先说明一些用到的小知识，变化过程以代码说明。
- 左位移
<<，右位移>>。
short i = 10;
short j = -10;
//右移操作
short a =(short) ( i >> 2);
//10->(二进制)0000000000001010->(右移2位)0000000000000010->(10进制)2
short b =(short) ( j >> 2);
//-10->(二进制)1000000000001010->(补码-反码+1)1111111111110110->(右移2位)1111111111111101->(-1 反码)100000000000011->(10进制)-3
//左移操作
short c = (short)( i << 2);
//10->(二进制)0000000000001010->(左移2位)0000000000101000->(10进制)40
short d = (short)( j << 2);
//-10->(二进制)1000000000001010->(补码，反码+1)1111111111110110->(左移2位)1111111111011000->(-1 反码 )1000000000101000->(10进制)-40
-
&0XFF操作。
byte i = 10;
byte j = -10;
short t = 0xff;
byte a = (byte)(i&t);
//10(10)->00001010(2)
//FF(16)->0000000011111111(2)
//00001010&0000000011111111->0000000000001010(2)->10(10)
short b = (short)(j&t);
//-10(10)->00001010(2)->11110110(补码)
//FF(16)->0000000011111111(2)
//11110110&0000000011111111->0000000011110110(2)->246(10)
很明显，-10二进制补码为11110110，&0xFF之后的原码为0000000011110110。所以此操作的效果就是保留了负数的二进制补码，便于后续的使用和计算。
时间
public class DateTool {
public static String[] presuppose = {"yyyy","MM","dd","HH","mm","ss","SSS"};
public static void main(String[] args) {
//当前时间的各种格式
{
Date date = new Date();
String patten = "yyyy/MM/dd HH:mm:ss";
SimpleDateFormat df = new SimpleDateFormat(patten);
String returnValue = df.format(date);
System.out.println(returnValue);
}
//时间戳转对象
{
//得到时间对象
long d = System.currentTimeMillis();
Date date = new Date(d);
System.out.println(date);
}
//时间字符串根据格式转时间对象
{
String d = "20190130144025";
String patten = "yyyyMMddHHmmss";
SimpleDateFormat df = new SimpleDateFormat(patten);
try {
Date date = df.parse(d);
System.out.println(date);
} catch (ParseException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
//当前时间往前，往后
{
Date date = new Date();
//当前时间
Calendar cal = Calendar.getInstance();
cal.setTime(date);
//后挪3天
cal.add(Calendar.DATE, 3);
//新的时间
Date newDate = cal.getTime();
String patten = "yyyy/MM/dd HH:mm:ss";
SimpleDateFormat df = new SimpleDateFormat(patten);
String returnValue = df.format(newDate);
System.out.println(returnValue);
}
//计算距今天数
{
Date date;
try {
String d = "20190129144025";
String patten = "yyyyMMddHHmmss";
SimpleDateFormat df = new SimpleDateFormat(patten);
date = df.parse(d);
long t = Calendar.getInstance().getTime().getTime();
Calendar c = Calendar.getInstance();
c.setTime(date);
long t1 = c.getTime().getTime();
int day = (int) (t / 1000 - t1 / 1000) / 3600 / 24;
System.out.println("距今" + day +"天");
} catch (ParseException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
//时间串变时间戳
{
String d = "20190129144025";
String patten = "yyyyMMdd";
try {
long timestamp = new SimpleDateFormat(patten).parse(d).getTime();
System.out.println("时间戳:" + timestamp);
} catch (ParseException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
//比较时间
{
String s = "20190129144025";
String e = "20190129184025";
String patten = "yyyyMMddHHmmss";
SimpleDateFormat df = new SimpleDateFormat(patten);
try {
Date sd = df.parse(s);
Date ed = df.parse(e);
if(sd.getTime() > ed.getTime()){
System.out.println("sd比较大");
}else{
System.out.println("ed比较大");
}
if(sd.before(ed)){
System.out.println("sd比较小");
}else{
System.out.println("ed比较小");
}
} catch (ParseException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
}
}
}
位图
image.png
位图目前我所知道是用于ISO8583协议的数据传输。根据位图可以知道哪些数据域有数据需要解析。下面提供64域的生成位图和解析位图的方法，如果是128域的，需要稍微扩展一下。
public class BitMapTool {
static byte[] mask = new byte[128];
static {
for (int i = 0; i <= 9; i++)
{
mask[i + 48] = (byte) i;
}
for (int i = 0; i <= 5; i++)
{
mask[i + 97] = (byte) (10 + i);
}
for (int i = 0; i <= 5; i++)
{
mask[i + 65] = (byte) (10 + i);
}
}
static String bitmap(int... vs){
//第一步 将域序号塞入byte域组中，每个byte标识一个域，此方法用于64域数据
byte[] map = new byte[64];
for (int i = 0; i < vs.length; i++)
{
if (0 != vs[i])
{
int b = vs[i];
map[b-1] = 1;
}
}
StringBuffer buff = new StringBuffer();
//64个字节分为8块，每块的8个字节转为16进制hex
for(int j = 0 ; j < map.length/8 ; j++){
byte[] temp = new byte[8];
System.arraycopy(map, j*8, temp, 0, 8);
StringBuffer sb = new StringBuffer();
for (byte b : temp) {
sb.append(b);
}
byte ret = (byte) (Integer.valueOf(sb.toString(), 2) & 0xff);
StringBuffer f = new StringBuffer();
int n = ret & 0xff;
if (n < 0x10)
f.append("0");
f.append(Integer.toHexString(n));
String t = f.toString();
if (null != t)
t = t.toUpperCase(Locale.getDefault());
buff.append(t);
}
return buff.toString();
}
//解析位图
public static int[] analysisBitmap(String va){
int len = (va.length() / 2);
byte[] hex = new byte[len];
char[] achar = va.toCharArray();
for (int i = 0; i < len; i++)
{
int pos = i * 2;
hex[i] = (byte) (mask[achar[pos]] << 4 | mask[achar[pos + 1]]);
}
int[] x = new int[64];
int p = 0;
for (int i = 0; i < hex.length; i++) {
if(0 != hex[i]){
String t = Integer.toBinaryString(hex[i]&0xFF);
StringBuffer sb = new StringBuffer();
for (int j = 0; j < 8-t.length(); j++) {
sb.append("0");
}
sb.append(t);
t = sb.toString();
char[] c = t.toCharArray();
for (int j = 0 ; j < c.length ; j++) {
if(c[j]=='1'){
x[p] = 8*i + j+1;
p++;
}
}
}
}
int[] re = new int[p];
System.arraycopy(x, 0, re, 0, p);
return re;
}
public static void main(String[] args) {
String re = bitmap(4,10,17,28,36,48,53,60);
System.out.println(re);
int[] s = analysisBitmap(re);
for (int i = 0; i < s.length; i++) {
System.out.println("解析位图之后:" + s[i]);
}
}
}
网友评论