在纯Java6中为非常大的文件创建SHA-1的最佳方法是什么?如何实现此方法:
public abstract String createSha1(java.io.File file);
使用MessageDigest
类并逐个提供数据。下面的示例忽略了将byte []转换为字符串并关闭文件等细节,但应该给出一般的想法。
public byte[] createSha1(File file) throws Exception {
MessageDigest digest = MessageDigest.getInstance("SHA-1");
InputStream fis = new FileInputStream(file);
int n = 0;
byte[] buffer = new byte[8192];
while (n != -1) {
n = fis.read(buffer);
if (n > 0) {
digest.update(buffer, 0, n);
}
}
return digest.digest();
}
操作系统要求函数返回SHA1的字符串,所以我接受了@jeffs的答案并将缺少的转换添加到String:
/**
* Read the file and calculate the SHA-1 checksum
*
* @param file
* the file to read
* @return the hex representation of the SHA-1 using uppercase chars
* @throws FileNotFoundException
* if the file does not exist, is a directory rather than a
* regular file, or for some other reason cannot be opened for
* reading
* @throws IOException
* if an I/O error occurs
* @throws NoSuchAlgorithmException
* should never happen
*/
private static String calcSHA1(File file) throws FileNotFoundException,
IOException, NoSuchAlgorithmException {
MessageDigest sha1 = MessageDigest.getInstance("SHA-1");
try (InputStream input = new FileInputStream(file)) {
byte[] buffer = new byte[8192];
int len = input.read(buffer);
while (len != -1) {
sha1.update(buffer, 0, len);
len = input.read(buffer);
}
return new HexBinaryAdapter().marshal(sha1.digest());
}
}
public static String computeFileSHA1( File file ) throws IOException
{
String sha1 = null;
MessageDigest digest;
try
{
digest = MessageDigest.getInstance( "SHA-1" );
}
catch ( NoSuchAlgorithmException e1 )
{
throw new IOException( "Impossible to get SHA-1 digester", e1 );
}
try (InputStream input = new FileInputStream( file );
DigestInputStream digestStream = new DigestInputStream( input, digest ) )
{
while(digestStream.read() != -1){
// read file stream without buffer
}
MessageDigest msgDigest = digestStream.getMessageDigest();
sha1 = new HexBinaryAdapter().marshal( msgDigest.digest() );
}
return sha1;
}