Question

我想使用意图将我的云存储中的（zip）文件下载到我的应用程序。我成功地访问了我的云：

        var activity = (Activity)Xamarin.Forms.Forms.Context;

        // ACTION_OPEN_DOCUMENT is the intent to choose a file via the system's file
        // browser.
        Intent intent = new Intent(Intent.ActionGetContent);

        // Filter to only show results that can be "opened", such as a
        // file (as opposed to a list of contacts or timezones)
        intent.AddCategory(Intent.CategoryOpenable);

        // Filter to show only images, using the image MIME data type.
        // If one wanted to search for ogg vorbis files, the type would be "audio/ogg".
        // To search for all documents available via installed storage providers,
        // it would be "*/*".
        intent.SetType("*/*");

        activity.StartActivityForResult(intent, 0);

我的Mainactivity.cs中有这样的OnActivityResult：

    protected override void OnActivityResult(int requestCode, Result resultCode, Intent data)
    {

        base.OnActivityResult(requestCode, resultCode, data);
        if (requestCode == 0)
        {

            System.Uri uri = new System.Uri(data.Data.ToString());



            WebClient myWebClient = new WebClient();
            myWebClient.DownloadFile(uri, "/root"));
        }
    }

但我认为我的uri地址并不好，因为它是一个内容uri但不是文件uri。那么我怎么能从我的内容uri获得一个有效的文件uri？

Answer 1

使用：

解决

System.Uri uri = new System.Uri（data.Data.ToString（））;

            ContentResolver rc = ContentResolver;

            var stream = rc.OpenInputStream(data.Data);

            try
            {
                using (var fileStream = System.IO.File.Create(filepath))
                {
                    stream.CopyTo(fileStream);
                }
            }
            catch (Exception ex)
            {
                System.Diagnostics.Debug.WriteLine("exception :" + ex.Message);
            }

如何从具有内容URI的云中下载我的应用程序中的文件

1 个答案: