用错误的转换将长日期转换为短日期

Question

这是我的长期约会：

Fri Dec 07 2018 05:47:22 GMT+0000

但是当我将其转换为短日期时，它返回了错误的日期：

var date = 'Fri Dec 07 2018 05:47:22 GMT+0000';
var convertedStartDate = new Date(date);
var year    = convertedStartDate.getFullYear();
var month   = convertedStartDate.getMonth();
var day     = convertedStartDate.getDay();
console.log(year+'/'+month+'/'+day)

浏览器的输出：

2018/11/5

我的Mac OS日期：

Answer 1

 var date = 'Fri Dec 07 2018 05:47:22 GMT+0000'; var convertedStartDate = new Date(date); var year = convertedStartDate.getFullYear(); // 2018 var month_index = convertedStartDate.getMonth(); // 11 month name index var weekday_index = convertedStartDate.getDay(); // 5 weekday index var day_date = convertedStartDate.getDate(); // 7 var day_names = ["Sun","Mon","Tue","Wed","Thu","Fri","Sat"]; var month_names = ["Jan","Feb","Mar","Apr", "May","Jun","Jul","Aug", "Sep","Oct","Nov","Dec"]; var month_name = month_names[month_index]; // Dec var day = day_names[weekday_index]; // Fri console.log(day+" "+day_date+"-"+month_name+"-"+year); // Fri 7-Dec-2018 console.log(year+"/"+(month_index+1)+"/"+day_date); // 2018/12/7 

var month = convertedStartDate.getMonth(); 

这将返回0到11之间的值。原因是您可能会在数组中使用此索引，例如：

  var month_names = ["jan","feb", .... "dec"];
  console.log(month_names[month]);   //dec

但是，如果您只对数值感兴趣，只需加1

  var month = convertedStartDate.getMonth() + 1; 

至于getDay() ，它将返回一个介于0和6之间的值，这样您就可以像这样使用它：

 var day     = convertedStartDate.getDay();
 var day_names = ['Sun', 'Mon',... ,'Sat'];
 console.log(day_names[day]);    

但是如果只需要数字值（介于1和28/29/30/31之间），则需要.getDate()而不是.getDay() 。 但是，在这种情况下，无需添加一个。

Answer 2

 var date = 'Fri Dec 07 2018 05:47:22 GMT+0000'; var convertedStartDate = new Date(date); var year = convertedStartDate.getFullYear(); var month = convertedStartDate.getMonth() + 1; var day = convertedStartDate.getDate(); console.log(year+'/'+month+'/'+day) 

将getDay()方法替换为getDate()并在月份中添加1，因为getMonth()将在0-11之间返回

看，我添加了工作示例，也希望它能解决问题。