I can't seem to find a way to determine the size of a file in Isolated Storage besides opening up the file stream and calling the "Length" property. Is there a more efficient way of doing this?
Thanks
I found a bit of a hack to make it work. What you have to do is use reflection to get the fully qualified file path to the file you want then create a new file info object:
//This is the private field name used for reflection
private const string IsolatedStoreRootDir = "m_RootDir";
//This method takes a file path relative to isolated storage
//and the current store
private static FileInfo GetFileInfo(string path, IsolatedStorageFile store)
{
return new FileInfo(GetFullyQualifiedFileName(path, store));
}
//This gets the fully qualified path of the root isolated storage directory
//then appends the relative path to it.
private static string GetFullyQualifiedFileName(string path, IsolatedStorageFile store)
{
return Path.Combine(store.GetType()
.GetField(IsolatedStorageFileSystem.IsolatedStoreRootDir,
System.Reflection.BindingFlags.NonPublic |
System.Reflection.BindingFlags.Instance).GetValue(store).ToString(), path);
}
//Here's how it's used
static void Main(string[] args)
{
var store = IsolatedStorageFile.GetUserStoreForAssembly();
var length = GetFileInfo("TestFile.txt", store).Length;
}
long Size = 0L;
using (IsolatedStorageFileStream stream = new IsolatedStorageFileStream(filePath, FileMode.Open, FileAccess.Read, isoFile))
{
Size = stream.Length;
}
Related
How one would write a Lucene 8.11 ByteBuffersDirectory to disk?
something similar to Lucene 2.9.4 Directory.copy(directory, FSDirectory.open(indexPath), true)
You can use the copyFrom method to do this.
For example:
You are using a ByteBuffersDirectory:
final Directory dir = new ByteBuffersDirectory();
Assuming you are not concurrently writing any new data to that dir, you can declare a target where you want to write the data - for example, a FSDirectory (a file system directory):
Directory to = FSDirectory.open(Paths.get(OUT_DIR_PATH));
Use whatever string you want for the OUT_DIR_PATH location.
Then you can iterate over all the files in the original dir object, writing them to this new to location:
IOContext ctx = new IOContext();
for (String file : dir.listAll()) {
System.out.println(file); // just for testing
to.copyFrom(dir, file, file, ctx);
}
This will create the new OUT_DIR_PATH dir and populate it with files, such as:
_0.cfe
_0.cfs
_0.si
segments_1
... or whatever files you happen to have in your dir.
Caveat:
I have only used this with a default IOContext object. There are other constructors for the context - not sure what they do. I assume they give you more control over how the write is performed.
Meanwhile I figured it out by myself and created a straight forward method for it:
#SneakyThrows
public static void copyIndex(ByteBuffersDirectory ramDirectory, Path destination) {
FSDirectory fsDirectory = FSDirectory.open(destination);
Arrays.stream(ramDirectory.listAll())
.forEach(fileName -> {
try {
// IOContext is null because in fact is not used (at least for the moment)
fsDirectory.copyFrom(ramDirectory, fileName, fileName, null);
} catch (IOException e) {
log.error(e.getMessage(), e);
}
});
}
So I'm trying out the cascading framework and I was able to run apps in local. As a next step I want to use Cascading to download files from S3 to local file system. I'm planning to use FileTap. When I google I found S3FS and looks like its deprecated. Can someone point me in the right direction. In other words, what is the right way to create a FileTap for Amazon S3 with access id and secret key like stuff.
