How to read .mp3 files in wcf? - wcf

How to read the .mp3 audio files ?
I wrote the following code.
public static byte[] ReadFully(Stream stream)
{
byte[] buffer = new byte[32768];
using (MemoryStream ms = new MemoryStream())
{
while (true)
{
int read = stream.Read(buffer, 0, buffer.Length);
if (read <= 0)
return ms.ToArray();
ms.Write(buffer, 0, read);
}
}
It accepts .3gp,.caf foramat streams .why it's not taking .mp3 files?
}

I used the below code to download a mp3 file from my REST WCF Service:
[WebGet]
public Stream GetMp3()
{
byte[] buffer;
WebOperationContext.Current.OutgoingResponse.ContentType = "audio/mp3";
using (FileStream ms = new FileStream(#"C:\Sample.mp3", FileMode.Open))
{
int length = (int)ms.Length;
WebOperationContext.Current.OutgoingResponse.ContentLength = length;
buffer = new byte[length];
int sum = 0;
int count;
while ((count = ms.Read(buffer, sum, length - sum)) > 0)
{
sum += count;
}
ms.Close();
}
return new MemoryStream(buffer);
}
UPDATE: Upload an mp3 file to Restful WCF service
Server side method:
[WebInvoke]
public string RecieveMp3(Stream mp3Stream)
{
byte[] buffer = new byte[5000000];
using (FileStream ms = new FileStream("C:\\Temp\\Test1.mp3", FileMode.OpenOrCreate))
{
int read = mp3Stream.Read(buffer, 0, buffer.Length);
ms.Write(buffer, 0, read);
}
return "Recieved Mp3 file";
}
NOTE:
1. Make sure that you have permissions on the folder to write the mp3 file for the user associated with the app pool under which the wcf service is running
2. For testing purpose i have hard coded the byte array size to be a large value.
Client to call the service:
private string UseRestSharpApproachForFiles(string serviceBaseUrl, string resourceUrl, Method method, string filepath)
{
var client = new RestClient();
client.BaseUrl = serviceBaseUrl;
var request = new RestRequest(method) { DateFormat = DataFormat.Xml.ToString(), Resource = resourceUrl };
request.AddFile("stream", filepath);
var response = client.Execute(request);
string responseString;
if (response.StatusCode == HttpStatusCode.OK)
{
responseString = HttpUtility.HtmlDecode(response.Content);
}
else
{
responseString = response.StatusDescription + " --------------------" + HttpUtility.HtmlDecode(response.Content);
}
return responseString;
}
Am using the 3rd party dll for invoking the rest service called RestSharp.

Related

C# How I can upload file to MinIO (AWS S3 compatible API) via gRPC without buffering data?

How can I upload large files to MinIO (AWS S3 compatible API) via gRPC service without buffering data?
I have gRPC service with following definition:
service MediaService {
rpc UploadMedia(stream UploadMediaRequest) returns (UploadMediaResponse);
}
message UploadMediaRequest {
oneof Data {
UploadMediaMetadata metadata = 1;
UploadMediaStream fileStream = 2;
}
}
message UploadMediaMetadata {
string bucket = 1;
string virtialDirectory = 2;
string fileName = 3;
string contentType = 4;
map<string, string> attributes = 6;
}
message UploadMediaStream {
bytes bytes = 1;
}
And implementation of UploadMedia:
public override async Task<UploadMediaResponse> UploadMedia(
IAsyncStreamReader<UploadMediaRequest> requestStream,
ServerCallContext context)
{
UploadMediaMetadata? metadata = null;
var token = context.CancellationToken;
var traceId = context.GetHttpContext().TraceIdentifier;
await using var memoryStream = new MemoryStream();
await foreach (var req in requestStream.ReadAllAsync(token))
{
if (req.DataCase == UploadMediaRequest.DataOneofCase.Metadata)
{
metadata = req.Metadata;
_logger.LogTrace("[Req: {TraceId}] Received metadata", traceId);
}
else
{
await memoryStream.WriteAsync(req.FileStream.Bytes.Memory, token);
_logger.LogTrace("[Req: {TraceId}] Received chunk of bytes", traceId);
}
}
if (metadata == null)
{
throw new RpcException(new Status(StatusCode.InvalidArgument, "Not found metadata."));
}
memoryStream.Seek(0L, SeekOrigin.Begin);
var uploadModel = _mapper.Map<UploadModel>(metadata);
uploadModel.FileStream = memoryStream;
var file = await _fileService.UploadFile(uploadModel, token);
await _eventsService.Notify(new MediaUploadedEvent(file.PublicId), token);
_logger.LogTrace("[Req: {TraceId}] File uploaded", traceId);
return new UploadMediaResponse { File = _mapper.Map<RpcFileModel>(file) };
}
At the method I read request stream and write data to MemoryStream. After that I upload file to storage:
var putObjectArgs = new PutObjectArgs()
.WithStreamData(fileStream)
.WithObjectSize(fileStream.Length)
.WithObject(virtualPath)
.WithBucket(bucket)
.WithContentType(contentType)
.WithHeaders(attributes);
return _storage.PutObjectAsync(putObjectArgs, token);
I want to upload files without buffering data in Memory.
I think I can write bytes from stream to disk and after that create FileStream, but I don't want one more dependency.

