例如：
String data1 = "20090909";
String data2 = "20100101";
data2 - data1 求出相差的月份！
------解决方案--------------------
送你个例子吧
- Java code
/** * 求月数差 * * @param strDate1 日期1 yyyy/MM/dd * @param strDate2 日期2 yyyy/MM/dd * @return 月数差 * @throws ParseException 日期格式错误 */public static int differenceMonth(String strDate1, String strDate2) throws ParseException { Date date1 = DateFormat.getDateInstance().parse(strDate1); Date date2 = DateFormat.getDateInstance().parse(strDate2); return differenceMonth(date1,date2);}public static int differenceMonth(Date date1, Date date2) { Calendar cal1 = Calendar.getInstance(); cal1.setTime(date1); cal1.set(Calendar.DATE, 1); Calendar cal2 = Calendar.getInstance(); cal2.setTime(date2); cal2.set(Calendar.DATE, 1); int count = 0; if (cal1.before(cal2)) { while (cal1.before(cal2)) { cal1.add(Calendar.MONTH, 1); count--; } } else { count--; while (!cal1.before(cal2)) { cal1.add(Calendar.MONTH, -1); count++; } } return count;}
------解决方案--------------------
这不是很难吧，直接写就能写出来
SimpleDateFormat df = new SimpleDateFormat("yyyyMMdd");
Date firstDate = firstDate = df.parse(data1);
Date secondDate = secondDate = df.parse(data2);
return secondDate.getMonth()-firstDate.getMonth()+12*(secondDate.getYear()-firstDate.getYear());
我没处理异常啊。
------解决方案--------------------
使用 joda-time
- Java code
import org.joda.time.*;import org.joda.time.format.*;DateTimeFormatter formatter = DateTimeFormat.forPattern("yyyyMMdd");DateTime d1 = formatter.parseDateTime("20090909");DateTime d2 = formatter.parseDateTime("20100101");Period p = new Period(d1,d2,PeriodType.months());PeriodFormat.getDefault().print(p); // 3 months