比较android中的两个drawable

Question

如何比较两个可绘制对象，我正在这样做但没有任何成功

public void MyClick(View view)
{
 Drawable fDraw = view.getBackground();
 Drawable sDraw = getResources().getDrawable(R.drawable.twt_hover);

  if(fDraw.equals(sDraw))
  {
   //Not coming
  }
}

Answer 1

更新 https://stackoverflow.com/a/36373569/1835650

getConstantState() 效果不佳

还有一种比较方式：

mRememberPwd.getDrawable().getConstantState().equals
            (getResources().getDrawable(R.drawable.login_checked).getConstantState());

mRemeberPwd 在此示例中是 ImageView。如果您使用的是TextView，请改用getBackground().getConstantState。

Answer 2

仅依靠getConstantState() 可以产生false negatives。

我采取的方法是在第一个实例中尝试比较 ConstantState，但如果检查失败则回退到位图比较。

这应该适用于所有情况（包括不是资源的图像），但请注意它会占用大量内存。

public static boolean areDrawablesIdentical(Drawable drawableA, Drawable drawableB) {
    Drawable.ConstantState stateA = drawableA.getConstantState();
    Drawable.ConstantState stateB = drawableB.getConstantState();
    // If the constant state is identical, they are using the same drawable resource.
    // However, the opposite is not necessarily true.
    return (stateA != null && stateB != null && stateA.equals(stateB))
            || getBitmap(drawableA).sameAs(getBitmap(drawableB));
}

public static Bitmap getBitmap(Drawable drawable) {
    Bitmap result;
    if (drawable instanceof BitmapDrawable) {
        result = ((BitmapDrawable) drawable).getBitmap();
    } else {
        int width = drawable.getIntrinsicWidth();
        int height = drawable.getIntrinsicHeight();
        // Some drawables have no intrinsic width - e.g. solid colours.
        if (width <= 0) {
            width = 1;
        }
        if (height <= 0) {
            height = 1;
        }

        result = Bitmap.createBitmap(width, height, Bitmap.Config.ARGB_8888);
        Canvas canvas = new Canvas(result);
        drawable.setBounds(0, 0, canvas.getWidth(), canvas.getHeight());
        drawable.draw(canvas);
    }
    return result;
}

Answer 3

我的问题是仅比较两个可绘制对象，我尝试但无法获得任何直接比较两个可绘制对象的方法，但是对于我的解决方案，我将可绘制对象更改为位图，然后比较两个位图，这是可行的。

Bitmap bitmap = ((BitmapDrawable)fDraw).getBitmap();
Bitmap bitmap2 = ((BitmapDrawable)sDraw).getBitmap();

if(bitmap == bitmap2)
    {
        //Code blcok
    }

Answer 4

适用于 SDK 21+

这适用于 SDK -21

mRememberPwd.getDrawable().getConstantState().equals
        (getResources().getDrawable(R.drawable.login_checked).getConstantState())

适用于 SDK +21 android 5。 使用标签将drawable id设置为imageview

img.setTag(R.drawable.xxx);

然后像这样比较

if ((Integer) img.getTag() == R.drawable.xxx)
{
....your code
}

此解决方案适用于想要将imageview 的drawable id 与drawable.xxx 的id 进行比较的人。

Answer 5

或许可以这样试试：

public void MyClick(View view)
{
 Drawable fDraw = view.getBackground();
 Drawable sDraw = getResources().getDrawable(R.drawable.twt_hover);

  if(fDraw.hashCode() == sDraw.hashCode())
  {
   //Not coming
  }
}

或者准备一个方法，它接受两个可绘制参数并返回布尔值。在该方法中，您可以将drawable转换为字节并进行比较，

public boolean compareDrawable(Drawable d1, Drawable d2){
    try{
        Bitmap bitmap1 = ((BitmapDrawable)d1).getBitmap();
        ByteArrayOutputStream stream1 = new ByteArrayOutputStream();
        bitmap1.compress(Bitmap.CompressFormat.JPEG, 100, stream1);
        stream1.flush();
        byte[] bitmapdata1 = stream1.toByteArray();
        stream1.close();

