blocks|key|277949|text|问题是Integer不够大，无法存储当前日期，您需要使用Long。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|277950|该日期在内部存储为自1970年1月1日以来的毫秒数。|277951|最大整数值为2147483648，而自1970年以来的毫秒数目前约为1345618537869|277952|将最大整数值放入日期中会得到1970年1月26日星期一的结果。|277953|编辑:按照下面的注释显示除以1000的代码：|277954|++++int+i+=+(int)+(new+Date().getTime()/1000);
++++System.out.println("Integer+:+"+%2B+i);
++++System.out.println("Long+:+"%2B+new+Date().getTime());
++++System.out.println("Long+date+:+"+%2B+new+Date(new+Date().getTime()));
++++System.out.println("Int+Date+:+"+%2B+new+Date(((long)i)*1000L));

Integer+:+1345619256
Long+:+1345619256308
Long+date+:+Wed+Aug+22+16:37:36+CST+2012
Int+Date+:+Wed+Aug+22+16:37:36+CST+2012|code-block|syntax|javascript|277955|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|T|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|U|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|V|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

The issue is that an Integer is not large enough to store a current date, you need to use a Long.

The date is stored internally as the number of milliseconds since 1/1/1970.

The maximum Integer value is 2147483648, whereas the number of milliseconds since 1970 is currently in the order of 1345618537869

Putting the maximum integer value into a date yields Monday 26th January 1970.

Edit: Code to display division by 1000 as per comment below:

<pre><code> int i = (int) (new Date().getTime()/1000);
 System.out.println("Integer : " + i);
 System.out.println("Long : "+ new Date().getTime());
 System.out.println("Long date : " + new Date(new Date().getTime()));
 System.out.println("Int Date : " + new Date(((long)i)*1000L));

Integer : 1345619256
Long : 1345619256308
Long date : Wed Aug 22 16:37:36 CST 2012
Int Date : Wed Aug 22 16:37:36 CST 2012
</code></pre>

blocks|key|278009|text|你需要这样的东西(没有时间)吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|278010|public+static+Integer+toJulianDate(Date+pDate)+{
if+(pDate+==+null)+{
++return+null;
}
Calendar+lCal+=+Calendar.getInstance();
lCal.setTime(pDate);
int+lYear+=+lCal.get(Calendar.YEAR);
int+lMonth+=+lCal.get(Calendar.MONTH)+%2B+1;
int+lDay+=+lCal.get(Calendar.DATE);
int+a+=+(14+-+lMonth)+/+12;
int+y+=+lYear+%2B+4800+-+a;
int+m+=+lMonth+%2B+12+*+a+-+3;
return+lDay+%2B+(153+*+m+%2B+2)+/+5+%2B+365+*+y+%2B+y+/+4+-+y+/+100+%2B+y+/+400+-+32045;
}|code-block|syntax|javascript|278011|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Do you need something like this(without time)?

<pre><code>public static Integer toJulianDate(Date pDate) {
if (pDate == null) {
 return null;
}
Calendar lCal = Calendar.getInstance();
lCal.setTime(pDate);
int lYear = lCal.get(Calendar.YEAR);
int lMonth = lCal.get(Calendar.MONTH) + 1;
int lDay = lCal.get(Calendar.DATE);
int a = (14 - lMonth) / 12;
int y = lYear + 4800 - a;
int m = lMonth + 12 * a - 3;
return lDay + (153 * m + 2) / 5 + 365 * y + y / 4 - y / 100 + y / 400 - 32045;
}
</code></pre>

blocks|key|278093|text|在我的Java7上，输出是不同的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|278094|Integer+:+1293732698
Long+:+1345618496346
Long+date+:+Wed+Aug+22+10:54:56+MSK+2012
Int+Date+:+Fri+Jan+16+02:22:12+MSK+1970|code-block|syntax|javascript|278095|这是一种预期的行为。|278096|不可能以毫秒为单位将当前日期显示为整数(10位数字)，因为最小可能的日期为Sun+Apr+26+20:46:39+MSK+1970|offset|length|style|CODE|278097|Date+d+=+new+Date(9999_9999_99L);
System.out.println("Date:+"+%2B+d);

Date:+Sun+Apr+26+20:46:39+MSK+1970|278098|您可能需要考虑以秒/分钟为单位显示它。|278099|entityMap^0|0|0|0|11|S|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@$K|Y|L|Z|M|N]]|9|@]|A|$]]|$1|O|3|P|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

On my Java 7, the output is different:

<pre><code>Integer : 1293732698
Long : 1345618496346
Long date : Wed Aug 22 10:54:56 MSK 2012
Int Date : Fri Jan 16 02:22:12 MSK 1970
</code></pre>

which is an expected behavior.

