How to download a dynamically generated XML file with ServiceStack and Angular? - c#

The following code works in theory, but it lacks error handling. The problem I have is that it starts downloading the XML file when a new window opens with the url created by the service stack. But now when an error occurs server side, you are on this new page with only the stack trace.
What is the right way to download a dynamic binary (not stored on disk) with service stack and Angular?
Angular with ServiceStack:
downloadExportXML(){
const request = new GetRatingsExportRequest(this.request);
const url = this.jsonServiceClient.createUrlFromDto("GET", request)
window.open(url);
}
WebAPI (C#) to collect the XML File.
public HttpResult Get(GetRatingsExportRequest request)
{
MemoryStream ms = new MemoryStream(Encoding.UTF8.GetBytes(xmlContent));
var result = new HttpResult(ms.ToArray());
var disposition = $"attachment;filename=myfilename;";
result.Headers.Add(HttpHeaders.ContentDisposition, disposition);
return result;
}


If you return the content with a Content-Disposition: attachment HTTP Header the browser will treat it as a separate File Download.
If you don't want that return it as a normal XML Response. But the way you're returning it is fairly inefficient, i.e. calling ToArray() defeats the purpose of using a Stream since you've forced loaded the entire contents of the Stream in memory, instead you should just return the stream so it gets asynchronously written to the HTTP Response as a Stream, e.g:
return new HttpResult(ms, MimeTypes.Xml);
But if you've already got the XML as a string you don't need the overhead of the MemoryStream wrapper either and can just return the XML string as-is, e.g:
return new HttpResult(xmlContent, MimeTypes.Xml);


I couldnt make it work. So i created this workaround: Generating the URL to download the XML with the ServiceStack method jsonServiceClient.createUrlFromDto() and than downloading the file via Angular HTTPClient.
downloadExportXML(){
const request = new GetRatingsExportRequest(this.request);
const url = this.jsonServiceClient.createUrlFromDto("GET", request)
this.httpClient.get(url, {observe: 'response', responseType: 'arraybuffer'})
.subscribe((response) => {
var contentDisposition = response.headers.get('content-disposition');
this.downloadFile(response.body)
},
error => this.handleError()
);
}
private downloadFile(blobData: ArrayBuffer)
{
let file = new Blob([blobData], { type: 'application/binary' });
let fileURL = URL.createObjectURL(file);
let fileLink = document.createElement('a');
fileLink.href = fileURL;
fileLink.download = "MYFILENAME.pdf";
fileLink.click();
}

Related

Passing a large file stream to MultipartFormDataContent with HttpClient

I'm experiencing a problem when trying to use MultipartFormDataContent with HttpClient with a stream of data.
Context
I'm trying to upload a large file to ASP.NET Core Web API. A client should send the file via POST request form-data to a front-end API, which in turn should forward the file to a back-end API.
Because the file can be large, I followed the Microsoft example, i.e. I don't want to use IFormFile type but instead read the Request.Body using MultipartReader. This is to avoid loading the entire file into memory on the server, or saving it in a temporary file on server's hard drive.
Problem
The back-end API controller action looks as follows (this is almost directly copied from the ASP.NET Core 5.0 sample app with just minor simplifications):
[HttpPost]
[DisableRequestSizeLimit]
public async Task<IActionResult> ReceiveLargeFile()
{
var request = HttpContext.Request;
if (!request.HasFormContentType
|| !MediaTypeHeaderValue.TryParse(request.ContentType, out var mediaTypeHeader)
|| string.IsNullOrEmpty(mediaTypeHeader.Boundary.Value))
{
return new UnsupportedMediaTypeResult();
}
var reader = new MultipartReader(mediaTypeHeader.Boundary.Value, request.Body);
/* This throws an IOException: Unexpected end of Stream, the content may have already been read by another component. */
var section = await reader.ReadNextSectionAsync();
while (section != null)
{
var hasContentDispositionHeader = ContentDispositionHeaderValue.TryParse(section.ContentDisposition,
out var contentDisposition);
if (hasContentDispositionHeader
&& contentDisposition!.DispositionType.Equals("form-data")
&& !string.IsNullOrEmpty(contentDisposition.FileName.Value))
{
/* Fake copy to nothing since it doesn't even get here */
await section.Body.CopyToAsync(Stream.Null);
return Ok();
}
section = await reader.ReadNextSectionAsync();
}
return BadRequest("No files data in the request.");
}
I managed to reduce the problem slightly by making an integration test using Microsoft.AspNetCore.Mvc.Testing NuGet package. The following test replaces the front-end API, so instead of reading Request.Body stream in a Web API, the test just tries to add StreamContent to MultipartFormDataContent and post it via HttpClient to the back-end API:
[Fact]
public async Task Client_posting_to_Api_returns_Ok()
{
/* Arrange */
await using var stream = new MemoryStream();
await using var writer = new StreamWriter(stream);
await writer.WriteLineAsync("FILE CONTENTS");
await writer.FlushAsync();
stream.Position = 0;
using var client = _factory.CreateDefaultClient();
/* Act */
using var response =
await client.PostAsync(
"Receive",
new MultipartFormDataContent
{
{
new StreamContent(stream),
"file",
"fileName"
}
});
/* Assert */
Assert.Equal(HttpStatusCode.OK, response.StatusCode);
}
The back-end API controller then throws an IOException at await reader.ReadNextSectionAsync(), saying "Unexpected end of Stream, the content may have already been read by another component".
GitHub Repository (Complete Example)
I uploaded a complete example of the problem (including back-end API and the test) a GitHub repo.
Question
I must be doing something wrong. How can I forward a file received in a request with form-data content type in one service (front-end API) to another service (back-end API) without loading the entire file into memory or hard-drive in the front-end API, i.e. to just forward the stream of data to the back-end API?
Thanks in advance for any help.

