I only get one uri from the file_picker when saving. But so that I can display the correct path, I need the absolute path of the file. Unfortunately I haven't found a working solution. The library flutter_absolute_path unfortunately doesn't seem to work with the new Android versions anymore.
Does anyone have any idea how I could get the absolute path of the file.
example:
content://com.android.providers.downloads.documents/document/297
convert to:
/data/user/0/xxx/cache/dummy1.png
Check out the toFilePath() method of the dart:core package to see if it solves your problem.
I thought so too. But unfortunately I get an exception.
test code:
//uri output from file picker (fileInfo.identifier)
var uriPath = "content://com.android.providers.downloads.documents/document/297";
var uri = Uri.parse(uriPath);
print(uri.toFilePath());
exception
-> Unsupported operation: Cannot extract a file path from a content URI
Related
I have a Dart code that need to take a path and work on it. For portability purposes I need the path to be relative and have Dart convert it to absolute path. How can I do this.
For example:
var samplePath = 'relative/path/file.txt';
// converted to 'C:/Users/XYZ/Desktop/relative/path/file.txt'
Flutter 2.7
import 'package:path/path.dart' as p;
var absPath = p.absolute('a/b', 'c');
// '...<current working dir>/a/b/c
Documentation
I have never programmed in Dart, but in this part of the documentation there is a property of the class File which does what you want.
Moreover, if you have the root path of your relative path (maybe your current directory) you can join them using the join function
Hope it helps you.
The following code runs without a hitch:
On the other hand, I get an access-denied error with this:
The destination is in my personal folder and I have full control. The directory is not read-only. Anyway, in either of those cases, the first code sample should not run either! I appreciate the help ...
In the second sample, you have two problems:
There are back slashes instead of forward slashes, so some of them may get interpreted as escape sequences.
You completely ignore the first parameter of write and specify what I assume is a folder as destination. You can't open a file stream on a folder, no wonder you get access denied.
This should work:
let write filename (ms:MemoryStream) =
let path = System.IO.Path.Combine( "C:/Users/<whatever>/signal_processor", filename )
use fs = new FileStream( path, FileMode.Create )
ms.WriteTo(fs)
I have a Cordova/PhoneGap app which uses the camera to capture a picture. I need to access this picture to send to my server.
For Android I simply use $cordovaFile.readAsDataURL(cordova.file.dataDirectory, fileName)
However; on iOS the filename does not come back as such. My return value for iOS shows the path as
assets-library://asset/asset.JPG?id=539425FE-43AA-48E7-9865-D76348208AC7&ext=JPG
How can I access and read this image? I'd prefer to stick with $cordovaFile but just want to get it to work.
I am using CordovaCameraPreview plugin and the best I can get back from the picture taker handler seems to be something formed similar to:
assets-library://asset/asset.JPG?id=990355E1-200A-4E35-AAA1-7D461E3999C8&ext=JPG
assets-library://asset/asset.JPG?id=C49FF0EB-CCCB-45B2-8577-B13868D8DB29&ext=JPG
How can I convert this to a filename and path that I can read with $cordovaFile? According to their documentation (http://ngcordova.com/docs/plugins/file/) it looks like I need to use one of their paths for File System Layout. It works fine on Android using cordova.file.dataDirectory but can't figure out how to access on iOS
I've also tried using window.resolveLocalFileSystemURL(path) but am getting undefined as my result.
Update
I feel like I'm getting closer... using window.resolveLocalFileSystemURL(path,resolveOnSuccess, resOnError); I get more information
name: "assets-library"
fullPath: "/asset/asset.JPG?id=711B4C9D-97D6-455A-BC43-C73059A5C3E8&ext=JPG"
name: "asset.JPG?id=711B4C9D-97D6-455A-BC43-C73059A5C3E8&ext=JPG"
nativeURL: "assets-library://asset/asset.JPG?id=711B4C9D-97D6-455A-BC43-C73059A5C3E8&ext=JPG"
It looks like I now need to use the fullPath but still can't figure out how to access with $cordovaFile
If I try to use $cordovaFile.readAsDataURL(cordova.file.dataDirectory, data.name) where data.name is asset.JPG?id=711B4C9D-97D6-455A-BC43-C73059A5C3E8&ext=JPG I get a file not found error
Update 2
I have tried using every single File System Layout available at http://ngcordova.com/docs/plugins/file/ and receive the same File Not Found error on each one. Still no clue how to access a file in assets-library using $cordovaFile
I had success with a simple substitution, like this
fullPath = fullPath.replace("assets-library://", "cdvfile://localhost/assets-library/")
It works in iOS 9.3
I was struggling with a similar issue for a while, but I figured it out today. I also thought that the FileSystem APIs didn't work with assets-libary:// URIs -- but that is not the case.
You should be able to get access to the File object (as well as the actual image name) by using the following code:
resolveLocalFileSystemURL('assets-library://asset/asset.JPG?id=711B4C9D-97D6-455A-BC43-C73059A5C3E8&ext=JPG', function(fileEntry) {
fileEntry.file(function(file) {
var reader = new FileReader();
reader.onloadend = function(event) {
console.log(event.target.result.byteLength);
};
console.log('Reading file: ' + file.name);
reader.readAsArrayBuffer(file);
});
});
My files are referenced like so (it's all relative):
// WHERE YOU KEEP THE PAGE TITLE XML
public static string myPageTitleXML = "xml/pagetitles.xml";
and
using (StreamReader r = new StreamReader(myPageTitleXML))
{ //etc.. . .etc....etc..
}
I get system.io.directorynotfound, and "this problem needs to be shut down", when I double click the executable. But running it from the console works like a charm. What's wrong here?
I played around with attempting to set Environment.CurrentDirectory but couldn't get anything to work. Why should I have to do that anyway? It defeats the purpose of a relative path no?
responding.. .
"application" does not exist in the current context, i'll keep trying what people have mentioned, this is not a windows.form
testing
Path.GetDirectoryName(Reflection.Assembly.GetExecutingAssembly().GetName().CodeBase), myPageTitleXML); gives error URI formats are not supported, as does Path.GetFullPath(). Server.MapPath results in an error as well, this is currently offline
Well assuming this directory is somewhere under the directory in which your code is executing, it sounds like you can use ..
Application.ExecutablePath()
or
Application.StartUpPath()
.. to get an idea as to what your application is seeing when it goes in search of an 'xml' directory with the 'pagetitles.xml' file in it.
If the directory returned by one of these methods does not point where you thought it did, you'll need to move the location of your application or the location of this folder so that it is within the same directory as the app.
Hope this gets you on the right path.
So, when you run it from double clicking the executable, is there a file named pagetitles.xml in a folder named xml, where xml is a folder in the same location as the executable?
It's certainly possible to use relative paths like this, but I wouldn't really recommend it. Instead, maybe use something like:
string fileToOpen = System.IO.Path.Combine(System.IO.Path.GetDirectoryName(System.Reflection.Assembly.GetExecutingAssembly().GetName().CodeBase), myPageTitleXML);
using (StreamReader r = new StreamReader(fileToOpen))
{
//etc.. . .etc....etc..
}
Is this ASP.NET code? If so then you probably need to do MapPath("xml/pagetitles.xml")
The message box crops up could not find part of the path of the file"file path" when i try to open a file that has space in its file path. I have used LocalPath instead of AbsolutePath and it works fine for me, but its only limited to WinApps, i needed a more generic solution. Some thing like Uri unescaped data path. I am not sure about the syntax.
In Java: URI uri = new File("spaces in file name").toURI();
If the file can be fetched depends on your implementation of the software. Try to replace the spaces with %20
What kind of development are we talking about here? JAVA GUI or WebApps? C/C++?
You should try enclosing your URI with quotes
new Uri("\"C:\some path\some file\"");