java压缩到二进制格式,然后解压缩
我有一个任务java压缩到二进制格式,然后解压缩,java,ems,Java,Ems,我有一个任务 将zip文件从本地读入二进制消息 以字符串形式通过EMS传输二进制消息(由java API完成) 以字符串形式接收传输的二进制消息(由java API完成) 解压缩二进制消息,然后将其打印出来 我面临的问题是解压缩消息时出现DataFormatException 我不知道哪一部分出了问题 我使用此命令将文件读入二进制消息: static String readFile_Stream(String fileName) throws IOException { File fil
static String readFile_Stream(String fileName) throws IOException {
File file = new File(fileName);
byte[] fileData = new byte[(int) file.length()];
FileInputStream in = new FileInputStream(file);
in.read(fileData);
String content = "";
System.out.print("Sent message: ");
for(byte b : fileData)
{
System.out.print(getBits(b));
content += getBits(b);
}
in.close();
return content;
}
static String getBits(byte b)
{
String result = "";
for(int i = 0; i < 8; i++)
result = ((b & (1 << i)) == 0 ? "0" : "1") + result;
return result;
}
检查后,二进制消息在传输后不会损坏。
请帮助解决问题。您是否尝试过使用InflaterInputStream?根据我的经验,直接使用充气机是相当棘手的。您可以使用此选项开始:
public static byte[] unzipByteArray(byte[] file) throws IOException {
InflaterInputStream iis = new InflaterInputStream(new ByteArrayInputStream(file));
ByteArrayOutputStream baos = new ByteArrayOutputStream();
byte[] buffer = new byte[512];
int length = 0;
while ((length = iis.read(buffer, 0, buffer.length) != 0) {
baos.write(buffer, 0, length);
}
iis.close();
baos.close();
return baos.toByteArray();
}
我终于解决了这个问题 问题是原始文件是一个.zip文件,所以在进一步处理之前,我应该使用zipInputStream解压该文件
public static byte[] unzipByteArray(byte[] file) throws IOException {
// create a buffer to improve copy performance later.
byte[] buffer = new byte[2048];
byte[] content ;
// open the zip file stream
InputStream theFile = new ByteArrayInputStream(file);
ZipInputStream stream = new ZipInputStream(theFile);
ByteArrayOutputStream output = new ByteArrayOutputStream();
try
{
ZipEntry entry;
while((entry = stream.getNextEntry())!=null)
{
//String s = String.format("Entry: %s len %d added %TD", entry.getName(), entry.getSize(), new Date(entry.getTime()));
//System.out.println(s);
// Once we get the entry from the stream, the stream is
// positioned read to read the raw data, and we keep
// reading until read returns 0 or less.
//String outpath = outdir + "/" + entry.getName();
try
{
//output = new FileOutputStream(outpath);
int len = 0;
while ((len = stream.read(buffer)) > 0)
{
output.write(buffer, 0, len);
}
}
finally
{
// we must always close the output file
if(output!=null) output.close();
}
}
}
finally
{
// we must always close the zip file.
stream.close();
}
content = output.toByteArray();
return content;
}
此代码适用于内部包含单个文件的zip文件。尝试过。但是问题变成了java.util.zip.ZipException:未知的压缩方法
public static byte[] unzipByteArray(byte[] file) throws IOException {
InflaterInputStream iis = new InflaterInputStream(new ByteArrayInputStream(file));
ByteArrayOutputStream baos = new ByteArrayOutputStream();
byte[] buffer = new byte[512];
int length = 0;
while ((length = iis.read(buffer, 0, buffer.length) != 0) {
baos.write(buffer, 0, length);
}
iis.close();
baos.close();
return baos.toByteArray();
}
public static byte[] unzipByteArray(byte[] file) throws IOException {
// create a buffer to improve copy performance later.
byte[] buffer = new byte[2048];
byte[] content ;
// open the zip file stream
InputStream theFile = new ByteArrayInputStream(file);
ZipInputStream stream = new ZipInputStream(theFile);
ByteArrayOutputStream output = new ByteArrayOutputStream();
try
{
ZipEntry entry;
while((entry = stream.getNextEntry())!=null)
{
//String s = String.format("Entry: %s len %d added %TD", entry.getName(), entry.getSize(), new Date(entry.getTime()));
//System.out.println(s);
// Once we get the entry from the stream, the stream is
// positioned read to read the raw data, and we keep
// reading until read returns 0 or less.
//String outpath = outdir + "/" + entry.getName();
try
{
//output = new FileOutputStream(outpath);
int len = 0;
while ((len = stream.read(buffer)) > 0)
{
output.write(buffer, 0, len);
}
}
finally
{
// we must always close the output file
if(output!=null) output.close();
}
}
}
finally
{
// we must always close the zip file.
stream.close();
}
content = output.toByteArray();
return content;
}