I expected the same issue as you and it turned out that the MediaTypeHeaderValue.TryParse method parses the boundary value wrong as it wraps the string with '"' characters, because HttpClient sends the content type header like this:
multipart/form-data; boundary="blablabla"
So for me the solution was to add a Trim() method to boundary like this and pass that to the MultipartReader
var boundary = mediaTypeHeader.Boundary.Value.Trim('"');
var reader = new MultipartReader(boundary, request.Body);

File and data in same HTTP request C# WebAPI [duplicate]

In an application I am developing RESTful API and we want the client to send data as JSON. Part of this application requires the client to upload a file (usually an image) as well as information about the image.
I'm having a hard time tracking down how this happens in a single request. Is it possible to Base64 the file data into a JSON string? Am I going to need to perform 2 posts to the server? Should I not be using JSON for this?
As a side note, we're using Grails on the backend and these services are accessed by native mobile clients (iPhone, Android, etc), if any of that makes a difference.

I asked a similar question here:
How do I upload a file with metadata using a REST web service?
You basically have three choices:
Base64 encode the file, at the expense of increasing the data size by around 33%, and add processing overhead in both the server and the client for encoding/decoding.
Send the file first in a multipart/form-data POST, and return an ID to the client. The client then sends the metadata with the ID, and the server re-associates the file and the metadata.
Send the metadata first, and return an ID to the client. The client then sends the file with the ID, and the server re-associates the file and the metadata.

You can send the file and data over in one request using the multipart/form-data content type:
In many applications, it is possible for a user to be presented with
a form. The user will fill out the form, including information that
is typed, generated by user input, or included from files that the
user has selected. When the form is filled out, the data from the
form is sent from the user to the receiving application.
The definition of MultiPart/Form-Data is derived from one of those
applications...
From http://www.faqs.org/rfcs/rfc2388.html:
"multipart/form-data" contains a series of parts. Each part is
expected to contain a content-disposition header [RFC 2183] where the
disposition type is "form-data", and where the disposition contains
an (additional) parameter of "name", where the value of that
parameter is the original field name in the form. For example, a part
might contain a header:
Content-Disposition: form-data; name="user"
with the value corresponding to the entry of the "user" field.
You can include file information or field information within each section between boundaries. I've successfully implemented a RESTful service that required the user to submit both data and a form, and multipart/form-data worked perfectly. The service was built using Java/Spring, and the client was using C#, so unfortunately I don't have any Grails examples to give you concerning how to set up the service. You don't need to use JSON in this case since each "form-data" section provides you a place to specify the name of the parameter and its value.
The good thing about using multipart/form-data is that you're using HTTP-defined headers, so you're sticking with the REST philosophy of using existing HTTP tools to create your service.

