android - 以编程方式保存照片联系人 - Android

Question

我实际上是这样做的:

    InputStream inputStream = ContactsContract.Contacts.openContactPhotoInputStream(context.getContentResolver(),
            ContentUris.withAppendedId(ContactsContract.Contacts.CONTENT_URI, Long.valueOf(contactId)));

    if (inputStream != null) {
        photo = BitmapFactory.decodeStream(inputStream);
    }else{
        // To open up a gallery browser
        Intent intent = new Intent();
        intent.setType("image/*");
        intent.setAction(Intent.ACTION_GET_CONTENT);
        startActivityForResult(Intent.createChooser(intent, "Seleccione una imagen para el contacto"),1);
    }
    return photo;

使用该代码，我可以通过电话号码检索联系人照片，当联系人没有照片时，我需要按 Intent 从图库中保存选定的照片:

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

    if (resultCode == RESULT_OK) {

        if (requestCode == 1) {
            currImageURI = data.getData();
            try {
                Bitmap bitmap = MediaStore.Images.Media.getBitmap(this.getContentResolver(), currImageURI);
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    }
}

有人知道如何在联系人中保存此位图吗？谢谢!

Answer 1

使用这段代码:

ByteArrayOutputStream stream = new ByteArrayOutputStream();
            if(mBitmap!=null){    // If an image is selected successfully
                mBitmap.compress(Bitmap.CompressFormat.PNG , 75, stream);

                // Adding insert operation to operations list
                // to insert Photo in the table ContactsContract.Data
                ops.add(ContentProviderOperation.newInsert(ContactsContract.Data.CONTENT_URI)
                        .withValueBackReference(ContactsContract.Data.RAW_CONTACT_ID, rawContactID)
                        .withValue(ContactsContract.Data.IS_SUPER_PRIMARY, 1)
                        .withValue(ContactsContract.Data.MIMETYPE,Photo.CONTENT_ITEM_TYPE)
                        .withValue(ContactsContract.CommonDataKinds.Photo.PHOTO,stream.toByteArray())
                        .build());

                try {
                    stream.flush();
                }catch (IOException e) {
                    e.printStackTrace();
                }
            }