I am working on a project that is reading the data from a Azure Blob and saving that data into an Object. I am currently running into a problem. The way my code is set up now - it will read all the .txt data within a container if there are no Virtual Folders present.
However, if there is a virtual folder structure present within a Azure Container
my code will error out, with a NullExceptionReference. My idea was to do a if check to see if there was Virtual Folders present within an Azure Container if so execute //some code. Is there a way to tell if there is a virtual folder is present?
ReturnBlobObject()
private List<Blob> ReturnBlobObject(O365 o365)
{
List<Blob> listResult = new List<Blob>();
string textToFindPattern = "(\\/)";
string fileName = null;
string content = null;
//Loop through all Blobs and split the container form the file name.
foreach (var blobItem in o365.Container.ListBlobs(useFlatBlobListing: true))
{
string containerAndFileName = blobItem.Parent.Uri.MakeRelativeUri(blobItem.Uri).ToString();
string[] subString = Regex.Split(containerAndFileName, textToFindPattern);
//subString[2] is the name of the file.
fileName = subString[2];
content = ReadFromBlobStream(o365.Container.GetBlobReference(subString[2]));
Blob blobObject = new Blob(fileName, content);
listResult.Add(blobObject);
}
return listResult;
}
ReadFromBlobStream
private string ReadFromBlobStream(CloudBlob blob)
{
Stream stream = blob.OpenRead();
using (StreamReader reader = new StreamReader(stream))
{
return reader.ReadToEnd();
}
}
I was able to solve this by refactoring my code. Instead of using Regex - which was returning some very odd behavior I decided to take a step back and think the problem. Below is the solution I came up with.
ReturnBlobObject()
private List<Blob> ReturnBlobObject(O365 o365)
{
List<Blob> listResult = new List<Blob>();
//Loop through all Blobs and split the container form the file name.
foreach (var blobItem in o365.Container.ListBlobs(useFlatBlobListing: true))
{
string fileName = blobItem.Uri.LocalPath.Replace(string.Format("/{0}/", o365.Container.Name), "");
string content = ReadFromBlobStream(o365.Container.GetBlobReference(fileName));
Blob blobObject = new Blob(fileName, content);
listResult.Add(blobObject);
}
return listResult;
}
Related
I am working on c# .Net 4.5
I have to upload some files on MongoDB and in other module, I have to get them back based on metadata.
for that I am doing like below,
static void uploadFileToMongoDB(GridFSBucket gridFsBucket)
{
if (Directory.Exists(_sourceFilePath))
{
if (!Directory.Exists(_uploadedFilePath))
Directory.CreateDirectory(_uploadedFilePath);
FileInfo[] sourceFileInfo = new DirectoryInfo(_sourceFilePath).GetFiles();
foreach (FileInfo fileInfo in sourceFileInfo)
{
string filePath = fileInfo.FullName;
string remoteFileName = fileInfo.Name;
string extension = Path.GetExtension(filePath);
double fileCreationDate = fileInfo.CreationTime.ToOADate();
GridFSUploadOptions gridUploadOption = new GridFSUploadOptions
{
Metadata = new BsonDocument
{{ "creationDate", fileCreationDate },
{ "extension", extension }}
};
using (Stream fileStream = File.OpenRead(filePath))
gridFsBucket.UploadFromStream(remoteFileName, fileStream, gridUploadOption);
}
}
}
and downloading,
static void getFileInfoFromMongoDB(GridFSBucket bucket, DateTime startDate, DateTime endDate)
{
double startDateDoube = startDate.ToOADate();
double endDateDouble = endDate.ToOADate();
var filter = Builders<GridFSFileInfo>.Filter.And(
Builders<GridFSFileInfo>.Filter.Gt(x => x.Metadata["creationDate"], startDateDoube),
Builders<GridFSFileInfo>.Filter.Lt(x => x.Metadata["creationDate"], endDateDouble));
IAsyncCursor<GridFSFileInfo> fileInfoList = bucket.Find(filter); //****
if (!Directory.Exists(_destFilePath))
Directory.CreateDirectory(_destFilePath);
foreach (GridFSFileInfo fileInfo in fileInfoList.ToList())
{
string destFile = _destFilePath + "\\" + fileInfo.Filename;
var fileContent = bucket.DownloadAsBytes(fileInfo.Id); //****
File.WriteAllBytes(destFile, fileContent);
}
}
in this code (working but) I have two problems which I am not sure how to fix.
If i have uploaded a file and I upload it again, it actually gets
uploaded. How to prevent it?
Ofcourse both uploaded files have different ObjectId but while uploading a file I will not be knowing that which files are already uploaded. So I want a mechanism which throws an exception if i upload already uploaded file. Is it possible? (I can use combination of filename, created date, etc)
If you have noticed in code, actually i am requesting to database server twice to get one file written on disk, How to do it in one shot?
