Java中String型与Date型数据的互相转换

    /**
     * Date类型转为指定格式的String类型
     *
     * @param source
     * @param pattern
     * @return
     */
    public static String DateToString(Date source, String pattern) {
        simpleDateFormat = new SimpleDateFormat(pattern);
        return simpleDateFormat.format(source);
    }

    /**
     *
     * 字符串转换为对应日期
     *
     * @param source
     * @param pattern
     * @return
     */
    public static Date stringToDate(String source, String pattern) {
        simpleDateFormat = new SimpleDateFormat(pattern);
        Date date = null;
        try {
            date = simpleDateFormat.parse(source);
        } catch (Exception e) {
        }
        return date;
    }

获取两个时间之间的分钟集合

public static List<String[]> findTimes(String dBegin, String dEnd) {
        List<String[]> timeList = new ArrayList<String[]>();
        Calendar calBegin = Calendar.getInstance();
        calBegin.setTime(DateUtils.stringToDate(dBegin, "yyyyMMddHH"));
        Calendar calEnd = Calendar.getInstance();
        calEnd.setTime(DateUtils.stringToDate(dEnd, "yyyyMMddHH"));
        String temps = null;
        String tempe = null;
        while (calBegin.getTime().compareTo(calEnd.getTime()) == -1) {
            String[] lDate = new String[2];
            temps = DateUtils.DateToString(calBegin.getTime(), "yyyyMMddHHmm");
            calBegin.add(Calendar.MINUTE, 1);
            tempe = DateUtils.DateToString(calBegin.getTime(), "yyyyMMddHHmm");
            lDate[0] = temps;
            lDate[1] = tempe;
            timeList.add(lDate);
            temps = tempe;
        }
        return timeList;
    }
 
List<String[]> listInfo=findTimes("2017080801","2017080802");
//结果数组
[["201708080100","201708080101"],["201708080101","201708080102"],.....,["201708080158","201708080159"],["201708080159","201708080200"]]