Blazor Server create link to download file from byte array

I have a method in my code behind to retrieve a get a pdf file from an API and return the byte[]
byte[] byteArray = response.Content.ReadAsByteArrayAsync().Result; ;
using (MemoryStream pdfStream = new MemoryStream())
{
pdfStream.Write(byteArray, 0, byteArray.Length);
pdfStream.Position = 0;
return new FileStreamResult(pdfStream, "application/pdf");
}
How in Blazor server to I create a link in my .razor component to consume this byte[] so that when the user clicks the link, it triggers the file download?
Your solution is close because you're creating the appropriate result, but you simply need the method that returns it.
Set up your API controller like the following:
[ApiController]
public class DownloadController : ControllerBase {
[HttpGet]
public ActionResult Get() {
byte[] byteArray = response.Content.ReadAsByteArrayAsync().Result; ;
using (MemoryStream pdfStream = new())
{
pdfStream.Write(byteArray, 0, byteArray.Length);
pdfStream.Position = 0;
var result = FileStreamResult(pdfStream, "application/pdf");
result.FileDownloadName = "sample.txt";
return result;
}
}
}

C# .NET Core ImageConverter

I am trying to use the ImageConverter class in my ASP.NET Core project to convert an Image to a byte[], but I can't seem to find the class.
I have installed the System.Drawing.Common package, but still can't find it.
I am using .NET Core 3.1.
If on .NET Core 3.1, ImageConverter requires the System.Windows.Extensions package:
https://learn.microsoft.com/en-us/dotnet/api/system.drawing.imageconverter?view=netcore-3.1
In .NET 5 it is included in System.Drawing.Common:
https://learn.microsoft.com/en-us/dotnet/api/system.drawing.imageconverter?view=net-5.0
You can convert image to byte easily.
protected virtual byte[] LoadPictureFromFile(string filePath)
{
if (!File.Exists(filePath))
return new byte[0];
return File.ReadAllBytes(filePath);
}
Extra help..
public byte[] ResizeImage(byte[] pictureBinary,int newWidth, int newHeight)
{
byte[] pictureBinaryResized;
using (var stream = new MemoryStream(pictureBinary))
{
Bitmap b = null;
try
{
//try-catch to ensure that picture binary is really OK. Otherwise, we can get "Parameter is not valid" exception if binary is corrupted for some reasons
b = new Bitmap(stream);
}
catch (ArgumentException exc)
{
// log error
}
if (b == null)
{
//bitmap could not be loaded for some reasons
return new byte[0];
}
using (var destStream = new MemoryStream())
{
ImageBuilder.Current.Build(b, destStream, new ResizeSettings
{
Width = newWidth,
Height = newHeight,
Scale = ScaleMode.Both,
Quality = _mediaSettings.DefaultImageQuality
});
pictureBinaryResized = destStream.ToArray();
b.Dispose();
}
}
return pictureBinaryResized;
}
Instead of ImageConverter, you can trying to look at this for speed up:
Save bitmap to stream:
bitmap.save(stream);
Or open image file:
FileStream stream = new FileStream(imageFilePath, FileMode.Open, FileAccess.Read);
Then simply use Stream2Bytes:
byte[] OO7b = Stream2Bytes(stream);
And this is the Stream2Bytes method:
public byte[] Stream2Bytes(Stream stream, int chunkSize = 1024)
{
if (stream == null)
{
throw new System.ArgumentException("Parameter cannot be null", "stream");
}
if (chunkSize < 1)
{
throw new System.ArgumentException("Parameter must be greater than zero", "chunkSize");
}
if (chunkSize > 1024 * 64)
{
throw new System.ArgumentException(String.Format("Parameter must be less or equal {0}", 1024 * 64), "chunkSize");
}
List<byte> buffers = new List<byte>();
using (BinaryReader br = new BinaryReader(stream))
{
byte[] chunk = br.ReadBytes(chunkSize);
while (chunk.Length > 0)
{
buffers.AddRange(chunk);
chunk = br.ReadBytes(chunkSize);
}
}
return buffers.ToArray();
}