I know that this thread is quite old, however, I am missing here one option. If you have metadata (in any format) that you want to send along with the data to upload, you can make a single multipart/related request.
The Multipart/Related media type is intended for compound objects consisting of several inter-related body parts.
You can check RFC 2387 specification for more in-depth details.
Basically each part of such a request can have content with different type and all parts are somehow related (e.g. an image and it metadata). The parts are identified by a boundary string, and the final boundary string is followed by two hyphens.
Example:
POST /upload HTTP/1.1
Host: www.hostname.com
Content-Type: multipart/related; boundary=xyz
Content-Length: [actual-content-length]
--xyz
Content-Type: application/json; charset=UTF-8
{
"name": "Sample image",
"desc": "...",
...
}
--xyz
Content-Type: image/jpeg
[image data]
[image data]
[image data]
...
--foo_bar_baz--

Here is my approach API (i use example) - as you can see, you I don't use any file_id (uploaded file identifier to the server) in API:
Create photo object on server:
POST: /projects/{project_id}/photos
body: { name: "some_schema.jpg", comment: "blah"}
response: photo_id
Upload file (note that file is in singular form because it is only one per photo):
POST: /projects/{project_id}/photos/{photo_id}/file
body: file to upload
response: -
And then for instance:
Read photos list
GET: /projects/{project_id}/photos
response: [ photo, photo, photo, ... ] (array of objects)
Read some photo details
GET: /projects/{project_id}/photos/{photo_id}
response: { id: 666, name: 'some_schema.jpg', comment:'blah'} (photo object)
Read photo file
GET: /projects/{project_id}/photos/{photo_id}/file
response: file content
So the conclusion is that, first you create an object (photo) by POST, and then you send second request with the file (again POST). To not have problems with CACHE in this approach we assume that we can only delete old photos and add new - no update binary photo files (because new binary file is in fact... NEW photo). However if you need to be able to update binary files and cache them, then in point 4 return also fileId and change 5 to GET: /projects/{project_id}/photos/{photo_id}/files/{fileId}.

I know this question is old, but in the last days I had searched whole web to solution this same question. I have grails REST webservices and iPhone Client that send pictures, title and description.
I don't know if my approach is the best, but is so easy and simple.
I take a picture using the UIImagePickerController and send to server the NSData using the header tags of request to send the picture's data.
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:[NSURL URLWithString:#"myServerAddress"]];
[request setHTTPMethod:#"POST"];
[request setHTTPBody:UIImageJPEGRepresentation(picture, 0.5)];
[request setValue:#"image/jpeg" forHTTPHeaderField:#"Content-Type"];
[request setValue:#"myPhotoTitle" forHTTPHeaderField:#"Photo-Title"];
[request setValue:#"myPhotoDescription" forHTTPHeaderField:#"Photo-Description"];
NSURLResponse *response;
NSError *error;
[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
At the server side, I receive the photo using the code:
InputStream is = request.inputStream
def receivedPhotoFile = (IOUtils.toByteArray(is))
def photo = new Photo()
photo.photoFile = receivedPhotoFile //photoFile is a transient attribute
photo.title = request.getHeader("Photo-Title")
photo.description = request.getHeader("Photo-Description")
photo.imageURL = "temp"
if (photo.save()) {
File saveLocation = grailsAttributes.getApplicationContext().getResource(File.separator + "images").getFile()
saveLocation.mkdirs()
File tempFile = File.createTempFile("photo", ".jpg", saveLocation)
photo.imageURL = saveLocation.getName() + "/" + tempFile.getName()
tempFile.append(photo.photoFile);
} else {
println("Error")
}
I don't know if I have problems in future, but now is working fine in production environment.

FormData Objects: Upload Files Using Ajax
XMLHttpRequest Level 2 adds support for the new FormData interface.
FormData objects provide a way to easily construct a set of key/value pairs representing form fields and their values, which can then be easily sent using the XMLHttpRequest send() method.
function AjaxFileUpload() {
var file = document.getElementById("files");
//var file = fileInput;
var fd = new FormData();
fd.append("imageFileData", file);
var xhr = new XMLHttpRequest();
xhr.open("POST", '/ws/fileUpload.do');
xhr.onreadystatechange = function () {
if (xhr.readyState == 4) {
alert('success');
}
else if (uploadResult == 'success')
alert('error');
};
xhr.send(fd);
}
https://developer.mozilla.org/en-US/docs/Web/API/FormData