Though this code is for the HadoopFlowConnector this will work with Local Flow connector if using the FileTap.
public class Main {
public void run(String[] args) {
Properties properties = new Properties();
String accessKey = args[0];
String secretKey = args[1];
// better put these keys to hadoop xml file
// for block file system
properties.setProperty("fs.s3.awsAccessKeyId", accessKey);
properties.setProperty("fs.s3.awsSecretAccessKey", secretKey);
// for s3 native file system
// properties.setProperty("fs.s3n.awsAccessKeyId", accessKey);
// properties.setProperty("fs.s3n.awsSecretAccessKey", secretKey);
// properties.setProperty("fs.defaultFS", "hdfs://localhost:8020/");
// properties.setProperty("fs.permissions.umask-mode", "007");
AppProps.setApplicationJarClass(properties, Main.class);
HadoopFlowConnector flowConnector = new HadoopFlowConnector(
properties);
String input = "s3://my-bucket/my-log.csv";
// If using the native S3
// String input = "s3n://my-bucket/my-log.csv";
Tap inTap = new Hfs(new TextDelimited(false, ";"), input);
Pipe copyPipe = new Pipe("copy");
Tap outTap = new Hfs(new TextDelimited(false, ";"),
"data/output");
FlowDef flowDef = FlowDef.flowDef()
.addSource(copyPipe, inTap)
.addTailSink(copyPipe, outTap);
flowConnector.connect(flowDef).complete();
}
public static void main(String[] args) {
new Main().run(args);
}
}
Code courtesy http://whiteboxdeveloper.blogspot.com/2015/01/processing-data-from-aws-s3-using.html.
I just added the S3N string input commented line. Since I was using S3N based accessing I uncommented the S3N features and used it.
I am building REST application. I want to upload a file and I want to save it for example in /WEB-INF/resource/uploads
How can I get path to this directory ? My Controller looks like this
#RequestMapping(value = "/admin/house/update", method = RequestMethod.POST)
public String updateHouse(House house, #RequestParam("file") MultipartFile file, Model model) {
try {
String fileName = null;
InputStream inputStream = null;
OutputStream outputStream = null;
if (file.getSize() > 0) {
inputStream = file.getInputStream();
fileName = "D:/" + file.getOriginalFilename();
outputStream = new FileOutputStream(fileName);
int readBytes = 0;
byte[] buffer = new byte[10000];
while ((readBytes = inputStream.read(buffer, 0, 10000)) != -1) {
outputStream.write(buffer, 0, readBytes);
}
outputStream.close();
inputStream.close();
}
} catch(Exception ex) {
ex.printStackTrace();
}
model.addAttribute("step", 3);
this.houseDao.update(house);
return "houseAdmin";
}
Second question...what is the best place to upload user files ?
/WEB-INF is a bad place to try to store file uploads. There's no guarantee that this is an actual directory on the disk, and even if it is, the appserver may forbid write access to it.
Where you should store your files depends on what you want to do with them, and what operating system you're running on. Just pick somewhere outside of the webapp itself, is my advice. Perhaps create a dedicated directory
Also, the process of transferring the MultipartFile to another location is much simpler than you're making it out to be:
#RequestMapping(value = "/admin/house/update", method = RequestMethod.POST)
public String updateHouse(House house, #RequestParam("file") MultipartFile srcFile, Model model) throws IOException {
File destFile = new File("/path/to/the/target/file");
srcFile.transferTo(destFile); // easy!
model.addAttribute("step", 3);
this.houseDao.update(house);
return "houseAdmin";
}
You shouldn't store files in /WEB-INF/resource/uploads. This directory is either inside your WAR (if packaged) or exploded somewhere inside servlet container. The first destination is read-only and the latter should not be used for user files.
There are usually two places considered when storing uploaded files:
Some dedicated folder. Make sure users cannot access this directory directly (e.g. anonymous FTP folder). Note that once your application runs on more than one machine you won't have access to this folder. So consider some form of network synchronization or a shared network drive.
Database. This is controversial since binary files tend to occupy a lot of space. But this approach is a bit simpler when distributing your application.
When I upload an image file to a blob, the image is uploaded apparently successfully (no errors). When I go to cloud storage studio, the file is there, but with a size of 0 (zero) bytes.
