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);
}
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/ .
var fs = require('fs');
var SQL = require('sql.js');
var filebuffer = fs.readFileSync('./resources/app.asar/app/data/sample.db');
var db = new SQL.Database(filebuffer);
function save_data(){
var name=document.getElementById('name').value;
var ip=document.getElementById('serverip').value;
var result=db.each("UPDATE Settings SET Name=$name, IP=$ip WHERE SettingsId=$set",{$name:name,$ip : ip,$set:1},function(row){console.log(row.name)});
var data = db.export();
var buffer = new Buffer(data);
fs.writeFileSync('./resources/app.asar/app/data/sample.db', buffer);
}
I was able to read the data from the database file inside the asar archive, but while writing the data into db file, it doesn't gets updated inside asar archive. So please help me crack this issue.
Asar is a read-only archive. It just concatenates all the files together into a single blob.
I would like to know how best to handle file upload and addtional information added to the file to be uploaded using ASP.NET Web API 2 without MVC components. I have google the net and I can tell you I am more confused than I expected.
The Additional info will be stored in db and the file on the disk.
So far the Web API app I am building does not support multipart/form-data. It only supports the default media types. I know I need to create a media formatter.
Pls help.
I had wrote Javascript split File and upload to WEB API . i think you can reference my backend codes
In front-end you need using below code to upload your File
var xhr = new self.XMLHttpRequest();
xhr.open('POST', url, false);
xhr.setRequestHeader('Content-Type', 'application/octet-stream');
xhr.send(chunk);
In backend use Request.InputStream.Read to catch your file bytes
[HttpPost]
[ValidateInput(false)]
public string fileUpload(string filename)
{
byte[] file = new byte[Request.InputStream.Length];
Request.InputStream.Read(file, 0, Convert.ToInt32(Request.InputStream.Length));
BinaryWriter binWriter = new BinaryWriter(new MemoryStream());
binWriter.Write(file);
StreamReader reader = new StreamReader(binWriter.BaseStream);
reader.BaseStream.Position = 0;
//This example is recevied text file
while ((line = reader.ReadLine()) != null)
{
};
}
You can just serialize your file data into BASE64 and send them as a string in case of multipart/from-data is not allowed for some reason.
So I've got an MVC 3 application that has a couple places where a text file gets generated and returned in an action using:
return File(System.Text.Encoding.UTF8.GetBytes(someString),
"text/plain", "Filename.extension");
and this works fabulously. Now i've got a situation where I'm trying to return a pair of files in a similar fashion. On the view, i have an action link like "Click here to get those 2 files" and i'd like both files to be downloaded much like the single file is downloaded in the above code snippet.
How can I achieve this? Been searching around quite a bit and haven't even seen this question posed anywhere...
Building on Yogendra Singh's idea and using DotNetZip:
var outputStream = new MemoryStream();
using (var zip = new ZipFile())
{
zip.AddEntry("file1.txt", "content1");
zip.AddEntry("file2.txt", "content2");
zip.Save(outputStream);
}
outputStream.Position = 0;
return File(outputStream, "application/zip", "filename.zip");
Update 2019/04/10:
As #Alex pointed out, zipping is supported natively since .NET Framework 4.5, from JitBit and others:
using (var memoryStream = new MemoryStream())
{
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
var file1 = archive.CreateEntry("file1.txt");
using (var streamWriter = new StreamWriter(file1.Open()))
{
streamWriter.Write("content1");
}
var file2 = archive.CreateEntry("file2.txt");
using (var streamWriter = new StreamWriter(file2.Open()))
{
streamWriter.Write("content2");
}
}
return File(memoryStream.ToArray(), "application/zip", "Images.zip")
}
Sorry for bumping an old question but...
Another alternative would be to initiate multiple file downloads using JavaScript, and serve files in two different Action Methods on ASP.NET's side.
You're saying you have a link:
On the view, i have an action link like "Click here to get those 2
files"
So make this link like this:
Click to get 2 files
<script src="download.js"></script>
I'm using download.js script found here but you can find plenty of different other options, see this SO question: starting file download with JavaScript for example
I would advice to create a zip file to include both the files using steps(ALGORITHM):
Create a Zip file and add the desired files into the zip
Return the zip file having all desired files from the action
Java Syntax (Just for understanding)
FileOutputStream fos = new FileOutputStream("downloadFile.zip");
ZipOutputStream zos = new ZipOutputStream(new BufferedOutputStream(fos));
zos.putNextEntry(new ZipEntry("Filename1.extension"+));
//write data in FileName1.extension
zos.write(contentBuffer1, 0, len);
zos.putNextEntry(new ZipEntry("Filename2.extension"));
//write data in FileName2.extension
zos.write(contentBuffer2, 0, len);
//write other files.....
zos.close();
Once zip file is created, return the newly created zip file to download.
return File("downloadFile.zip");
.DOT Net Equivalent using DotNetZip
var os = new MemoryStream();
using (var zip = new ZipFile())
{
//write the first file into the zip
zip.AddEntry("file1.txt", "content1");
//write the second file into the zip
zip.AddEntry("file2.txt", "content2");
//write other files.....
zip.Save(os);
}
outputStream.Position = 0;
return File(outputStream, "application/zip", "filename.zip");
Hope this helps!
Look at this SO solution: MVC Streaming Zip File
The advantage of this solution is that it streams the file to the client.
I just implemented this solution a couple of days ago and it worked fantastic.
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);