如何使用 Intent.ACTION_CREATE_DOCUMENT 编写文件答案

【问题标题】：How to write a file using Intent.ACTION_CREATE_DOCUMENT如何使用 Intent.ACTION_CREATE_DOCUMENT 编写文件
【发布时间】：2019-09-30 10:32:06
【问题描述】：

简单来说，我想将视图组转换为 jpg 图像文件。由于Environment.getExternalStorageDirectory 已弃用，我使用此意图Intent.ACTION_CREATE_DOCUMENT

private void createFile(String mimeType, String fileName) {
        Intent intent = new Intent(Intent.ACTION_CREATE_DOCUMENT);
        intent.addCategory(Intent.CATEGORY_OPENABLE);
        intent.setType(mimeType);
        intent.putExtra(Intent.EXTRA_TITLE, fileName);
        startActivityForResult(intent, WRITE_REQUEST_CODE);
    }

在onActivityResult(); 中，我得到了结果返回的 Uri。 我的问题是getExternalStorage() 我会使用

Bitmap bitmap = Bitmap.createBitmap(
                    containerLayout.getWidth(),
                    containerLayout.getHeight(), Bitmap.Config.ARGB_8888);
            Canvas canvas = new Canvas(bitmap);
            containerLayout.draw(canvas);
            FileOutputStream fileOutupStream = null;


            try {
                fileOutupStream = new FileOutputStream(fileName);
                bitmap.compress(Bitmap.CompressFormat.JPEG, 100, fileOutupStream);
                fileOutupStream.flush();
                fileOutupStream.close();
                Toast.makeText(this, "saved " + fileName, Toast.LENGTH_LONG).show();
            } catch (Exception e) {
                Toast.makeText(this, "something went wrong" + e.getMessage(), Toast.LENGTH_SHORT).show();
                e.printStackTrace();
            }

现在我得到了结果返回的 Uri，但是我不知道如何将所需的位图写入这个 Uri

@Override
        protected void onActivityResult(int requestCode, int resultCode, @Nullable Intent data) {
            if (resultCode == RESULT_OK && requestCode == WRITE_REQUEST_CODE) {
                Uri resultUri = data.getData();
//need help

}
}

【问题讨论】：

标签： android android-intent bitmap fileoutputstream

【解决方案1】：

你需要使用getContentResolver().openOutputStream

        @Override
        protected void onActivityResult(int requestCode, int resultCode, @Nullable Intent data) {
            if (resultCode == RESULT_OK && requestCode == WRITE_REQUEST_CODE) {
                FileOutputStream fileOutupStream = getContentResolver().openOutputStream(data.getData());
            try {
                bitmap.compress(Bitmap.CompressFormat.JPEG, 100, fileOutupStream);
                fileOutupStream.flush();
                fileOutupStream.close();
                Toast.makeText(this, "saved " + fileName, Toast.LENGTH_LONG).show();
            } catch (Exception e) {
                Toast.makeText(this, "something went wrong" + e.getMessage(), Toast.LENGTH_SHORT).show();
                e.printStackTrace();
            }

}
}

【讨论】：

在第 4 行分配 fileOutupStream 后，为什么要在第 6 行重新创建它？
我认为这是复制粘贴错误，我只是将其从问题中的代码中保留下来
getContentResolver().openOutputStream(data.getData()) 这将创建一个 OutputStream，而不是 FileOutputStream

【解决方案2】：

这是在旧 API（API 21 之前）上使用 Intent.ACTION_CREATE_DOCUMENT 将文件从一个位置复制到另一个位置的最小示例。如有必要，您应该添加请求代码和结果代码处理。

    ...
    // Intent creation
    int REQUEST_CODE_ARBITRARY = 1;
    Intent intent = new Intent(Intent.ACTION_CREATE_DOCUMENT);
    // will trigger exception if no  appropriate category passed
    intent.addCategory(Intent.CATEGORY_OPENABLE);
    // or whatever mimeType you want
    intent.setType("*text/plain");
    intent.putExtra(Intent.EXTRA_TITLE, "file_name_to_save_as");
    startActivityForResult(intent, REQUEST_CODE_ARBITRARY);

    // Handling result
    @Override
    protected void onActivityResult(int requestCode, int resultCode, Intent data) {
        super.onActivityResult(requestCode, resultCode, data);

        // Example file to copy
        File sourceFile = getDatabasePath("MyDatabaseName");

        InputStream is = null;
        OutputStream os = null;

        // Note: you may use try-with resources if your API is 19+
        try {
            // InputStream constructor takes File, String (path), or FileDescriptor
            is = new FileInputStream(sourceFile);
            // data.getData() holds the URI of the path selected by the picker
            os = getContentResolver().openOutputStream(data.getData());

            byte[] buffer = new byte[1024];
            int length;
            while ((length = is.read(buffer)) > 0) {
                os.write(buffer, 0, length);
            }
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                is.close();
                os.close();
            } catch (IOException e) {
                //
            }
        }
    }

API 19+ 上更短的 try-catch：

            try (InputStream is = new FileInputStream(sourceFile); OutputStream os = getContentResolver().openOutputStream(data.getData())) {
                byte[] buffer = new byte[1024];
                int length;
                while ((length = is.read(buffer)) > 0) {
                    os.write(buffer, 0, length);
                }
            } catch (IOException e) {
                e.printStackTrace();
            }

【讨论】：