WebJun 1, 2016 · I have code that convert word document or excel document to html files then open it using the code below. However, the code below only opens html file for word document for excel file, it keeps looking for something else at a different location sheet.html instead of the main html. I did notice that with word document, there is not tab. WebDec 24, 2011 · using (FileStream file = new FileStream("file.bin", FileMode.Open, FileAccess.Read)) { byte[] bytes = new byte[file.Length]; file.Read(bytes, 0, (int)file.Length); ms.Write(bytes, 0, (int)file.Length); } If the files are large, then it's worth noting that the reading operation will use twice as much memory as the total file size. …
c# read byte array from resource - Stack Overflow
WebJan 15, 2015 · 22. You can add resources to your application by going in the properties of the project, "Resources" tab (create one if needed), Add Resource (existing file). When your file is added, you can set its FileType (in its properties) to Binary. Documentation. After that, you can access to your file as a byte [] easily: var myByteArray = Properties ... WebOpens the file if it exists and seeks to the end of the file, or creates a new file. This would look something like: public static void AppendAllBytes (string path, byte [] bytes) { //argument-checking here. using (var stream = new FileStream (path, FileMode.Append)) { stream.Write (bytes, 0, bytes.Length); } } Share Improve this answer psychology university of michigan
How to create ZipArchive from files in memory in C#?
WebUse File.ReadAllBytes to easily read the entire file into a byte array. void Test () { var dialog = new OpenFileDialog (); var result = dialog.ShowDialog (); if (result != DialogResult.OK) return; byte [] buffer = File.ReadAllBytes (dialog.FileName); // Do whatever you want here with buffer } You're currently calling ShowDialog () twice. WebApr 23, 2024 · The only thing you'd have to do is: This: using (var stream = File.Open (filePath, FileMode.Open, FileAccess.Read)) Becomes this: using (var stream = new MemoryStream (yourByte []) And if you just want to open … WebOct 21, 2011 · byte [] fileBytes = new byte [myStream.Length]; myStream.Read (fileBytes,0,mystream.Length); obj.UploadFile (...) Share Improve this answer Follow answered Oct 21, 2011 at 12:40 jlew 10.4k 1 34 58 2 That's currently assuming the whole file will be read in a single call. That's a dangerous assumption. – Jon Skeet Oct 21, … hosting in github