Note lines of code which I have marked with "//****" comment. First I am querying into database to get fileInfo (GridFSFileInfo). I was expecting that I could get actual content of file from this objects only. But I didnot find any related property or method in that object. so I had to do var fileContent = bucket.DownloadAsBytes(fileInfo.Id); to get content. M I missing something basic here ?
I am trying to browse and upload a file from client to server using Angular Js and WEB API.I used Input file type for user to select file and post the file to WEB API. In web API, I am getting following error "This method or property is not supported after HttpRequest.GetBufferlessInputStream has been invoked."
I am using the following code:-
public IHttpActionResult UploadForm()
{
HttpResponseMessage response = new HttpResponseMessage();
var httpRequest = System.Web.HttpContext.Current.Request;
if (httpRequest.Files.Count > 0)
{
foreach (string file in httpRequest.Files)
{
var postedFile = httpRequest.Files[file];
var filePath = System.Web.HttpContext.Current.Server.MapPath("~/UploadFile/" + postedFile.FileName);
postedFile.SaveAs(filePath);
}
}
return Json("Document Saved");
}
I get this error when i tried to get files from HTTP request... should I update anything in web config??
Please help me to resolve this issue..
try this it work fine for me.
//get the root folder where file will be store
string root = HttpContext.Current.Server.MapPath("~/UploadFile");
// Read the form data.
var provider = new MultipartFormDataStreamProvider(root);
await Request.Content.ReadAsMultipartAsync(provider);
if (provider.FileData.Count > 0 && provider.FileData[0] != null)
{
MultipartFileData file = provider.FileData[0];
//clean the file name
var fileWithoutQuote = file.Headers.ContentDisposition.FileName.Substring(1, file.Headers.ContentDisposition.FileName.Length - 2);
//get current file directory on the server
var directory = Path.GetDirectoryName(file.LocalFileName);
if (directory != null)
{
//generate new random file name (not mandatory)
var randomFileName = Path.Combine(directory, Path.GetRandomFileName());
var fileExtension = Path.GetExtension(fileWithoutQuote);
var newfilename = Path.ChangeExtension(randomFileName, fileExtension);
//Move file to rename existing upload file name with new random filr name
File.Move(file.LocalFileName, newfilename);
}
}
I also had the same problem. And the solution by #Jean did not work for me.
I need to upload some CSV file and had to use it in the controller.
In Javascript, I used Fetch API to upload the csv file.
But, in the controller, I used this code:
[HttpPost]
[CatchException]
public bool ImportBundlesFromCsv()
{
var a = Request.Content.ReadAsByteArrayAsync();
//convert to Stream if needed
Stream stream = new MemoryStream(a.Result); // a.Result is byte[]
// convert to String if needed
string result = System.Text.Encoding.UTF8.GetString(a.Result);
// your code
return true;
}
This worked for me. Hope this helps!
Hi I'm writing a c# code where in there is a string sent as a parameter input to the method. And then the inputString has to be searched in the file and the result has to be returned. Currently I know how do I do this in the regular way(using the file IO).
[HttpPost]
public string UsernameValidation(string username)
{
string text = username;
string userExists = usernameNotAvailable;
string line;
System.IO.StreamReader file = new System.IO.StreamReader("~/UserData/usernameslist.txt");
while ((line = file.ReadLine()) != null)
{
if (line.Contains(text))
{
userExists = usernameAvailable;
}
}
return userExists;
}
But here is the twist, my project is in MVC. I'm able to get the path of file using string userDataFile = Server.MapPath("~/UserData/usernameslist.txt");.
But I'm unable to know how can I get the functionality of searching a string in a file.
Please let me know how can I do this.
Thanks
If the the file usernameslist.txt really exists inside a subfolder named UserData from your root folder then you just need to pass the output of Server.MapPath to your StreamReader constructor
string fileName = Server.MapPath("~/UserData/usernameslist.txt");
using(StreamReader file = new System.IO.StreamReader(fileName))
{
....
}
And don't forget to use the using statement around a Stream object
I need to upload a file using Stream (Azure Blobstorage), and just cannot find out how to get the stream from the object itself. See code below.
I'm new to the WebAPI and have used some examples. I'm getting the files and filedata, but it's not correct type for my methods to upload it. Therefore, I need to get or convert it into a normal Stream, which seems a bit hard at the moment :)
I know I need to use ReadAsStreamAsync().Result in some way, but it crashes in the foreach loop since I'm getting two provider.Contents (first one seems right, second one does not).
