I am getting the following error while uploading the file from my local drive.
The given path's format is not supported.
The code is given.
Please tell me what changes I have to make.
string file0 = MapPathReverse(FileUpload1.PostedFile.FileName);// Get virtual path
string conversationFileSource = Server.MapPath(file0);
StreamReader file = new StreamReader(conversationFileSource);
If you want to access the input stream of the uploaded file:
using (StreamReader reader = new StreamReader(FileUpload1.PostedFile.InputStream))
{
...
}
If you want to save the uploaded file on some folder on your server:
var uploadsFolder = Server.MapPath("~/uploads");
var file = Path.Combine(uploadsFolder, Path.GetFileName(FileUpload1.PostedFile.FileName));
FileUpload1.PostedFile.SaveAs(file);
Related
I am developing an application for Android using Xamarin.
I have created a JsonData folder in the Android project and created a Setting.json file.
\MyApp\MyApp.Android\JsonData\Setting.json
In the properties, we set the Copy when new.
The following folders in the local environment contain the files.
\MyApp\MyApp.Android\bin\Debug\JsonData\Setting.json
I want to load this file in the actual Android device.
When I do this, it tells me that the file is missing.
Could not find a part of the path "/JsonData/Setting.json."
Try
{
var text = File.ReadAllText("JsonData/Setting.json", Encoding.UTF8);
var setting = JsonConvert.DeserializeObject<Setting>(text);
}
catch(Exception exception)
{
var error = exception.Message;
}
What is the path of the file in Android?
I think you're using File Handling in Xamarin.Forms incorrectly.
From the parameter of function File.ReadAllText, the app will access the file system to getSetting.json from folder JsonData in your android device.
The path of the file on each platform can be determined from a .NET Standard library by using a value of the Environment.SpecialFolder enumeration as the first argument to the Environment.GetFolderPath method. This can then be combined with a filename with the Path.Combine method:
string fileName = Path.Combine(Environment.GetFolderPath(Environment.SpecialFolder.LocalApplicationData), "temp.txt");
And you can read the file by code:
string text = File.ReadAllText(fileName);
In addition, from your code,I guess you want to Load your Embedded file( Setting.json) as Resources,right?
In this case,we should make sure the Build Action of your Setting.json is Embedded Resource.
And GetManifestResourceStream is used to access the embedded file using its Resource ID.
You can refer to the following code:
var assembly = IntrospectionExtensions.GetTypeInfo(typeof(LoadResourceText)).Assembly;
Stream stream = assembly.GetManifestResourceStream("YourAppName.JsonData.Setting.json");
string text = "";
using (var reader = new System.IO.StreamReader (stream))
{
text = reader.ReadToEnd ();
}
For more , you can check document : File Handling in Xamarin.Forms.
And you can also check the sample code here: https://learn.microsoft.com/en-us/samples/xamarin/xamarin-forms-samples/workingwithfiles/ .
newbie here,I could not find any example on Xamarin Forms read a local json file and display it. I need to do a local testing to read the local Json file.
1) Where do I save the json file for reading? in Android and iOS Projects or just in PCL project?
2) How to read the file?
here the code but it is not complete as I dont how to read the file.
using (var reader = new System.IO.StreamReader(stream))
{
var json = reader.ReadToEnd();
var rootobject = JsonConvert.DeserializeObject<Rootobject>(json);
whateverArray = rootobject.Whatever;
}
The code miss the Path and others which required.
You can directly add your JSON file in PCL. Then change build action to Embedded Resource
Now you can read Json data by:
var assembly = typeof("<ContentPageName>").GetTypeInfo().Assembly;
Stream stream = assembly.GetManifestResourceStream("Your_File.json");
using (var reader = new System.IO.StreamReader(stream))
{
var json = reader.ReadToEnd();
var data= JsonConvert.DeserializeObject<Model>(json);
}
I tried to upload file to blob. But I'm getting error like this:
"'C:\Program Files (x86)\IIS
Express\Nominative-Officers-Entry-Form-Stu.docx'."
I don't use HttpPostedFileBase in my code. I just pass object to my controller with files to be uploaded. Lease tell me what I'm doing wrong?
I want to know wt this line means :
"blockBlob.Properties.ContentType = "
This is my code:
public static SaveResponses CreateFile(Blob_Storage_Header docDetails)
{
string storageConnectionString = ConfigurationManager.ConnectionStrings["StorageConnectionString"].ConnectionString.ToString();
CloudStorageAccount storageAccount = CloudStorageAccount.Parse(storageConnectionString);
CloudBlobClient blobClient = storageAccount.CreateCloudBlobClient();
CloudBlobContainer container = blobClient.GetContainerReference("test2");
ICollection<Blob_Storage_Details> BlobStorageDetails = docDetails.Blob_Storage_Details;
if (BlobStorageDetails.Count > 0) {
foreach (Blob_Storage_Details item in BlobStorageDetails)
{
string DocUUID = Guid.NewGuid().ToString();
CloudBlockBlob blockBlob = container.GetBlockBlobReference(DocUUID + item.Blob_Name);
var fileName = Path.GetFileName(item.Blob_Name);
//blockBlob.Properties.ContentType = item.ContentType;
// Create or overwrite the "myblob" blob with contents from a local file.
using (var fileStream = File.OpenRead(fileName))
{
blockBlob.UploadFromStream(fileStream);
}
}
}
SaveResponses saveResponse = new SaveResponses();
saveResponse.saveStatus = "true";
saveResponse.messageType = "success";
saveResponse.message = "File Create message";
return (saveResponse);
}
"'C:\Program Files (x86)\IIS Express\Nominative-Officers-Entry-Form-Stu.docx'."