        Bitmap bitmap2 = ((BitmapDrawable)d2).getBitmap();
        ByteArrayOutputStream stream2 = new ByteArrayOutputStream();
        bitmap2.compress(Bitmap.CompressFormat.JPEG, 100, stream2);
        stream2.flush();
        byte[] bitmapdata2 = stream2.toByteArray();
        stream2.close();

        return bitmapdata1.equals(bitmapdata2);
    }
    catch (Exception e) {
        // TODO: handle exception
    }
    return false;
}

Answer 6

Android 5 的解决方案：

 if(image.getDrawable().getConstantState().equals(image.getContext().getDrawable(R.drawable.something).getConstantState()))

Answer 7

getDrawable(int) 现已弃用。使用 getDrawable(context,R.drawable.yourimageid)

比较两个背景

Boolean Condition1=v.getBackground().getConstantState().equals(
ContextCompat.getDrawable(getApplicationContext(),R.drawable.***).getConstantState());

Answer 8

好的，我想我已经找到了最终解决方案。由于 AppCompat 和朋友的原因，提供的 drawable 有时会以不同的形式膨胀，所以做getResources().getBitmap(R.drawable.my_awesome_drawable) 是不够的。

因此，为了获得与视图提供的相同类型和形式的可绘制实例，可以这样做：

public static Drawable drawableFrom(View view, @DrawableRes int drawableId) {
    Context context = view.getContext();
    try {
        View dummyView = view.getClass().getConstructor(Context.class).newInstance(context);
        dummyView.setBackgroundResource(drawableId);
        return dummyView.getBackground();
    } catch (Exception e) {
      return ResourcesCompat.getDrawable(context.getResources(), drawableId, null);
    }
}

这在进行测试时很有用。但是，我不建议在生产中这样做。如果需要，需要额外的缓存以避免进行过多的反射。

对于 Expresso 测试，您可以很好地使用它：

onView(withDrawable(R.drawable.awesome_drawable))
  .check(matches(isDisplayed()));

或

onView(withId(R.id.view_id))
  .check(matches(withDrawable(R.drawable.awesome_drawable)));

在你必须声明这个助手类之前：

public class CustomMatchers {

  public static Matcher<View> withDrawable(@DrawableRes final int drawableId) {
     return new DrawableViewMatcher(drawableId);
  }
  private static class DrawableViewMatcher extends TypeSafeMatcher<View> {

     private final int expectedId;
     private String resourceName;

     private enum DrawableExtractionPolicy {
        IMAGE_VIEW {
          @Override
          Drawable findDrawable(View view) {
             return view instanceof ImageView ? ((ImageView) view).getDrawable() : null;
          }
        },
        TEXT_VIEW_COMPOUND {
          @Override
          Drawable findDrawable(View view) {
             return view instanceof TextView ? findFirstCompoundDrawable((TextView) view) : null;
          }
        },
        BACKGROUND {
          @Override
          Drawable findDrawable(View view) {
             return view.getBackground();
          }
        };

        @Nullable
        private static Drawable findFirstCompoundDrawable(TextView view) {
          for (Drawable drawable : view.getCompoundDrawables()) {
             if (drawable != null) {
                return drawable;
             }
          }
          return null;
        }

        abstract Drawable findDrawable(View view);

     }

     private DrawableViewMatcher(@DrawableRes int expectedId) {
        this.expectedId = expectedId;
     }

     @Override
     protected boolean matchesSafely(View view) {
        resourceName = resources(view).getResourceName(expectedId);
        return haveSameState(actualDrawable(view), expectedDrawable(view));
     }

     private boolean haveSameState(Drawable actual, Drawable expected) {
        return actual != null && expected != null && areEqual(expected.getConstantState(), actual.getConstantState());
     }

     private Drawable actualDrawable(View view) {
        for (DrawableExtractionPolicy policy : DrawableExtractionPolicy.values()) {
          Drawable drawable = policy.findDrawable(view);
          if (drawable != null) {
             return drawable;
          }
        }
        return null;
     }

     private boolean areEqual(Object first, Object second) {
        return first == null ? second == null : first.equals(second);
     }

     private Drawable expectedDrawable(View view) {
        return drawableFrom(view, expectedId);
     }

     private static Drawable drawableFrom(View view, @DrawableRes int drawableId) {
        Context context = view.getContext();
        try {
          View dummyView = view.getClass().getConstructor(Context.class).newInstance(context);
          dummyView.setBackgroundResource(drawableId);
          return dummyView.getBackground();
        } catch (Exception e) {
          return ResourcesCompat.getDrawable(context.getResources(), drawableId, null);
        }
     }