How to encrypt payload file streamingly via WSO2 ESB

I have to implement a scenario by using WSO2 ESB, as encrypting the binary payload streamingly while response to the client side (I assume the content-type in the case is Application/Octet-Stream), below is some details by my thought:
An Endpoint like "http://myhost/backend/" which provides business functionality;
A proxy which pass messages through the endpoint;
I attempt to write an OutSequence to check the Content-type: if the Content-Type matches Application/Octet-Stream, invoke my customized class mediator to encrypt the fileStream Streamingly and response.
I have no idea on how to write the class mediator to make it implemented? How could I get/read the file stream from the message as well as how to put the outputStream back to the response while I could only see mc.getEnvelope().getBody() in mediation method? Below is my current mediator which doesn't work.
public boolean mediate(MessageContext mc) {
org.apache.axis2.context.MessageContext amc = ((Axis2MessageContext) mc).getAxis2MessageContext();
try {
String contentID = amc.getAttachmentMap().getAllContentIDs()[0];
DataHandler dh = amc.getAttachment(contentID);
dh.getDataSource().getName();
InputStream is = null;
try {
is = dh.getInputStream();
BufferedReader br = new BufferedReader(new InputStreamReader(is));
String line = null;
while ((line = br.readLine()) != null) {
System.out.println("client read:" + line);
}
} catch (IOException e) {
e.printStackTrace();
}
}
}
return true;
}
Many thanks if anybody with experience would kindly help.
Pasted my current solution for anyone else who confronts similar issue.
In the mediator, I read the file content from response stream via OMText.InputStream and use net.lingala.zip4j package to write a zip file(in memory) with the raw file encrypted; Finally I write the zip file content as ByteArray back to the OMElement of the soap message.
public boolean mediate(MessageContext mc) {
System.out.println("========================Mediator log start================================");
org.apache.axis2.context.MessageContext amc = ((Axis2MessageContext) mc).getAxis2MessageContext();
try {
#SuppressWarnings("unchecked")
Map<String, String> responseHeaders = (Map<String, String>) amc.getProperty("TRANSPORT_HEADERS");
String rawFileName = "";
String[] contentDisps = responseHeaders.get("Content-Disposition").split(";");
for (String item : contentDisps) {
System.out.println("item::" + item);
if (item.trim().startsWith(CONTENT_DISPOSITION_FILENAME)) {
rawFileName = item.substring(item.indexOf("\"") + 1, item.length() - 1);
break;
}
}
responseHeaders.put(
"Content-Disposition",
responseHeaders.get("Content-Disposition").replace(rawFileName,
rawFileName.substring(0, rawFileName.lastIndexOf(".")) + ".myzip"));
OMElement binaryPayload =
amc.getEnvelope().getBody()
.getFirstChildWithName(new QName("http://ws.apache.org/commons/ns/payload", "binary"));
OMText binaryNode = (OMText) binaryPayload.getFirstOMChild();
DataHandler dataHandler = (DataHandler) binaryNode.getDataHandler();
InputStream is = dataHandler.getInputStream();
ByteArrayOutputStream responseOutputStream = new ByteArrayOutputStream();
ZipOutputStream zipOutputStream = getZipOutputStreamInstance(responseOutputStream, rawFileName);
// write to zipOutputStream
byte data[] = new byte[BUFFER_SIZE];
int count;
while ((count = is.read(data, 0, BUFFER_SIZE)) != -1) {
zipOutputStream.write(data, 0, count);
zipOutputStream.flush();
}
zipOutputStream.closeEntry();
zipOutputStream.finish();
InputStream in = new ByteArrayInputStream(responseOutputStream.toByteArray());
DataHandler zipDataHandler = new DataHandler(new StreamingOnRequestDataSource(in));
OMFactory factory = OMAbstractFactory.getOMFactory();
OMText zipData = factory.createOMText(zipDataHandler, true);
zipData.setBinary(true);
binaryPayload.getFirstOMChild().detach();
binaryPayload.addChild(zipData);
amc.setProperty("TRANSPORT_HEADERS", responseHeaders);
System.out.println("========================Mediator end==================================");
} catch (Exception ex) {
System.out.println("exception found here:");
ex.printStackTrace();
}
return true;
}

Using wcf how to upload a image

using wcf/wcf web services to upload a images give me with example?
In my project i want to upload image by using WCF
Basically you should use WCF streaming.
[ServiceContract]
public interface ITransferService
{
[OperationContract]
void UploadFile(RemoteFileInfo request);
}
public void UploadFile(RemoteFileInfo request)
{
FileStream targetStream = null;
Stream sourceStream = request.FileByteStream;
string uploadFolder = #"C:\\upload\\";
string filePath = Path.Combine(uploadFolder, request.FileName);
using (targetStream = new FileStream(filePath, FileMode.Create,
FileAccess.Write, FileShare.None))
{
//read from the input stream in 65000 byte chunks
const int bufferLen = 65000;
byte[] buffer = new byte[bufferLen];
int count = 0;
while ((count = sourceStream.Read(buffer, 0, bufferLen)) > 0)
{
// save to output stream
targetStream.Write(buffer, 0, count);
}
targetStream.Close();
sourceStream.Close();
}
}
The easiest way is to convert the image to a byte array before sending it, and then converting it back to an image on the destination site.
Here are two methods for doing just that:
public byte[] ImageToByteArray( Image image)
{
var ms = new MemoryStream();
image.Save(ms, ImageFormat.Png);
return ms.ToArray();
}
public static Image ByteArrayToImage(byte[] byteArray)
{
var ms = new MemoryStream(byteArray);
return Image.FromStream(ms);
}
That means your web service can have a method something like this:
public void UploadImage( byte[] imageData )
{
var image = ByteArrayToImage( imageData );
//save image to disk here, or do whatever you like with it
}