It is impossible to display the current date in milliseconds as an integer (10 digit number), because the latest possible date is <code>Sun Apr 26 20:46:39 MSK 1970</code>:

<pre><code>Date d = new Date(9999_9999_99L);
System.out.println("Date: " + d);

Date: Sun Apr 26 20:46:39 MSK 1970
</code></pre>

You might want to consider displaying it in seconds/minutes.

blocks|key|495054|text|也许你不能，Long是比Integer更高的数据类型。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|495055|或者这个链接可能会帮助你|495056|http://www.velocityreviews.com/forums/t142373-convert-date-to-integer-and-back.html|offset|length|495057|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|2B|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@$F|R|G|S|1|T]]|A|$]]|$1|H|3|-4|5|6|7|U|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|E]]]]

Probably You can not, Long is higher datatype than Integer.

or this link might help you

<a href="http://www.velocityreviews.com/forums/t142373-convert-date-to-integer-and-back.html" rel="nofollow">http://www.velocityreviews.com/forums/t142373-convert-date-to-integer-and-back.html</a>

blocks|key|495117|text|为了获得整数形式的当前日期(10位数字)，需要将从new+Date().getTime()返回的长整型除以1000。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|495118|这将在int范围内，并在2038年1月18日之前有效。|495119|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

In order to get current date as integer(10 digit number), you need to divide the long returned from new Date().getTime() by 1000.

This will be in int range and is good until 18 Jan 2038.

blocks|key|495163|text|如果您只需要一个表示自1970年1月1日以来经过的天数的整数，您可以尝试以下方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|495164|//+magic+number=
//+millisec+*+sec+*+min+*+hours
//+1000+*+60+*+60+*+24+=+86400000
public+static+final+long+MAGIC=86400000L;

public+int+DateToDays+(Date+date){
++++//++convert+a+date+to+an+integer+and+back+again
++++long+currentTime=date.getTime();
++++currentTime=currentTime/MAGIC;
++++return+(int)+currentTime;+
}

public+Date+DaysToDate(int+days)+{
++++//++convert+integer+back+again+to+a+date
++++long+currentTime=(long)+days*MAGIC;
++++return+new+Date(currentTime);
}|code-block|syntax|javascript|495165|较短但可读性较差(稍微快一点？)：|495166|public+static+final+long+MAGIC=86400000L;

public+int+DateToDays+(Date+date){
++++return+(int)+(date.getTime()/MAGIC);
}

public+Date+DaysToDate(int+days)+{
++++return+new+Date((long)+days*MAGIC);
}|495167|希望这能有所帮助。|495168|编辑:这可能会持续到周五7月11日01:00:00+CET+5881580|495169|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

If you need only an integer representing elapsed days since Jan. 1, 1970, you can try these:

<pre><code>// magic number=
// millisec * sec * min * hours
// 1000 * 60 * 60 * 24 = 86400000
public static final long MAGIC=86400000L;

public int DateToDays (Date date){
 // convert a date to an integer and back again
 long currentTime=date.getTime();
 currentTime=currentTime/MAGIC;
 return (int) currentTime; 
}

public Date DaysToDate(int days) {
 // convert integer back again to a date
 long currentTime=(long) days*MAGIC;
 return new Date(currentTime);
}
</code></pre>

Shorter but less readable (slightly faster?):

<pre><code>public static final long MAGIC=86400000L;

public int DateToDays (Date date){
 return (int) (date.getTime()/MAGIC);
}

public Date DaysToDate(int days) {
 return new Date((long) days*MAGIC);
}
</code></pre>

Hope this helps.