[System.Web.Http.HttpPost]
public async Task<HttpResponseMessage> Upload()
{
if (!Request.Content.IsMimeMultipartContent())
{
this.Request.CreateResponse(HttpStatusCode.UnsupportedMediaType);
}
var provider = GetMultipartProvider();
var result = await Request.Content.ReadAsMultipartAsync(provider);
// On upload, files are given a generic name like "BodyPart_26d6abe1-3ae1-416a-9429-b35f15e6e5d5"
// so this is how you can get the original file name
var originalFileName = GetDeserializedFileName(result.FileData.First());
// uploadedFileInfo object will give you some additional stuff like file length,
// creation time, directory name, a few filesystem methods etc..
var uploadedFileInfo = new FileInfo(result.FileData.First().LocalFileName);
// Remove this line as well as GetFormData method if you're not
// sending any form data with your upload request
var fileUploadObj = GetFormData<UploadDataModel>(result);
Stream filestream = null;
using (Stream stream = new MemoryStream())
{
foreach (HttpContent content in provider.Contents)
{
BinaryFormatter bFormatter = new BinaryFormatter();
bFormatter.Serialize(stream, content.ReadAsStreamAsync().Result);
stream.Position = 0;
filestream = stream;
}
}
var storage = new StorageServices();
storage.UploadBlob(filestream, originalFileName);**strong text**
private MultipartFormDataStreamProvider GetMultipartProvider()
{
var uploadFolder = "~/App_Data/Tmp/FileUploads"; // you could put this to web.config
var root = HttpContext.Current.Server.MapPath(uploadFolder);
Directory.CreateDirectory(root);
return new MultipartFormDataStreamProvider(root);
}
This is identical to a dilemma I had a few months ago (capturing the upload stream before the MultipartStreamProvider took over and auto-magically saved the stream to a file). The recommendation was to inherit that class and override the methods ... but that didn't work in my case. :( (I wanted the functionality of both the MultipartFileStreamProvider and MultipartFormDataStreamProvider rolled into one MultipartStreamProvider, without the autosave part).
This might help; here's one written by one of the Web API developers, and this from the same developer.
Hi just wanted to post my answer so if anybody encounters the same issue they can find a solution here itself.
here
MultipartMemoryStreamProvider stream = await this.Request.Content.ReadAsMultipartAsync();
foreach (var st in stream.Contents)
{
var fileBytes = await st.ReadAsByteArrayAsync();
string base64 = Convert.ToBase64String(fileBytes);
var contentHeader = st.Headers;
string filename = contentHeader.ContentDisposition.FileName.Replace("\"", "");
string filetype = contentHeader.ContentType.MediaType;
}
I used MultipartMemoryStreamProvider and got all the details like filename and filetype from the header of content.
Hope this helps someone.
I have a folder called data/ in my project that contains txt files.
I configured Build Action to resources to all files.
I tried these different ways:
method 1
var resource = Application.GetResourceStream(new Uri(fName, UriKind.Relative));
StreamReader streamReader = new StreamReader(resource.Stream);
Debug.WriteLine(streamReader.ReadToEnd());
method 2
IsolatedStorageFile myIsolatedStorage = IsolatedStorageFile.GetUserStoreForApplication();
string[] fileNames = myIsolatedStorage.GetFileNames("*.txt");
method 3
using (IsolatedStorageFile isf = IsolatedStorageFile.GetUserStoreForApplication())
{
using (StreamReader fileReader = new StreamReader(new IsolatedStorageFileStream(fName, FileMode.Open, isf)))
{
while (!fileReader.EndOfStream)
{
string line = fileReader.ReadLine();
al.Add(line);
Debug.WriteLine(line);
}
}
}
Now, i tried different ways to read files without success, why?
Where is the problem?
What's wrong with these methods?
fName is the name of the file.
It's necessary the full path data/filename.txt? It's indifferent...
please help me with this stupid issue,
thanks.
Your 2nd & 3rd approaches are wrong. When you include a text file locally in your app, you can't refer it via the IS. Instead, use this function, it will return the file content if found else it will return "null". It works for me, hope it works for you.
Note, if the file is set as content, the filePath = "data/filename.txt" but if it is set as resource it should be referred like this filePath = "/ProjectName;component/data/filename.txt". That may be why your 1st approach might have failed.
private string ReadFile(string filePath)
{
//this verse is loaded for the first time so fill it from the text file
var ResrouceStream = Application.GetResourceStream(new Uri(filePath, UriKind.Relative));
if (ResrouceStream != null)
{
Stream myFileStream = ResrouceStream.Stream;
if (myFileStream.CanRead)
{
StreamReader myStreamReader = new StreamReader(myFileStream);
//read the content here
return myStreamReader.ReadToEnd();
}
}
return "NULL";
}