Based on your code, I ran it on my side, then I got the following error:
Remark: In development environment, you could add “<customErrors mode="Off"/>” within the system.web node in your Web.config file, then you could view the detailed errors.
According to this error, I checked and found the specified file path was not existed.
After some trials, I fixed the issue on my side. Please follow the descriptions below to check your code to see whether it works:
a) Please pay attention to the code “Path.GetFileName”, it returns the file name and extension of the specified path string (e.g. settings.job).
b) Make sure that the filename that is used in “File.OpenRead(fileName)” is an absolute file path and the file is existed in your environment.
"blockBlob.Properties.ContentType = "
Azure Storage Client Library for .NET is based on Storage Service REST API,
From the official document we could find that “blockBlob.Properties.ContentType” represents the MIME content type of the blob, the default type is application/octet-stream.
MIME is a way to identity files on internet according to their nature and format. For example, using the "Content-type" header value defined in a HTTP response, the browser can open the file with the proper extension/plugin. For more details about MIME, please refer to this link: http://www.freeformatter.com/mime-types-list.html
Im trying to search and download the files uploaded to the database,I can able to retrieve the files from the database.
Downloading single file works but cant download multiple files at a time,I have read about downloading as a zip file,Can anyone help??????
using (sampledbase dbcontext = new sampledbase ())
{
ZipFile zip = new ZipFile();
long id = Convert.ToInt64(Request.QueryString["id"]);
System.Nullable<long> fileid = id;
var query = dbcontext.searchfile(ref fileid );
foreach (var i in query)
{
byte[] binarydata = i.Data.ToArray();
Response.AddHeader("content-disposition", string.Format("attachment; filename =\"{0}\"", i.Name));
Response.BinaryWrite(binarydata);
Response.ContentType = i.ContentType;
Response.End();
}
}
I see that you are using ZipFile so you are in the right direction. In order to zip your files, you can save your files in a directory and zip the directory using this code :
string startPath = #"c:\example\start";
string zipPath = #"c:\example\result.zip";
ZipFile.CreateFromDirectory(startPath, zipPath);
In your case, look at the data type returned by dbcontext.searchfile() and save the content to a directory using something like StreamWriter. Looking at your code, the type seems to be a byte[], you can see this link to learn how to save the bytes in a file : Write bytes to file
Another solution is to zip directly the bytes[] : Create zip file from byte[]
I am trying to create a upload servlet that handles enctype="multipart/form-data" from a form. The file I am trying to upload is a zip. However, I can upload and read the file on localhost, but when I upload to the server, I get a "File not found" error when I want to upload a file. Is this due to the Struts framework that I am using? Thanks for your help. Here is part of my code, I am using FileUpload from http://commons.apache.org/fileupload/using.html
I have changed to using ZipInputStream, however, how to I reference to the ZipFile zip without using a local disk address (ie: C://zipfile.zip). zip is null because its not instantiated. I will need to unzip and read the zipentry in memory, without writing to the server.
For the upload servlet:
>
private ZipFile zip;
private CSVReader reader;
boolean isMultipart = ServletFileUpload.isMultipartContent(request);
if(isMultipart){
DiskFileItemFactory factory = new DiskFileItemFactory();
ServletFileUpload upload = new ServletFileUpload(factory);
List <FileItem> items = upload.parseRequest(request);
Iterator iter = items.iterator();
while (iter.hasNext()) {
//Iterating through the uploaded zip file and reading the content
FileItem item = (FileItem) iter.next();
ZipInputStream input = new ZipInputStream(item.getInputStream());
ZipEntry entry = null;
while (( entry= input.getNextEntry()) != null) {
ZipEntry entry = (ZipEntry) e.nextElement();
if(entry.getName().toString().equals("file.csv")){
//unzip(entry)
}
}
}
public static void unzip(ZipEntry entry){
try{
InputStream inputStream = **zip**.getInputStream(entry);
InputStreamReader inputStreamReader = new InputStreamReader(inputStream);
reader = new CSVReader(inputStreamReader);
}
catch(Exception e){
e.printStackTrace();
}
}
<
Here,
zip = new ZipFile(new File(fileName));
You're assuming that the local disk file system at the server machine already contains the file with exactly the same name as it is at the client side. This is a wrong assumption. That it worked at localhost is obviously because both the webbrowser and webserver "by coincidence" runs at physically the same machine with the same disk file system.
Also, you seem to be using Internet Explorer as browser which incorrectly includes the full path in the filename like C:/full/path/to/file.ext. You shouldn't be relying on this browser specific bug. Other browsers like Firefox correctly sends only the file name like file.ext, which in turn would have caused a failure with new File(fileName) (which should have helped you to spot your mistake much sooner).
To fix this "problem", you need to obtain the file contents as InputStream by item.getInputStream():
ZipInputStream input = new ZipInputStream(item.getInputStream());
// ...
Or to write it to disk by item.write(file) and reference it in ZipFile:
File file = File.createTempFile("temp", ".zip");
item.write(file);
ZipFile zipFile = new ZipFile(file);
// ...
Note: don't forget to check the file extension beforehand, else this may choke.