EDIT: This could work until Fri Jul 11 01:00:00 CET 5881580

blocks|key|495209|text|Simple确实创建了一个代表程序默认开始日期的long变量，将该日期转换为另一个long变量。然后扣除长的开始日期并转换为整数，这样只需添加而不是减法即可读取并转换回来。显然，这取决于您需要的日期范围有多大。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|495210|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Simple really create a long variable that represents a default start date for your program
Get the date to another long variable.
Then deduct the long start date and convert to a integer voila
To read and convert back just add rather than subtract.
obviously this is dependant on how large a date range you require.

blocks|key|495241|text|我已经解决了这个问题，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|495242|++++long+year+=+calendar.get(Calendar.YEAR);
++++long+month+=+calendar.get(Calendar.MONTH)+%2B+1;
++++long+day+=+calendar.get(Calendar.DAY_OF_MONTH);
++++long+calcDate+=+year+*+100+%2B+month;
++++calcDate+=+calcDate+*+100+%2B+day;
++++System.out.println("int:+"+%2B+calcDate);|code-block|syntax|javascript|495243|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I've solved this as is shown below:

<pre><code> long year = calendar.get(Calendar.YEAR);
 long month = calendar.get(Calendar.MONTH) + 1;
 long day = calendar.get(Calendar.DAY_OF_MONTH);
 long calcDate = year * 100 + month;
 calcDate = calcDate * 100 + day;
 System.out.println("int: " + calcDate);
</code></pre>

blocks|key|278349|text|尝尝这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|278350|Calendar+currentDay=+Calendar.getInstance();
int+currDate=+currentDay.get(Calendar.DATE);
int+currMonth=+currentDay.get(Calendar.MONTH);
int+currYear=+currentDay.get(Calendar.YEAR);
System.out.println(currDate+%2B+"-"+%2B++currMonth+%2B+"-"+%2B+currYear);|code-block|syntax|javascript|278351|另一种使用LocalDate的方法。|278352|LocalDate+today+=+LocalDate.now();
int+currentDate=+today.getDayOfMonth();
int+currentMonth=+today.getMonthValue();
int+currentYear=+today.getYear()|278353|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Try This 

<pre><code>Calendar currentDay= Calendar.getInstance();
int currDate= currentDay.get(Calendar.DATE);
int currMonth= currentDay.get(Calendar.MONTH);
int currYear= currentDay.get(Calendar.YEAR);
System.out.println(currDate + "-" + currMonth + "-" + currYear);
</code></pre>

an alternative way using LocalDate.

<pre><code>LocalDate today = LocalDate.now();
int currentDate= today.getDayOfMonth();
int currentMonth= today.getMonthValue();
int currentYear= today.getYear()
</code></pre>

blocks|key|495287|text|Java+Date到int的转换：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|495288|public+static+final+String+DATE_FORMAT_INT+=+"yyyyMMdd";

public+static+String+format(Date+date,+String+format)+{
++++return+isNull(date)+?++
++++null+:+new+SimpleDateFormat(format).format(date);
}

public+static+Integer+getDateInt(Date+date)+{
++++if+(isNull(date))+{
++++++++throw+new+IllegalArgumentException("Date+must+not+be+NULL");
++++}

++++return+parseInt(format(date,+DATE_FORMAT_INT));
}|code-block|syntax|javascript|495289|entityMap^0|5|4|A|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Java <code>Date</code> to <code>int</code> conversion: 

<pre><code>public static final String DATE_FORMAT_INT = "yyyyMMdd";

public static String format(Date date, String format) {
 return isNull(date) ? 
 null : new SimpleDateFormat(format).format(date);
}

public static Integer getDateInt(Date date) {
 if (isNull(date)) {
 throw new IllegalArgumentException("Date must not be NULL");
 }

 return parseInt(format(date, DATE_FORMAT_INT));
}
</code></pre>

I want to convert the current date to integer value. By default, it returns long. When I try to convert long to integer, and afterwards I convert the integer value to date, means it shows 1970's date,

<pre><code> int i = (int) new Date().getTime();
 System.out.println("Integer : " + i);
 System.out.println("Long : "+ new Date().getTime());
 System.out.println("Long date : " + new Date(new Date().getTime()));
 System.out.println("Int Date : " + new Date(i));
</code></pre>

the output as follows:

<pre><code>Integer : 1292838124
Long : 1345617601771
Long date : Wed Aug 22 12:10:01 IST 2012
Int Date : Fri Jan 16 04:37:18 IST 1970
</code></pre>

Any one please help me out, how to convert current date to integer(10 digit number)?

Convert Current date to integer

Java

我想将当前日期转换为整数值。默认情况下，它返回long。当我尝试将long转换为整数，然后将整数值转换为date，这意味着它显示了1970年的日期， int i = (int) new Date().getTime(); System.out.println("Integer : " + i); System.out....

问将当前日期转换为整数
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问将当前日期转换为整数EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问将当前日期转换为整数
EN