The following is the code that I am using:
// These two methods belong to the ContentService class used to upload
// files in the storage.
public void SetContent(HttpPostedFileBase file, string filename, bool overwrite)
{
CloudBlobContainer blobContainer = GetContainer();
var blob = blobContainer.GetBlobReference(filename);
if (file != null)
{
blob.Properties.ContentType = file.ContentType;
blob.UploadFromStream(file.InputStream);
}
else
{
blob.Properties.ContentType = "application/octet-stream";
blob.UploadByteArray(new byte[1]);
}
}
public string UploadFile(HttpPostedFileBase file, string uploadPath)
{
if (file.ContentLength == 0)
{
return null;
}
string filename;
int indexBar = file.FileName.LastIndexOf('\\');
if (indexBar > -1)
{
filename = DateTime.UtcNow.Ticks + file.FileName.Substring(indexBar + 1);
}
else
{
filename = DateTime.UtcNow.Ticks + file.FileName;
}
ContentService.Instance.SetContent(file, Helper.CombinePath(uploadPath, filename), true);
return filename;
}
// The above code is called by this code.
HttpPostedFileBase newFile = Request.Files["newFile"] as HttpPostedFileBase;
ContentService service = new ContentService();
blog.Image = service.UploadFile(newFile, string.Format("{0}{1}", Constants.Paths.BlogImages, blog.RowKey));
Before the image file is uploaded to the storage, the Property InputStream from the HttpPostedFileBase appears to be fine (the size of the of image corresponds to what is expected! And no exceptions are thrown).
And the really strange thing is that this works perfectly in other cases (uploading Power Points or even other images from the Worker role). The code that calls the SetContent method seems to be exactly the same and file seems to be correct since a new file with zero bytes is created at the correct location.
Does any one have any suggestion please? I debugged this code dozens of times and I cannot see the problem. Any suggestions are welcome!
Thanks
The Position property of the InputStream of the HttpPostedFileBase had the same value as the Length property (probably because I had another file previous to this one - stupid I think!).
All I had to do was to set the Position property back to 0 (zero)!
I hope this helps somebody in the future.
Thanks Fabio for bringing this up and solving your own question. I just want to add code to whatever you have said. Your suggestion worked perfectly for me.
var memoryStream = new MemoryStream();
// "upload" is the object returned by fine uploader
upload.InputStream.CopyTo(memoryStream);
memoryStream.ToArray();
// After copying the contents to stream, initialize it's position
// back to zeroth location
memoryStream.Seek(0, SeekOrigin.Begin);
And now you are ready to upload memoryStream using:
blockBlob.UploadFromStream(memoryStream);
How can I add a string as a file on amazon s3? From whaterver I searched, I got to know that we can upload a file to s3. What is the best way to upload data without creating file?
There is an overload for the AmazonS3.putObject method that accepts the bucket string, a key string, and a string of text content. I hadn't seen mention of it on stack overflow so I'm putting this here. It's going to be similar #Jonik's answer, but without the additional dependency.
AmazonS3 s3client = AmazonS3ClientBuilder.standard().withRegion(Regions.US_EAST_1).build();
s3client.putObject(bucket, key, contents);
Doesn't look as nice, but here is how you can do it using Amazons Java client, probably what JetS3t does behind the scenes anyway.
private boolean putArtistPage(AmazonS3 s3,String bucketName, String key, String webpage)
{
try
{
byte[] contentAsBytes = webpage.getBytes("UTF-8");
ByteArrayInputStream contentsAsStream = new ByteArrayInputStream(contentAsBytes);
ObjectMetadata md = new ObjectMetadata();
md.setContentLength(contentAsBytes.length);
s3.putObject(new PutObjectRequest(bucketname, key, contentsAsStream, md));
return true;
}
catch(AmazonServiceException e)
{
log.log(Level.SEVERE, e.getMessage(), e);
return false;
}
catch(Exception ex)
{
log.log(Level.SEVERE, ex.getMessage(), ex);
return false;
}
}
What is the best way to upload data
without creating file?
