Java:将字符串转换为压缩十进制
新来的 情况: 我正在从事一个需要与AS/400服务器通信的项目。我的任务基本上是处理发送到AS/400服务器的请求。为此,所有用户输入都应以EDCDIC字节为单位 问题:Java:将字符串转换为压缩十进制,java,byte,ascii,ibm-midrange,ebcdic,Java,Byte,Ascii,Ibm Midrange,Ebcdic,新来的 情况: 我正在从事一个需要与AS/400服务器通信的项目。我的任务基本上是处理发送到AS/400服务器的请求。为此,所有用户输入都应以EDCDIC字节为单位 问题: 我已成功地使用以下代码将压缩小数转换为字符串,发现: public类PackedDecimal{ 公共静态长解析(字节[]pdIn)引发异常{ //将压缩小数转换为长小数 final int PlusSign=0x0C;//加号 final int MinusSign=0x0D;//减号 final int NoSign=0
我已成功地使用以下代码将压缩小数转换为字符串,发现:
public类PackedDecimal{
公共静态长解析(字节[]pdIn)引发异常{
//将压缩小数转换为长小数
final int PlusSign=0x0C;//加号
final int MinusSign=0x0D;//减号
final int NoSign=0x0F;//未签名
final int DropHO=0xFF;//和用于丢弃HO符号位的掩码
final int GetLO=0x0F;//仅获取LO位
long val=0;//要返回的值
对于(int i=0;i>4;//第一个获取数字
val=val*10+位;
//System.out.println(“digit=“+digit+”,val=“+val”);
int sign=aByte&GetLO;//现在获取符号
如果(符号==最小符号)
val=-val;
否则{
//我们是否关心是否有无效标志?
if(sign!=PlusSign&&sign!=NoSign)
抛出新异常(“OC7”);
}
}否则{
int digit=aByte>>4;//HO优先
val=val*10+位;
//System.out.println(“digit=“+digit+”,val=“+val”);
digit=aByte&GetLO;//现在是LO
val=val*10+位;
//System.out.println(“digit=“+digit+”,val=“+val”);
}
}
返回val;
}//end parse()
//测试上述内容
公共静态void main(字符串[]args)引发异常{
byte[]pd=新字节[]{0x19,0x2C};//192
System.out.println(PackedDecimal.parse(pd));
pd=新字节[]{(字节)0x98、0x44、0x32、0x3D};//-9844323
System.out.println(PackedDecimal.parse(pd));
pd=新字节[]{(字节)0x98、0x44、0x32};//符号无效
System.out.println(PackedDecimal.parse(pd));
}
}
我现在的问题是,我必须将这些字符串值再次转换为EBCDIC字节,以便AS/400服务器能够理解它。我计划使用Silverlake文档中指定的格式构造一个请求(原始字节)。一旦构建了请求,我计划使用一个POJO手动更改该请求中的值,该POJO存储我的请求(带有setter和getter),这样我就可以像request.setField1(“Stuff”.getBytes(Charset.forName(“Cp1047”))
我对比特、字节和半字节没有那么多经验。我希望有人能帮助我
在我们的代码中,我们发现了一个压缩的十进制数,它由5个字节组成。它有点像={000f}。我使用从上面代码中得到的方法来转换它,得到的值是0。现在,我想将0转换回其原始形式,其原始字节大小为5。该库提供的类正是用于此目的。这是我的长到压缩十进制方法版本
public class PackedDecimal {
public static byte[] format(long number, int bytes) {
byte[] b = new byte[bytes];
final byte minusSign = 0x0D; // Minus
final byte noSign = 0x0F; // Unsigned
String s = Long.toString(number);
int length = s.length();
boolean isNegative = false;
if (s.charAt(0) == '-') {
isNegative = true;
s = s.substring(1);
length--;
}
int extraBytes = length - bytes + 1;
if (extraBytes < 0) {
// Pad extra byte positions with zero
for (int i = 0; i < -extraBytes; i++) {
b[i] = 0x00;
}
} else if (extraBytes > 0) {
// Truncate the high order digits of the number to fit
s = s.substring(extraBytes);
length -= extraBytes;
extraBytes = 0;
}
// Translate the string digits into bytes
for (int i = 0; i < length; i++) {
String digit = s.substring(i, i + 1);
b[i - extraBytes] = Byte.valueOf(digit);
}
// Add the sign byte
if (isNegative) {
b[bytes - 1] = minusSign;
} else {
b[bytes - 1] = noSign;
}
return b;
}
public static void main(String[] args) {
long number = -456L;
byte[] b = PackedDecimal.format(number, 5);
System.out.println("Number: " + number + ", packed: " + byteToString(b));
number = 0L;
b = PackedDecimal.format(number, 5);
System.out.println("Number: " + number + ", packed: " + byteToString(b));
number = 5823L;
b = PackedDecimal.format(number, 5);