Since the only missing example is the ANDROID example, I'll add it.
This technique uses a custom AsyncTask that should be declared inside your Activity class.
private class UploadFile extends AsyncTask<Void, Integer, String> {
#Override
protected void onPreExecute() {
// set a status bar or show a dialog to the user here
super.onPreExecute();
}
#Override
protected void onProgressUpdate(Integer... progress) {
// progress[0] is the current status (e.g. 10%)
// here you can update the user interface with the current status
}
#Override
protected String doInBackground(Void... params) {
return uploadFile();
}
private String uploadFile() {
String responseString = null;
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://example.com/upload-file");
try {
AndroidMultiPartEntity ampEntity = new AndroidMultiPartEntity(
new ProgressListener() {
#Override
public void transferred(long num) {
// this trigger the progressUpdate event
publishProgress((int) ((num / (float) totalSize) * 100));
}
});
File myFile = new File("/my/image/path/example.jpg");
ampEntity.addPart("fileFieldName", new FileBody(myFile));
totalSize = ampEntity.getContentLength();
httpPost.setEntity(ampEntity);
// Making server call
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
int statusCode = httpResponse.getStatusLine().getStatusCode();
if (statusCode == 200) {
responseString = EntityUtils.toString(httpEntity);
} else {
responseString = "Error, http status: "
+ statusCode;
}
} catch (Exception e) {
responseString = e.getMessage();
}
return responseString;
}
#Override
protected void onPostExecute(String result) {
// if you want update the user interface with upload result
super.onPostExecute(result);
}
}
So, when you want to upload your file just call:
new UploadFile().execute();

I wanted send some strings to backend server. I didnt use json with multipart, I have used request params.
#RequestMapping(value = "/upload", method = RequestMethod.POST)
public void uploadFile(HttpServletRequest request,
HttpServletResponse response, #RequestParam("uuid") String uuid,
#RequestParam("type") DocType type,
#RequestParam("file") MultipartFile uploadfile)
Url would look like
http://localhost:8080/file/upload?uuid=46f073d0&type=PASSPORT
I am passing two params (uuid and type) along with file upload.
Hope this will help who don't have the complex json data to send.

You could try using https://square.github.io/okhttp/ library.
You can set the request body to multipart and then add the file and json objects separately like so:
MultipartBody requestBody = new MultipartBody.Builder()
.setType(MultipartBody.FORM)
.addFormDataPart("uploadFile", uploadFile.getName(), okhttp3.RequestBody.create(uploadFile, MediaType.parse("image/png")))
.addFormDataPart("file metadata", json)
.build();
Request request = new Request.Builder()
.url("https://uploadurl.com/uploadFile")
.post(requestBody)
.build();
try (Response response = client.newCall(request).execute()) {
if (!response.isSuccessful()) throw new IOException("Unexpected code " + response);
logger.info(response.body().string());

#RequestMapping(value = "/uploadImageJson", method = RequestMethod.POST)
public #ResponseBody Object jsongStrImage(#RequestParam(value="image") MultipartFile image, #RequestParam String jsonStr) {
-- use com.fasterxml.jackson.databind.ObjectMapper convert Json String to Object
}

Please ensure that you have following import. Ofcourse other standard imports
import org.springframework.core.io.FileSystemResource
void uploadzipFiles(String token) {
RestBuilder rest = new RestBuilder(connectTimeout:10000, readTimeout:20000)
def zipFile = new File("testdata.zip")
def Id = "001G00000"
MultiValueMap<String, String> form = new LinkedMultiValueMap<String, String>()
form.add("id", id)
form.add('file',new FileSystemResource(zipFile))
def urld ='''http://URL''';
def resp = rest.post(urld) {
header('X-Auth-Token', clientSecret)
contentType "multipart/form-data"
body(form)
}
println "resp::"+resp
println "resp::"+resp.text
println "resp::"+resp.headers
println "resp::"+resp.body
println "resp::"+resp.status
}

