为什么会出现AmazonS3Exception？

huangapple
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英文:

Why am I getting a AmazonS3Exception?

问题

我的任务是将上传到S3的文件进行加密。在加密之前，上传工作正常，但是在我对文件进行加密后，出现了以下异常。

所提供的 XML 不符合格式要求，或未根据我们发布的架构进行验证

我将以下代码添加到现有代码中。

  1. 		final AwsCrypto crypto = new AwsCrypto();
  2. 		
  3. 		try (
  4. 			final FileInputStream in = new FileInputStream(encryptfile);
  5.             final FileOutputStream out = new FileOutputStream(file);
  6.             final CryptoOutputStream<?> encryptingStream = crypto.createEncryptingStream(crypt, out)) 
  7. 		{
  8.             IOUtils.copy(in, encryptingStream);
  9. 		}

我的想法是，为什么AmazonS3希望是XML文件？而不是普通的文本文档？
是否有选项可以通过存储桶策略进行更改？

编辑
这是上传代码，可能存在问题。我不明白为什么没有加密时它能正常工作。

  1.              File uploaffile = encryptFile(file);
  2. 			 List<PartETag> partETags = new ArrayList<PartETag>();
  3. 			 String filename = String.valueOf(System.currentTimeMillis());
  4. 	
  5. 	         InitiateMultipartUploadRequest initRequest = new InitiateMultipartUploadRequest(awss3bucket, filename);
  6. 	         InitiateMultipartUploadResult initResponse = amazons3.initiateMultipartUpload(initRequest);
  7. 	         long partSize = 5 * 1024 * 1024; 
  8. 	         long contentLength = uploaffile.length();
  9. 	         long filePosition = 0;
  10. 	         for (int i = 1; filePosition < contentLength; i++) {
  11. 	             partSize = Math.min(partSize, (contentLength - filePosition));
  12. 	
  13. 	             UploadPartRequest uploadRequest = new UploadPartRequest()
  14. 	                     .withBucketName(awss3bucket)
  15. 	                     .withKey(filename)
  16. 	                     .withUploadId(initResponse.getUploadId())
  17. 	                     .withPartNumber(i)
  18. 	                     .withFileOffset(filePosition)
  19. 	                     .withFile(uploaffile)
  20. 	                     .withPartSize(partSize);
  21. 	
  22. 	             PartETag petag = new PartETag(amazons3.uploadPart(uploadRequest).getPartNumber(), amazons3.uploadPart(uploadRequest).getETag());
  23. 	             partETags.add(petag);
  24. 	
  25. 	             filePosition += partSize;
  26. 	         }
  27. 	
  28. 	         CompleteMultipartUploadRequest compRequest = new CompleteMultipartUploadRequest(awss3bucket, filename,
  29. 	                 																		initResponse.getUploadId(), partETags);
  30. 	        
  31. 	         amazons3.completeMultipartUpload(compRequest);
英文:

my task is to encrypt a file that is uploaded to S3. The upload worked fine before the encryption but now after I encrypted the file I get this exception.<br/><br/>
The XML you provided was not well-formed or did not validate against our published schema
<br/><br/>
I added this to the existing Code <br/><br/>

  1. 		final AwsCrypto crypto = new AwsCrypto();
  2. 		
  3. 		try (
  4. 			final FileInputStream in = new FileInputStream(encryptfile);
  5.             final FileOutputStream out = new FileOutputStream(file);
  6.             final CryptoOutputStream&lt;?&gt; encryptingStream = crypto.createEncryptingStream(crypt, out)) 
  7. 		{
  8.             IOUtils.copy(in, encryptingStream);
  9. 		}

<br/>
My thoughts, Why does AmazonS3 expect a XML-File ? Why not a normal text document ?<br/>
Is there a Option to change this maybe with the Bucket Policy ?

EDIT <br/>
That is the upload code, maybe there is a Issue. I dont understand why it´s working without the encryption.

  1.              File uploaffile = encryptFile(file);
  2. 			 List&lt;PartETag&gt; partETags = new ArrayList&lt;PartETag&gt;();
  3. 			 String filename = String.valueOf(System.currentTimeMillis());
  4. 	
  5. 	         InitiateMultipartUploadRequest initRequest = new InitiateMultipartUploadRequest(awss3bucket, filename);
  6. 	         InitiateMultipartUploadResult initResponse = amazons3.initiateMultipartUpload(initRequest);
  7. 	         long partSize = 5 * 1024 * 1024; 
  8. 	         long contentLength = uploaffile.length();
  9. 	         long filePosition = 0;
  10. 	         for (int i = 1; filePosition &lt; contentLength; i++) {
  11. 	             partSize = Math.min(partSize, (contentLength - filePosition));
  12. 	
  13. 	             UploadPartRequest uploadRequest = new UploadPartRequest()
  14. 	                     .withBucketName(awss3bucket)
  15. 	                     .withKey(filename)
  16. 	                     .withUploadId(initResponse.getUploadId())
  17. 	                     .withPartNumber(i)
  18. 	                     .withFileOffset(filePosition)
  19. 	                     .withFile(uploaffile)
  20. 	                     .withPartSize(partSize);
  21. 	
  22. 	             PartETag petag = new PartETag(amazons3.uploadPart(uploadRequest).getPartNumber(), amazons3.uploadPart(uploadRequest).getETag());
  23. 	             partETags.add(petag);
  24. 	
  25. 	             filePosition += partSize;
  26. 	         }
  27. 	
  28. 	         CompleteMultipartUploadRequest compRequest = new CompleteMultipartUploadRequest(awss3bucket, filename,
  29. 	                 																		initResponse.getUploadId(), partETags);
  30. 	        
  31. 	         amazons3.completeMultipartUpload(compRequest);

答案1

得分: 0

也许下次我应该停止复制互联网上的随机代码。<br/>
你应该使用FileOutputStream来写，而不是反过来。所以文件是空的，这导致了异常。

  1. CryptoInputStream&lt;KmsMasterKey&gt; encryptingStream = crypto.createEncryptingStream(crypt, in);
  2. FileOutputStream out = null;
  4. try {
  5.     out = new FileOutputStream(encryptfile);
  6.     IOUtils.copy(encryptingStream, out);
  7.     encryptingStream.close();
  8.     out.close();
  9. } catch (IOException e)
英文:

Maybe next time I should stop to copy some random Code from the Internet. <br/>
You use The FileoutputStream to write not the other way. So the File so was Empty which created the Exception.

  1.         CryptoInputStream&lt;KmsMasterKey&gt; encryptingStream = crypto.createEncryptingStream(crypt, in);
  2.         FileOutputStream out = null;
  3.         
  4. 		try {
  5. 			out = new FileOutputStream(encryptfile);
  6. 			IOUtils.copy(encryptingStream, out);
  7. 	        encryptingStream.close();
  8. 	        out.close();
  9. 		} catch (IOException e) 
  11. </details>

huangapple
  • 本文由 发表于 2020年5月4日 23:18:42
  • 转载请务必保留本文链接：https://go.coder-hub.com/61595637.html
  • amazon-s3
  • amazon-web-services
  • aws-java-sdk
  • java
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