System.out.println("Number: " + number + ", packed: " + byteToString(b));
number = 123456L;
b = PackedDecimal.format(number, 5);
System.out.println("Number: " + number + ", packed: " + byteToString(b));
}
public static String byteToString(byte[] b) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < b.length; i++) {
sb.append("0x");
sb.append(Integer.toHexString((int) b[i]).toUpperCase());
sb.append(" ");
}
return sb.toString();
}
}
我遇到了一个类似的问题 该类构成了解码/解析PackedDecimals的第一篇文章,工作得很好。。。但是Gilbert Le Blancs答案中的代码没有产生有效的输出 所以我修正了他的密码
public class PackedDecimal {
private static final int PlusSign = 0x0C; // Plus sign
private static final int MinusSign = 0x0D; // Minus
private static final int NoSign = 0x0F; // Unsigned
private static final int DropHO = 0xFF; // AND mask to drop HO sign bits
private static final int GetLO = 0x0F; // Get only LO digit
public static long parse(byte[] pdIn) throws Exception {
long val = 0; // Value to return
for (int i = 0; i < pdIn.length; i++) {
int aByte = pdIn[i] & DropHO; // Get next 2 digits & drop sign bits
if (i == pdIn.length - 1) { // last digit?
int digit = aByte >> 4; // First get digit
val = val * 10 + digit;
log("digit=" + digit + ", val=" + val);
int sign = aByte & GetLO; // now get sign
if (sign == MinusSign)
val = -val;
else {
// Do we care if there is an invalid sign?
if (sign != PlusSign && sign != NoSign) {
System.out.println();
for (int x = 0; x < pdIn.length; x++) {
System.out.print(Integer.toString(pdIn[x] & 0x000000ff, 16));
}
System.out.println();
throw new Exception("OC7");
}
}
} else {
int digit = aByte >> 4; // HO first
val = val * 10 + digit;
log("digit=" + digit + ", val=" + val);
digit = aByte & GetLO; // now LO
val = val * 10 + digit;
log("digit=" + digit + ", val=" + val);
}
}
return val;
}
public static byte[] format(long number, int byteCount) {
byte[] bytes = new byte[byteCount];
String data = Long.toString(number);
int length = data.length();
boolean isNegative = false;
if (data.charAt(0) == '-') {
isNegative = true;
data = data.substring(1);
length--;
}
if (length % 2 == 0) {
data = "0" + data;
length++;
}
int neededBytes = (int) (((length + 1) / 2f) + 0.5f);
int extraBytes = neededBytes - byteCount;
if (extraBytes < 0) {
// Pad extra byte positions with zero
for (int i = 0; i < -extraBytes; i++) {
bytes[i] = 0x00;
}
} else if (extraBytes > 0) {
// Truncate the high order digits of the number to fit
data = data.substring(extraBytes);
length -= extraBytes;
extraBytes = 0;
}
// Translate the string digits into bytes
for (int pos = 0; pos <= length - 1; pos++) {
String digit = data.substring(pos, pos + 1);
int now = (pos / 2) - extraBytes;
if (pos % 2 == 0) { // High
bytes[now] = (byte) (Byte.valueOf(digit) << 4);
log("HIGH " + digit);
} else { // Low
bytes[now] = (byte) (bytes[now] | (Byte.valueOf(digit) & 0x0f));
log("LOW " + digit);
}
}
// Add the sign byte
if (isNegative) {
bytes[byteCount - 1] = (byte) (bytes[byteCount - 1] | MinusSign);
} else {
bytes[byteCount - 1] = (byte) (bytes[byteCount - 1] | PlusSign);
}
return bytes;
}
private static void log(String string) {