.NET Core 2.0 Web API for Video Streaming from FileStream

I have found a bunch of examples that use objects not available to me within my application and don't seem to match up to my version of .NET Core web API. In essence I am working on a project that will have <video> tags on a web page and want to load the videos using a stream from the server rather than directly serving the files via a path. One reason is the source of the files may change and serving them via path isn't what my customer wants. So I need to be able to open a stream and async write the video file.
This for some reason produces JSON data so that's wrong. But I just don't understand what I need to do to send a streamed video file to a <video> tag in HTML.
Current Code:
[HttpGet]
public HttpResponseMessage GetVideoContent()
{
if (Program.TryOpenFile("BigBuckBunny.mp4", FileMode.Open, out FileStream fs))
{
using (var file = fs)
{
var range = Request.Headers.GetCommaSeparatedValues("Range").FirstOrDefault();
if (range != null)
{
var msg = new HttpResponseMessage(HttpStatusCode.PartialContent);
var body = GetRange(file, range);
msg.Content = new StreamContent(body);
msg.Content.Headers.Add("Content-Type", "video/mp4");
//msg.Content.Headers.Add("Content-Range", $"0-0/{fs.Length}");
return msg;
}
else
{
var msg = new HttpResponseMessage(HttpStatusCode.OK);
msg.Content = new StreamContent(file);
msg.Content.Headers.Add("Content-Type", "video/mp4");
return msg;
}
}
}
else
{
return new HttpResponseMessage(HttpStatusCode.BadRequest);
}
}

HttpResponseMessage is not used as a return type in asp.net-core it will read that as an object model and serialize it in the response by design, as you have already observed.
Luckily in ASP.NET Core 2.0, you have
Enhanced HTTP header support
If an application visitor requests content with a Range Request header, ASP.NET will recognize that and handle that header. If the requested content can be partially delivered, ASP.NET will appropriately skip and return just the requested set of bytes. You don't need to write any special handlers into your methods to adapt or handle this feature; it's automatically handled for you.
So now all you have to do is return the file stream
[HttpGet]
public IActionResult GetVideoContent() {
if (Program.TryOpenFile("BigBuckBunny.mp4", FileMode.Open, out FileStream fs)) {
FileStreamResult result = File(
fileStream: fs,
contentType: new MediaTypeHeaderValue("video/mp4").MediaType,
enableRangeProcessing: true //<-- enable range requests processing
);
return result;
}
return BadRequest();
}
Making sure to enable range requests processing. Though, as stated in the documentation, that should be handled based on the request headers and whether that data can be partially delivered.
From there it is now a simple matter of pointing to the endpoint from the video client and let it do its magic

API is getting hit twice while returning audio file

i am returning an audio file from web api. requirement is to play the media file instead of downloading,i am using this code.
[HttpGet]
[Route("audiofile/download", Name = "GetAudioFile")]
public HttpResponseMessage GetAudioFile(string q)
{
if (string.IsNullOrWhiteSpace(q))
{
return new HttpResponseMessage { StatusCode = HttpStatusCode.BadRequest };
}
String path = HttpContext.Current.Server.MapPath("~/AudioUploads/");
string filePath = Path.Combine(path, q);
var response = new HttpResponseMessage(HttpStatusCode.OK)
{
Content = new StreamContent(File.OpenRead(filePath))
};
var contentType = MimeMapping.GetMimeMapping(Path.GetExtension(filePath));
response.Content.Headers.ContentType = new MediaTypeHeaderValue(contentType);
return response;
}
i noticed that this action method is being hit twice as like
Can anyone suggest why its happening? why my api method is being called twice?
P.S I am using Url.Link in order to make uploaded file url. when i hit that, api method is called twice.

Servers only respond to requests. They can't initiate a communication with a client, without initial request.
That said, your client code is to blame here, as it's sending two requests instead of one, and the server correctly responds to both.

How to send post url request and get response return file download into local directory in C# .net

Does anyone know how can I send a post request url and get response return file download into local directory in C# .net? I have the following code which manage to send a post url request and response with the zip file. how can I save the file into the directory?
var client = new WebClient();
var values = new NameValueCollection();
values["StartDate"] = "18-Sep-2016 00:00";
values["EndDate"] = "23-Sep-2016 00:00";
values["CampaignName"] = "null";
values["SearchBy"] = "Evaluation";
values["CustomFilter"] = "rep.CALLDNIS like '#0400000000%'";
var response = client.UploadValues("http://test.com", values);
client.DownloadFile(response, #"C:\myfile.zip");

UploadValues returns a type of byte[].
You can use
System.IO.File.WriteAllBytes(filePath, response) to save it as a file.
OR
Check here How to zip (and unzip) byte[] in C#? on how to save it as a zip file.

Categories