在 Android 中保存位图

发布于 2024-12-18 05:48:38 字数 320 浏览 6 评论 0原文

我正在尝试为 Android 实现一个隐写术项目。我已经操纵了像素值并创建了一个新的位图。现在，当我使用 getPixels() 将位图存储到手机内存或存储卡中

//fo denotes File output Stream
Bitmap.compress(Bitmap.CompressFormat.JPEG,100,fo);
//OR
Bitmap.compress(Bitmap.CompressFormat.PNG,100,fo);

并尝试访问像素时；

这些值将恢复为原始位图，即而不是操作后的位图。有人能弄清楚这是为什么吗？

原文

I am trying to implement a Steganography project for Android. I have manipulated the pixel values and created a new bitmap. Now when i store the bitmap into the phone memory or memory card using

//fo denotes File output Stream
Bitmap.compress(Bitmap.CompressFormat.JPEG,100,fo);
//OR
Bitmap.compress(Bitmap.CompressFormat.PNG,100,fo);

and try accessing the pixels back using getPixels();

the values are reverted back to the original bitmap i.e. rather than the manipulated bitmap.
Can anybody figure why this is?

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嗫嚅 2024-12-25 05:48:38

File myDir=new File("/sdcard/saved_images");
myDir.mkdirs();
Random generator = new Random();
int n = 10000;
n = generator.nextInt(n);
String fname = "Image-"+ n +".jpg";
File file = new File (myDir, fname);
if (file.exists ()) file.delete (); 
try {
       FileOutputStream out = new FileOutputStream(file);
       finalBitmap.compress(Bitmap.CompressFormat.JPEG, 90, out);
       out.flush();
       out.close();

} catch (Exception e) {
       e.printStackTrace();
}

看看这个答案 https://stackoverflow.com/a/7887114/964741

File myDir=new File("/sdcard/saved_images");
myDir.mkdirs();
Random generator = new Random();
int n = 10000;
n = generator.nextInt(n);
String fname = "Image-"+ n +".jpg";
File file = new File (myDir, fname);
if (file.exists ()) file.delete (); 
try {
       FileOutputStream out = new FileOutputStream(file);
       finalBitmap.compress(Bitmap.CompressFormat.JPEG, 90, out);
       out.flush();
       out.close();

} catch (Exception e) {
       e.printStackTrace();
}

look at this answer https://stackoverflow.com/a/7887114/964741

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