// System.out.println(string);
}
public static void main(String[] args) throws Exception {
long price;
byte[] format;
price = 44981;
format = PackedDecimal.format(price, 5);
System.out.println("Input: " + price);
System.out.println("Bytes: " + byteToString(format));
System.out.println("Result: " + PackedDecimal.parse(format));
System.out.println("---------");
price = 4498;
format = PackedDecimal.format(price, 4);
System.out.println("Input: " + price);
System.out.println("Bytes: " + byteToString(format));
System.out.println("Result: " + PackedDecimal.parse(format));
System.out.println("---------");
price = 1337;
format = PackedDecimal.format(price, 3);
System.out.println("Input: " + price);
System.out.println("Bytes: " + byteToString(format));
System.out.println("Result: " + PackedDecimal.parse(format));
System.out.println("---------");
}
public static String byteToString(byte[] b) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < b.length; i++) {
int curByte = b[i] & 0xFF;
sb.append("0x");
if (curByte <= 0x0F) {
sb.append("0");
}
sb.append(Integer.toString(curByte, 16));
sb.append(" ");
}
return sb.toString().trim();
}
}
尽情享受吧这是我的简单解决方案:它将字符串转换为正压缩数 实用程序类:
public class StringToPacked {
public static byte[] stringToPacked(String number){
int length = number.length();
int remainder = length % 2;
if(remainder==0)number ="0" + number;
int quo = length / 2;
byte result[] = new byte[quo+1];
for(int i=0,j=0;i<quo;i++,j=j+2){
int a = Integer.parseInt(number.substring(j,j+1));
int b = Integer.parseInt(number.substring(j+1,j+2));
a = a<<4;
int c = a|b;
result[i]=(byte)c;
}
int a = Integer.parseInt(number.substring(number.length()-1,number.length()));
a = a<<4;
int b = 12;
int c = a|b;
result[quo]=(byte)c;
return result;
}
public static String displayHex(byte number[]){
char array[] ="0123456789ABCDF".toCharArray();
char result[]=new char[2*number.length];
for(int i=0;i<number.length;i++){
result[2*i]=array[number[i]>>4 & 0x0f];
result[2*i+1]=array[number[i] & 0x0f];
}
String okay = new String(result);
return okay;
}
}
public class Packedtest {
public static void main(String[] args) {
// TODO Auto-generated method stub
String number = "123456";
System.out.println(StringToPacked.stringToPacked(number));
byte b1[]=StringToPacked.stringToPacked(number);
String C1 =StringToPacked.displayHex(b1);
System.out.println(number);
System.out.println(C1);
number = "12";
System.out.println(StringToPacked.stringToPacked(number));
b1=StringToPacked.stringToPacked(number);
C1 =StringToPacked.displayHex(b1);
System.out.println(number);
System.out.println(C1);
number = "123";
System.out.println(StringToPacked.stringToPacked(number));
b1=StringToPacked.stringToPacked(number);
C1 =StringToPacked.displayHex(b1);
System.out.println(number);
System.out.println(C1);
number = "1234567";
System.out.println(StringToPacked.stringToPacked(number));
b1=StringToPacked.stringToPacked(number);
C1 =StringToPacked.displayHex(b1);
System.out.println(number);
System.out.println(C1);
number = "12345678";
System.out.println(StringToPacked.stringToPacked(number));