If you meant without creating a file on S3, well, you can't really do that. On Amazon S3, the only way to store data is as files, or using more accurate terminology, objects. An object can contain from 1 byte zero bytes to 5 terabytes of data, and is stored in a bucket. Amazon's S3 homepage lays out the basic facts quite clearly. (For other data storing options on AWS, you might want to read e.g. about SimpleDB.)
If you meant without creating a local temporary file, then the answer depends on what library/tool you are using. (As RickMeasham suggested, please add more details!) With the s3cmd tool, for example, you can't skip creating temp file, while with the JetS3t Java library uploading a String directly would be easy:
// (First init s3Service and testBucket)
S3Object stringObject = new S3Object("HelloWorld.txt", "Hello World!");
s3Service.putObject(testBucket, stringObject);
There is a simple way to do it with PHP, simply send the string as the body of the object, specifying the name of the new file in the key -
$s3->putObject(array(
'Bucket' => [Bucket name],
'Key' => [path/to/file.ext],
'Body' => [Your string goes here],
'ContentType' => [specify mimetype if you want],
));
This will create a new file according to the specified key, which has a content as specified in the string.
If you're using java, check out https://ivan-site.com/2015/11/interact-with-s3-without-temp-files/
import com.amazonaws.services.s3.AmazonS3;
import com.amazonaws.services.s3.AmazonS3Client;
import com.amazonaws.services.s3.model.GetObjectRequest;
import com.amazonaws.services.s3.model.ObjectMetadata;
import com.amazonaws.services.s3.model.PutObjectRequest;
import com.amazonaws.services.s3.model.S3Object;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.ObjectMapper;
import java.io.*;
import java.nio.charset.StandardCharsets;
class S3StreamJacksonTest {
private static final String S3_BUCKET_NAME = "bucket";
private static final String S3_KEY_NAME = "key";
private static final String CONTENT_TYPE = "application/json";
private static final AmazonS3 AMAZON_S3 = new AmazonS3Client();
private static final ObjectMapper OBJECT_MAPPER = new ObjectMapper();
private static final TestObject TEST_OBJECT = new TestObject("test", 123, 456L);
public void testUploadWithStream() throws JsonProcessingException {
String fileContentString = OBJECT_MAPPER.writeValueAsString(TEST_OBJECT);
byte[] fileContentBytes = fileContentString.getBytes(StandardCharsets.UTF_8);
InputStream fileInputStream = new ByteArrayInputStream(fileContentBytes);
ObjectMetadata metadata = new ObjectMetadata();
metadata.setContentType(CONTENT_TYPE);
metadata.setContentLength(fileContentBytes.length);
PutObjectRequest putObjectRequest = new PutObjectRequest(
S3_BUCKET_NAME, S3_KEY_NAME, fileInputStream, metadata);
AMAZON_S3.putObject(putObjectRequest);
}
}
This works for me:
public static PutObjectResult WriteString(String bucket, String key, String stringToWrite, AmazonS3Client s3Client) {
ObjectMetadata meta = new ObjectMetadata();
meta.setContentMD5(new String(com.amazonaws.util.Base64.encode(DigestUtils.md5(stringToWrite))));
meta.setContentLength(stringToWrite.length());
InputStream stream = new ByteArrayInputStream(stringToWrite.getBytes(StandardCharsets.UTF_8));
return s3Client.putObject(bucket, key, stream, meta);
}
The sample code at https://docs.aws.amazon.com/AmazonS3/latest/dev/UploadObjSingleOpJava.html works for me.
s3Client.putObject(bucketName, stringObjKeyName, "Uploaded String Object");
Looks like this was added around 1.11.20, so make sure you are using that or new version of SDK.
https://javadoc.io/doc/com.amazonaws/aws-java-sdk-s3/1.11.20/com/amazonaws/services/s3/AmazonS3.html#putObject-java.lang.String-java.lang.String-java.lang.String-