如何在javascript中捕获无效输入

Question

如何在javascript中实现以下逻辑：

根据限制级影片说明: 17岁以下儿童需有父母或成年监护人( 21周岁及以上)陪同，25周岁及以下成人需出示身份证，6岁以下儿童下午6点后不得入内。

《死穴》是一部限制级电影。

编写一个名为canIWatch的JavaScript函数，该函数将使用年龄作为参数。

如果年龄小于6岁，则返回您在下午6点后不能观看Deadpool。

如年龄在6岁或以上但不足17岁，则须由21岁或以上的监护人陪同。

如果年龄在17岁或以上但小于25岁，请在您出示身份证后立即返回您可以观看Deadpool。

如果年龄大于或等于25，则返回Yay！你可以在没有任何附加条件的情况下观看Deadpool！

如果age无效，则返回invalid age。

我已经写了下面的代码，但仍然不能通过测试，我做错了什么？

function canIWatch(age){
  if(age<6){
    return 'You are not allowed to watch Deadpool after 6.00pm.';
  }
  else if(age>=6 && age<17){
    return 'You must be accompanied by a guardian who is 21 or older.';
  }
  else if(age>=17 && age<25){
    return 'You are allowed to watch Deadpool, right after you show some         ID.';
  }
  else if(age>=25){
    return 'Yay! You can watch Deadpool with no strings attached!';
  }
  else if(!age){
    return 'Invalid age.';
  }
}

我收到一个错误:如果提供的年龄无效，canIWatch测试应该返回相应的消息

预期是“下午6点以后你不能看死亡池”。设置为等于“无效年龄”。

Answer 1

删除else..if处的else，包括isNaN检查是否存在无效的时间。

​

function canIWatch(age) {
  if (!age || isNaN(age) || !isNaN(age) && age <= 0) {
    return 'Invalid age.';
  }
  if (age < 6) {
    return 'You are not allowed to watch Deadpool after 6.00pm.';
  }
  if (age >= 6 && age < 17) {
    return 'You must be accompanied by a guardian who is 21 or older.';
  }
  if (age >= 17 && age < 25) {
    return 'You are allowed to watch Deadpool, right after you show some         ID.';
  }
  if (age >= 25) {
    return 'Yay! You can watch Deadpool with no strings attached!';
  }
}

[5, 6, 16, 17, 25, void 0, null, -7, 0].forEach(function(age) {
  console.log(canIWatch(age))
});

​

Answer 2

我建议您添加一个检查，以确保年龄是一个数字：

function canIWatch(age){
  if (typeof age !== 'number' || age < 0 || age % 1 > 0) {
    return 'Invalid age.';
  }

  if(age<6){
    return 'You are not allowed to watch Deadpool after 6.00pm.';
  }
  else if(age>=6 && age<17){
    return 'You must be accompanied by a guardian who is 21 or older.';
  }
  else if(age>=17 && age<25){
    return 'You are allowed to watch Deadpool, right after you show some         ID.';
  }
  else if(age>=25){
    return 'Yay! You can watch Deadpool with no strings attached!';
  }
}

这里我检查数字的类型是否为年龄，年龄是否不是十进制数。

另请注意:如果您的每条语句都是只返回语句，那么您可能不会每次都编写else if。

所以我重写了你的代码如下：

function canIWatch(age){
  if (typeof age !== 'number' || age < 0 || age % 1 > 0) {
    return 'Invalid age.';
  }

  if(age<6){
    return 'You are not allowed to watch Deadpool after 6.00pm.';
  }

  if(age>=6 && age<17){
    return 'You must be accompanied by a guardian who is 21 or older.';
  }

  if(age>=17 && age<25){
    return 'You are allowed to watch Deadpool, right after you show some         ID.';
  }

  if(age>=25){
    return 'Yay! You can watch Deadpool with no strings attached!';
  }
}

我也格式化了你的代码。

所以最终的版本是：

function canIWatch(age) {
  if (typeof age !== 'number' || age < 0 || age % 1 > 0) {
    return 'Invalid age.';
  }

  if (age < 6) {
    return 'You are not allowed to watch Deadpool after 6.00pm.';
  }

  if (age >= 6 && age < 17) {
    return 'You must be accompanied by a guardian who is 21 or older.';
  }

  if (age >= 17 && age < 25) {
    return 'You are allowed to watch Deadpool, right after you show some         ID.';
  }

  if (age >= 25) {
    return 'Yay! You can watch Deadpool with no strings attached!';
  }
}

Answer 3

像这样的东西应该可以很好地工作。我只是在最后删除了else if条件。

​

function canIWatch(age){
  if(age<6){
    return 'You are not allowed to watch Deadpool after 6.00pm.';
  }
  else if(age>=6 && age<17){
    return 'You must be accompanied by a guardian who is 21 or older.';
  }
  else if(age>=17 && age<25){
    return 'You are allowed to watch Deadpool, right after you show some         ID.';
  }
  else if(age>=25){
    return 'Yay! You can watch Deadpool with no strings attached!';
  }
  else {
    return 'Invalid age.';
  }
}

console.log(canIWatch(4));
console.log(canIWatch(10));
console.log(canIWatch(22));
console.log(canIWatch(29));
console.log(canIWatch("invalid type"));

​