picking file for async-http file upload (uri to file path)

Question

I know my problem has been tackled multiple times already, but I couldn't find anything that would help me with my specific problem.

I have an intent to pick a file anywhere from the system:

private void showFileChooser() {
    Intent intent = new Intent(Intent.ACTION_GET_CONTENT);
    intent.setType("*/*");
    intent.addCategory(Intent.CATEGORY_OPENABLE);

    try {
        startActivityForResult(
                Intent.createChooser(intent, "Select a File to Upload"),
                FILE_SELECT_CODE);
    } catch (android.content.ActivityNotFoundException ex) {
        // Potentially direct the user to the Market with a Dialog
        Toast.makeText(this, "Please install a File Manager.",
                Toast.LENGTH_SHORT).show();
    }


}

I pick up this intent here:

@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {

    switch (requestCode) {
        case FILE_SELECT_CODE:

            if (resultCode == RESULT_OK) {
                // Get the Uri of the selected file
                Uri uri = data.getData();

}

and I have an upload method like so:

private void upload(File file){
    RequestParams params = new RequestParams();
    try {
        params.put("fileToUpload", file);
    } catch (FileNotFoundException e) {
        e.printStackTrace();
    }
    AsyncHttpClient client = new AsyncHttpClient();
    client.post("http://...", params, new AsyncHttpResponseHandler() {

        @Override
        public void onSuccess(int statusCode, Header[] headers, byte[] responseBody) {
            System.out.println("statusCode "+statusCode);//statusCode 200
        }

        @Override
        public void onFailure(int statusCode, Header[] headers, byte[] responseBody, Throwable error) {

        }
    });
}

the problem is that I don't know how to "marry" the two methods onActivity and upload because I don't know how to process the information I get from the intent so that the AsyncHttpClient can use it.

I tried converting the Uri to an absolute path, but couldn't manage (the solutions online that seemed to work where specifically for images). I also can't "convert" the uri to a file either.

Any way I could do this? Am I missing something?

CommonsWare · Accepted Answer · 2017-01-24 21:09:37Z

4

The ideal solution is to use ContentResolver and openInputStream() to get an InputStream on the content identified by the Uri, then pass that InputStream to your HTTP client API for use in uploading.

If your HTTP client API does not support the use of InputStream or something that could be derived from one (e.g., Reader), and so you need a File, use that InputStream and a FileOutputStream on some file that you control to make a local copy of the content (e.g., in getCacheDir()). Then, use your local copy for the file upload operation, and delete the local copy when you are done.

answered Jan 24, 2017 at 21:09

CommonsWare

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1 Comment

hachel Over a year ago

Yes, thank you, Async-Http allows for an input stream instead of a File.

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