Question

＆＃13;

var customerData = { 
  'Joe': {visits: 1},
  'Carol': {visits: 2},
  'Howard': {visits: 3},
  'Carrie': {visits: 4}
};

function greetCustomer(firstName) {
  var greeting = '';
  var token = customerData[firstName];
  var value = token['visits'];
  if (firstName in customerData === false) {
    greeting = 'Welcome! Is this your first time?';
  } else if (value === 1) {
    greeting = 'Welcome back, ' + firstName + ', We\'re glad you liked us the first time!';
  } else if (value > 1) {
    greeting = 'Welcome back, ' + firstName + '! So glad to see you again!'
  }
  return greeting;
}

console.log(greetCustomer('Nobody'));

＆＃13;

此if / else函数中的三个条件。没有通过测试的唯一条件是第一个。真的不知道为什么，因为逻辑看起来合法。我怀疑它与访问嵌套对象位有关，但无法弄清楚如何。

Answer 1

好像你只需要重新安排你的逻辑。如果对象中不存在该名称，那么您的令牌变量将为null，并且当您尝试引用该null对象的属性时，您的应用程序将中断。在尝试获取您的令牌值后，如果它为null，则欢迎新人。如果不为null，则继续为现有人员提供逻辑。

var customerData = { 
  'Joe': {visits: 1},
  'Carol': {visits: 2},
  'Howard': {visits: 3},
  'Carrie': {visits: 4}
};

function greetCustomer(firstName) {
  var token = customerData[firstName];
  if (token == null) {
    return 'Welcome! Is this your first time?';
  } else {
    var value = token['visits'];
    if (value === 1) {
      return 'Welcome back, ' + firstName + ', We\'re glad you liked us the first time!';
    } else if (value > 1) {
      return 'Welcome back, ' + firstName + '! So glad to see you again!'
    }
  }
  return '';
}

console.log(greetCustomer('Nobody'));

if / else函数中的TypeError访问嵌套对象

1 个答案: