Android Question (SOLVED)File path to URI path

Ferdari · Feb 25, 2020

Hi everyone

Im creating an app that need to save URI path to a JSON file, is there a function to return the URI from a file?

for example my json should contain uri like this:
file:///data/user/0/com.myapp/files/share/1/0.webp
My files are stored on the safe dir share created

XML:

CreateResource(xml, provider_paths,
   <files-path name="name" path="shared" />
)

All folders are ok and work, but i still not realize how to extract the URI from a file, for example the file sound.mp3 saved at shared folder

B4X:

safefolder = Starter.Provider.SharedFolder
mp.Load(safefolder,"sound.mp3")

I want to get his URI path and save it to file for later working with it.

peacemaker · Feb 25, 2020

If you copied each needed file into "safefolder" - each file path will be "safefolder/filename", i.e. "shared" subfolder in the internal folder of your app, something like "data/user/0/com.myapp/files/shared/sound.mp3"

Ferdari · Feb 25, 2020

peacemaker said:
If you copied each needed file into "safefolder" - each file path will be "safefolder/filname", i.e. "shared" subfolder in the internal folder of your app

Ok, but how to the get the real URI:
file:///data/user/0/com.myapp/files/share/1/sound.mp3

peacemaker · Feb 25, 2020

Strange question... URI of the file in your app folders is just file's path

B4X:

Dim URI as String = "file:///" & safefolder & "/sound.mp3"

Ferdari · Feb 25, 2020

peacemaker said:
Strange question... URI of the file in your app folders is just "file:///" path to the file

B4X:

Dim URI as String = "file:///" & safefolder & "/sound.mp3"

Thanks i will try

Ferdari · Feb 25, 2020

I just found this function on my FileProvider class, if someone want to know the URI of a file use this function:

B4X:

'Returns the file uri.
Public Sub GetFileUri (FileName As String) As Object
   
    If UseFileProvider = False Then
        Dim uri As JavaObject
        Return uri.InitializeStatic("android.net.Uri").RunMethod("parse", Array("file://" & File.Combine(SharedFolder, FileName)))
    Else
        Dim f As JavaObject
        f.InitializeNewInstance("java.io.File", Array(SharedFolder, FileName))
        Dim fp As JavaObject
        Dim context As JavaObject
        context.InitializeContext
        fp.InitializeStatic("android.support.v4.content.FileProvider")
        Return fp.RunMethod("getUriForFile", Array(context, Application.PackageName & ".provider", f))
    End If
End Sub

it returns the uri like this:
GetFileUri(File.DirAssets & "/file.mp3")
content://com.app.provider/name/AssetsDir/file.mp3

Android Question (SOLVED)File path to URI path

Ferdari

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peacemaker

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Ferdari

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peacemaker

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Ferdari

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Ferdari

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