更改日期格式（Javascript）

发布于 2024-11-09 03:12:17 字数 336 浏览 0 评论 0原文

我有这个代码：

var fd=1+self.theDate.getMonth() +'/'+ today+'/'+self.theDate.getFullYear();

它有效，但它的格式是月、日、年。

我需要将其更改为：日、月、年。

所以，我尝试了以下方法：

var fd=1+today +'/'+ self.theDate.getMonth()+'/'+self.theDate.getFullYear();

现在，我的更改不起作用。是我没做对还是我的改变对了？

谢谢

原文

I have this code:

var fd=1+self.theDate.getMonth() +'/'+ today+'/'+self.theDate.getFullYear();

It works, but it's format is Month, Day, Year.

I need to change it to: Day, Month Year.

So, I tried this:

var fd=1+today +'/'+ self.theDate.getMonth()+'/'+self.theDate.getFullYear();

Now, my change does not work. Is it that I have not done it properly or is my change right?

Thanks

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⊕婉儿 2024-11-16 03:12:17

我预计正确的答案是这样的：

var fd=today +'/'+ (self.theDate.getMonth() + 1) +'/'+self.theDate.getFullYear();

这将单独保留今天，并将月份分组，以便它进行正确的数字加法而不是字符串连接。

I expect the correct answer is this:

var fd=today +'/'+ (self.theDate.getMonth() + 1) +'/'+self.theDate.getFullYear();

This leaves today alone, and groups Month so that it does a proper number addition instead of string concatenation.

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以可爱出名 2024-11-16 03:12:17

var theDate = new Date();
var today = theDate.getDate();
var month = theDate.getMonth()+1; // js months are 0 based
var year = theDate.getFullYear();
var fd=today +'/'+ month +'/'+year

或者您可能更喜欢 22/05/2011

var theDate = new Date();
var today = theDate.getDate();
if (today<10) today="0"+today;
var month = theDate.getMonth()+1; // js months are 0 based
if (month < 10) month = "0"+month;
var year = theDate.getFullYear();
var fd=""+today +"/"+ month +"/"+year

var theDate = new Date();
var today = theDate.getDate();
var month = theDate.getMonth()+1; // js months are 0 based
var year = theDate.getFullYear();
var fd=today +'/'+ month +'/'+year

or perhaps you prefer 22/05/2011

var theDate = new Date();
var today = theDate.getDate();
if (today<10) today="0"+today;
var month = theDate.getMonth()+1; // js months are 0 based
if (month < 10) month = "0"+month;
var year = theDate.getFullYear();
var fd=""+today +"/"+ month +"/"+year

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