Java 时间操作

方式一：获取时间

Calendar calendar = Calendar.getInstance();
        int year = calendar.get(Calendar.YEAR);
        int month = calendar.get(Calendar.MONTH)+1;
        int day = calendar.get(Calendar.DAY_OF_MONTH);
        int hour = calendar.get(Calendar.HOUR_OF_DAY);
        int  min = calendar.get(Calendar.MINUTE);
        int second = calendar.get(Calendar.SECOND);

方式二：获取格式化后的时间

public String getThisToday(){
        SimpleDateFormat sdf = new SimpleDateFormat("yyyy:MM:dd:HH:mm:ss");
        Calendar cal = Calendar.getInstance();
        String start = sdf.format(cal.getTime());
        //取出具体年月日
        String[] times=time.split(":");
        String year = times[0];
        String month = times[1];
        String day = times[2];
        String hour = times[3];
        String min = times[4];
        String second = times[5];
        return start;
    }

如果是用第一种方式获取时间，需要个位数前面补0的话，可以进行如下操作：

//这样小时和分钟就会在1位时往前补0，其他以此类推
        if ((hour/10)==0)
            hourS="0"+hour;
        if ((min/10)==0)
            minS="0"+min;
        if ((second/10)==0)
            secondS="0"+secondS;

获取时间戳

​​date2long(2022+"", 8+"", 26+"", 8+"", 1+"", 1+"")​​

public static long date2long(String year, String month, String day, String hour, String minutes, String sec) {
        StringBuilder stringBuilder = new StringBuilder();
        stringBuilder.append(year);
        stringBuilder.append("/");
        stringBuilder.append(month);
        stringBuilder.append("/");
        stringBuilder.append(day);
        stringBuilder.append(" ");
        stringBuilder.append(hour);
        stringBuilder.append(":");
        stringBuilder.append(minutes);
        stringBuilder.append(":");
        stringBuilder.append(sec);

        String sDt = stringBuilder.toString();

        SimpleDateFormat sdf = new SimpleDateFormat("yyyy/MM/dd HH:mm:ss");
        Date dt = null;
        try {
            dt = sdf.parse(sDt);
            if (dt != null) {
                return dt.getTime();
            }
        } catch (ParseException e) {
            e.printStackTrace();
        }
        return -1L;
    }