如何第二次正确保存我的位图? [英] How can I save my bitmap correctly for second time?

问题描述

我想将位图保存到缓存目录. 我使用以下代码:

I want to save my bitmap to cache directory. I use this code:

try {
        File file_d = new File(dir+"screenshot.jpg");
        @SuppressWarnings("unused")
        boolean deleted = file_d.delete();
    } catch (Exception e) {
        // TODO: handle exception
    }

    imagePath = new File(dir+"screenshot.jpg");
    FileOutputStream fos;
    try {
        fos = new FileOutputStream(imagePath);
        bitmap.compress(CompressFormat.JPEG, 100, fos);
        fos.flush();
        fos.close();
    } catch (FileNotFoundException e) {
        Log.e("GREC", e.getMessage(), e);
    } catch (IOException e) {
        Log.e("GREC", e.getMessage(), e);
    }
}

工作正常.但是，如果我想将不同的img保存到同一路径，则会出问题.我的意思是将其保存到相同的路径，但是看到的是旧图像，但是当我单击该图像时，可以看到我第二次保存的正确图像.

it s working fine. But if I want to save different img to same path, something goes wrong. I mean it is saved to same path but I see it old image, but when I click the image I can see the correct image which I saved second time.

也许它来自缓存，但是我不想看到旧图像，因为当我想与所看到的whatsapp旧图像共享该图像时，如果我发送该图像似乎是正确的.

Maybe its come from cache but I do not want to see old image because when I want to share that image with whatsapp old image seen , if i send the image it seems correct.

我想像这样的代码在whatsapp上共享保存的图像:

I want to share saved image on whatsapp like this code:

Intent shareIntent = new Intent();
shareIntent.setAction(Intent.ACTION_SEND);
shareIntent.putExtra(Intent.EXTRA_STREAM, Uri.fromFile(imagePath));
shareIntent.setType("image/jpeg");
startActivityForResult(Intent.createChooser(shareIntent, getResources().getText(R.string.title_share)),whtsapp_result);

我该如何解决?

提前谢谢.

推荐答案

最后，我这样解决了我的问题:

Finally I solved my problem like this:

Intent shareIntent = new Intent();
shareIntent.setAction(Intent.ACTION_SEND);
shareIntent.putExtra(Intent.EXTRA_STREAM, getImageUri(context,bitmap_));
shareIntent.setType("image/jpeg");
startActivityForResult(Intent.createChooser(shareIntent, getResources().getText(R.string.title_share)),whtsapp_result);

v

public Uri getImageUri(Context inContext, Bitmap inImage) {
          ByteArrayOutputStream bytes = new ByteArrayOutputStream();
          inImage.compress(Bitmap.CompressFormat.JPEG, 100, bytes);
          String path = Images.Media.insertImage(inContext.getContentResolver(), inImage, "TitleC1", null);
          return Uri.parse(path);
        }

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