 0) {
    var Username = num[PickerUN];
  } else {
    var Username = alpha[PickerUA];
  }

  //append each new choice to a string of previous choices
  //when string = UNlength <= 10 chosen ranbdomly above, stop recursion.
  // 'Username' declaration outside scope to allow loop access
  Username = String(Username) + String(Chosen);
  return Username;

  counter += 1;
  do {
    UsernameFn()
  } while (Username.length <= UNlength.length);
}

Answer 1

循环应该在函数之外。然后，您需要将返回值连接到 Username 变量。

此外，UNlength.length 应该只是 UNlength。

// dataset for username 

const alpha = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'];
const num = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '0'];
var counter = 0;
var Username = '';

// determine random length of the username
var UNlength = Math.random() * 11;
UNlength = UNlength.toFixed(0);

do {
  Username += UsernameFn();
  counter++;
} while (Username.length <= UNlength);

console.log(Username, counter);

function UsernameFn() {
  // generate random nums as pickers
  var PickerN = Math.random() * 10;
  PickerN = PickerN.toFixed(0);
  var PickerA = Math.random() * 26;
  PickerA = PickerA.toFixed(0);
  var PickerG = Math.random() * 10;
  PickerG = PickerG.toFixed(0);

  // Pickers for the first character in "Username"
  var PickerUN = Math.random() * 10;
  PickerUN = PickerUN.toFixed(0);
  var PickerUA = Math.random() * 26;
  PickerUA = PickerUA.toFixed(0);
  var PickerUG = Math.random() * 10;
  PickerUG = PickerUG.toFixed(0);

  // choose randomly btw num and alpha. If Math.random 
  // is an even num, option = alpha. If odd num, option = num
  if (PickerG % 2 == 0) {
    var Chosen = num[PickerN];
  } else {
    var Chosen = alpha[PickerA];
  }

  if (PickerUG % 2 == 0) {
    var Username = num[PickerUN];
  } else {
    var Username = alpha[PickerUA];
  }

  //append each new choice to a string of previous choices
  //when string = UNlength <= 10 chosen ranbdomly above, stop recursion.
  // 'Username' declaration outside scope to allow loop access
  Username = String(Username) + String(Chosen);
  return Username;
}

如果在循环中不满足某个条件，则返回的循环函数

Recurring fuction which reruns if a certain condition is not met in a loop

javascript

recursion