     @NonNull
     private Resources resources(View view) {
        return view.getContext().getResources();
     }

     @Override
     public void describeTo(Description description) {
        description.appendText("with drawable from resource id: ");
        description.appendValue(expectedId);
        if (resourceName != null) {
          description.appendValueList("[", "", "]", resourceName);
        }
     }
  }

}

Answer 9

比较 2 个可绘制对象：

drawable1.constantState == drawable2.constantState
            || drawable1.toBitmap().sameAs(drawable2.toBitmap())

如果你找不到Drawable.toBitmap(...)这里是Drawable.kt

Answer 10

使用 getTag() 和 setTag() 进行比较

Answer 11

我已经在这里回答了类似的话题：Get the ID of a drawable in ImageView。 该方法基于在自定义 LayoutInflater 中使用指定资源 id 标记视图。整个过程由一个简单的库TagView自动化。

因此，您可以仅通过 id 比较两个可绘制对象：

TagViewUtils.getTag(view, ViewTag.VIEW_BACKGROUND.id) == R.drawable.twt_hover

Answer 12

扩展来自@vaughandroid 的答案，以下 Matcher 适用于着色的 Vector Drawable。您必须提供用于 Drawable 的色调。

public static Matcher<View> compareVectorDrawables(final int imageId, final int tintId) {
        return new TypeSafeMatcher<View>() {

        @Override
        protected boolean matchesSafely(View target) {
            if (!(target instanceof ImageView)) {
                return false;
            }
            ImageView imageView = (ImageView) target;
            if (imageId < 0) {
                return imageView.getDrawable() == null;
            }
            Resources resources = target.getContext().getResources();
            Drawable expectedDrawable = resources.getDrawable(imageId, null);
            if (expectedDrawable == null) {
                return false;
            }

            Drawable imageDrawable = imageView.getDrawable();
            ColorFilter imageColorFilter = imageDrawable.getColorFilter();

            expectedDrawable.setColorFilter(imageColorFilter);
            expectedDrawable.setTintList(target.getResources()
                    .getColorStateList(tintId, null));

            boolean areSame = areDrawablesIdentical(imageDrawable, expectedDrawable);
            return areSame;
        }

        public boolean areDrawablesIdentical(Drawable drawableA, Drawable drawableB) {
            Drawable.ConstantState stateA = drawableA.getConstantState();
            Drawable.ConstantState stateB = drawableB.getConstantState();
            // If the constant state is identical, they are using the same drawable resource.
            // However, the opposite is not necessarily true.
            return (stateA != null && stateB != null && stateA.equals(stateB))
                    || getBitmap(drawableA).sameAs(getBitmap(drawableB));
        }

        public Bitmap getBitmap(Drawable drawable) {
            Bitmap result;
            if (drawable instanceof BitmapDrawable) {
                result = ((BitmapDrawable) drawable).getBitmap();
            } else {
                int width = drawable.getIntrinsicWidth();
                int height = drawable.getIntrinsicHeight();
                // Some drawables have no intrinsic width - e.g. solid colours.
                if (width <= 0) {
                    width = 1;
                }
                if (height <= 0) {
                    height = 1;
                }

                result = Bitmap.createBitmap(width, height, Bitmap.Config.ARGB_8888);
                Canvas canvas = new Canvas(result);
                drawable.setBounds(0, 0, canvas.getWidth(), canvas.getHeight());
                drawable.draw(canvas);
            }
            return result;
        }

        @Override
        public void describeTo(Description description) {

        }
    };
}

Answer 13

如果您想直接比较两个可绘制对象，请使用以下代码

可绘制 fDraw = getResources().getDrawable(R.drawable.twt_hover);

可绘制 sDraw = getResources().getDrawable(R.drawable.twt_hover);

if (fDraw.getConstantState().equals(sDraw.getConstantState())) {
    //write your code.
} else {
    //write your code.
}

Answer 14

当您使用equals() 方法时，它用于比较内容。你应该尝试== 来比较两个对象。

public void MyClick(View view)
{
 Drawable fDraw = view.getBackground();
 Drawable sDraw = getResources().getDrawable(R.drawable.twt_hover);

  if( fDraw == sDraw )
  {
   // Coming
  }
}