I'm attempting to cache XML files, using .NET Core 5.0.
I'm adapting the example from this page
https://learn.microsoft.com/en-us/aspnet/core/fundamentals/change-tokens?view=aspnetcore-5.0
I have the following code, that successfully caches the file (it doesn't load from disk every time), but when the file changes, the file content is not re-cached.
public XmlDocument loadXML(string strFileName) {
XmlDocument xml;
// Try to obtain the file contents from the cache.
if (_cache.TryGetValue(strFileName, out xml)) {
return xml;
}
xml = this.createNewDocument();
xml.Load(strFileName);
if (xml != null) {
var changeToken = _fileProvider.Watch(strFileName);
var cacheEntryOptions = new MemoryCacheEntryOptions()
.SetSlidingExpiration(TimeSpan.FromMinutes(5))
.AddExpirationToken(changeToken);
// Put the file contents into the cache.
_cache.Set(strFileName, xml, cacheEntryOptions);
}
return xml;
}
}
I figured it out, hope this helps someone in the future.
I pass in a fully qualified file name like "d:\website\file.xml" as "strFileName". This is needed to properly load the XML.
However _fileProvider.Watch() method requires a relative URL, not a qualified file name, so in this case it needs to be "/file.xml".
So I convert "d:\website\file.xml" to "/file.xml" and call _fileProvider.Watch( "/file.xml" );
Related
When I receive a file from C# to iformfile, how do I know the file path of that file?
If I don't know, how do I set the path?
public async Task<JObject> files(IFormFile files){
string filePath = "";
var fileStream = System.IO.File.OpenRead(filePath)
}
In this document, some instructions on IFormFile are introduced:
Files uploaded using the IFormFile technique are buffered in memory or on disk on the server before processing. Inside the action method, the IFormFile contents are accessible as a Stream.
So for IFormFile, you need to save it locally before using the local path.
For example:
// Uses Path.GetTempFileName to return a full path for a file, including the file name.
var filePath = Path.GetTempFileName();
using (var stream = System.IO.File.Create(filePath))
{
// The formFile is the method parameter which type is IFormFile
// Saves the files to the local file system using a file name generated by the app.
await formFile.CopyToAsync(stream);
}
After this, you can get the local file path, namely filePath.
all examples show how to read an xml from an local file. But how do I read a xml from a url or a stream and process it further?
Example: http://www.oreillynet.com/xml/blog/2006/03/hello_saxon_on_net_an_aspnet_i.html
thanks in advance
Look for XsltExamples.cs in the saxon-resources download available on both Sourceforge and www.saxonica.com. The very first example seems to do what you are asking for.
public static void ExampleSimple1(String sourceUri, String xsltUri) {
// Create a Processor instance.
Processor processor = new Processor();
// Load the source document
XdmNode input = processor.NewDocumentBuilder().Build(new Uri(sourceUri));
// Create a transformer for the stylesheet.
XsltTransformer transformer = processor.NewXsltCompiler().Compile(new Uri(xsltUri)).Load();
// Set the root node of the source document to be the initial context node
transformer.InitialContextNode = input;
// Create a serializer
Serializer serializer = new Serializer();
serializer.SetOutputWriter(Console.Out);
// Transform the source XML to System.out.
transformer.Run(serializer);
}
Are you using an XmlDocument object for reading the XML? If so, you'll want XMLDocument.Load() method, which can take a file path or URL, TextReader or Stream as input.
Likewise, XDocument.Load()(msdn.microsoft.com/en-us/library/system.xml.linq.xdocument.load(v=vs.110).aspx) has a similar set of overloads.
How do I open an existing file on the server when a user clicks an actionlink? The following code works for downloading a file but I want to open a new browser window, or tab, and display the file contents.
public ActionResult Download()
{
return File(#"~\Files\output.txt", "application/text", "blahblahblah.txt");
}
You must add "inline" for a new tab.
byte[] fileBytes = System.IO.File.ReadAllBytes(contentDetailInfo.ContentFilePath);
Response.AppendHeader("Content-Disposition", "inline; filename=" + contentDetailInfo.ContentFileName);
return File(fileBytes, contentDetailInfo.ContentFileMimeType);
The way you're using the File() method is to specify a file name in the third argument, which results in a content-disposition header being sent to the client. This header is what tells a web browser that the response is a file to be saved (and suggests a name to save it). A browser can override this behavior, but that's not controllable from the server.
One thing you can try is to not specify a file name:
return File(#"~\Files\output.txt", "application/text");
The response is still a file, and ultimately it's still up to the browser what to do with it. (Again, not controllable from the server.) Technically there's no such thing as a "file" in HTTP, it's just headers and content in the response. By omitting a suggested file name, the framework in this case may omit the content-disposition header, which is your desired outcome. It's worth testing the result in your browser to see if the header is actually omitted.
Use a target of blank on your link to open it in a new window or tab:
Download File
However, forcing the browser to display the contents is out of your control, as it entirely depends on how the user has configured their browser to deal with files that are application/text.
If you are dealing with text, you can create a view and populate the text on that view, which is then returned to the user as a regular HTML page.
please try this and replace your controller name and action name in html action link
public ActionResult ShowFileInNewTab()
{
using (var client = new WebClient()) //this is to open new webclient with specifice file
{
var buffer = client.DownloadData("~\Files\output.txt");
return File(buffer, "application/text");
}
}
OR
public ActionResult ShowFileInNewTab()
{
var buffer = "~\Files\output.txt"; //bytes form this
return File(buffer, "application/text");
}
this is action link which show in new blank tab
<%=Html.ActionLink("Open File in New Tab", "ShowFileInNewTab","ControllerName", new { target = "_blank" })%>
I can´t vote your answered as is useful, follow dow. Thanks very much !
public FileResult Downloads(string file)
{
string diretorio = Server.MapPath("~/Docs");
var ext = ".pdf";
file = file + extensao;
var arquivo = Path.Combine(diretorio, file);
var contentType = "application/pdf";
using (var client = new WebClient())
{
var buffer = client.DownloadData(arquivo);
return File(buffer, contentType);
}
}
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);