b1=StringToPacked.stringToPacked(number);
C1 =StringToPacked.displayHex(b1);
System.out.println(number);
System.out.println(C1);
}
[B@19e0bfd
123456
0123456C
[B@139a55
12
012C
[B@1db9742
123
123C
[B@106d69c
1234567
1234567C
[B@52e922
12345678
012345678C
结果:
public class StringToPacked {
public static byte[] stringToPacked(String number){
int length = number.length();
int remainder = length % 2;
if(remainder==0)number ="0" + number;
int quo = length / 2;
byte result[] = new byte[quo+1];
for(int i=0,j=0;i<quo;i++,j=j+2){
int a = Integer.parseInt(number.substring(j,j+1));
int b = Integer.parseInt(number.substring(j+1,j+2));
a = a<<4;
int c = a|b;
result[i]=(byte)c;
}
int a = Integer.parseInt(number.substring(number.length()-1,number.length()));
a = a<<4;
int b = 12;
int c = a|b;
result[quo]=(byte)c;
return result;
}
public static String displayHex(byte number[]){
char array[] ="0123456789ABCDF".toCharArray();
char result[]=new char[2*number.length];
for(int i=0;i<number.length;i++){
result[2*i]=array[number[i]>>4 & 0x0f];
result[2*i+1]=array[number[i] & 0x0f];
}
String okay = new String(result);
return okay;
}
}
public class Packedtest {
public static void main(String[] args) {
// TODO Auto-generated method stub
String number = "123456";
System.out.println(StringToPacked.stringToPacked(number));
byte b1[]=StringToPacked.stringToPacked(number);
String C1 =StringToPacked.displayHex(b1);
System.out.println(number);
System.out.println(C1);
number = "12";
System.out.println(StringToPacked.stringToPacked(number));
b1=StringToPacked.stringToPacked(number);
C1 =StringToPacked.displayHex(b1);
System.out.println(number);
System.out.println(C1);
number = "123";
System.out.println(StringToPacked.stringToPacked(number));
b1=StringToPacked.stringToPacked(number);
C1 =StringToPacked.displayHex(b1);
System.out.println(number);
System.out.println(C1);
number = "1234567";
System.out.println(StringToPacked.stringToPacked(number));
b1=StringToPacked.stringToPacked(number);
C1 =StringToPacked.displayHex(b1);
System.out.println(number);
System.out.println(C1);
number = "12345678";
System.out.println(StringToPacked.stringToPacked(number));
b1=StringToPacked.stringToPacked(number);
C1 =StringToPacked.displayHex(b1);
System.out.println(number);
System.out.println(C1);
}
[B@19e0bfd
123456
0123456C
[B@139a55
12
012C
[B@1db9742
123
123C
[B@106d69c
1234567
1234567C
[B@52e922
12345678
012345678C
您希望压缩的十进制数像您的输入一样作为字节数组,还是希望实际的n个字节作为实际的压缩十进制数,哪一个是Java中的字符串?@GilbertLeBlanc:我希望输出也是一个字节数组,因为我将把它传递给as/400服务器。我真的不明白您是如何尝试与as/400通信的。我不完全确定压缩小数,但到目前为止,我可以避免在java或php和as/400之间进行通信时,使用诸如ILE调用或SQL之类的东西进行位/字节操作,甚至避免一些代码页问题。你如何把你的信息放在as/400上?我团队中的一些人已经在做这个了。我的小组被指派在不使用JTOpen和IBM工具箱等的情况下完成这项工作:PI将尝试一下。谢谢您的回复:)哦,顺便问一下,如果没有
int bytes
参数,这可能吗?我的意思是,我是否可以从转换后的字符串本身检索字节数组大小?@Miguel-pulation:可以,但IBM计算机上通常压缩的十进制字段大小固定,与内容无关。压缩十进制字段几乎总是偶数字节。@GilbertLeBlanc:在byteToString方法中,示例代码使用0x加字节整数值的表示法。0x通常表示十六进制值,因此任何大于9的值都会产生误导(例如,输出中的0x13和0x15不正确)。将第二个append更改为sb.append(Integer.toHexString(b[i]);更合适。@jr:你说得对。我不记得三年前我在想什么。谢谢道奇。如何将十进制数(带浮点值)转换为COBOL字节数组,反之亦然?@HaMi对不起,我不知道。而且我已经很久没有谈这个话题了。