Android 如何获取和替换图像'；onTouch事件中的背景色
基本上有以下两个障碍：Android 如何获取和替换图像'；onTouch事件中的背景色,android,Android,基本上有以下两个障碍： 我想在单击光标的坐标处获取图像的背景色。代码第一次获取颜色和坐标。但当我再次单击时，会发生错误 我想用十六进制而不是RGB整数值获取图像的颜色，然后在找到颜色的特定区域用我的颜色更改该特定颜色 代码如下： private ImageView mImageView; private Button button; @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate
private ImageView mImageView;
private Button button;
@Override
protected void onCreate(Bundle savedInstanceState)
{
super.onCreate(savedInstanceState);
setContentView(R.layout.vp_view_img);
if(getIntent().hasExtra("BitmapImage"))
{
final Bitmap bitmap = (Bitmap) getIntent().getParcelableExtra("BitmapImage");
mImageView = (ImageView) findViewById(R.id.canvas_image);
mImageView.setImageBitmap(bitmap);
mImageView.setOnTouchListener(new ImageView.OnTouchListener()
{
@Override
public boolean onTouch(View v, MotionEvent event)
{
int x = (int)event.getX();
int y = (int)event.getY();
/*Toast.makeText(getBaseContext(), "Touch coordinates : "
+ String.valueOf(event.getX()) + "x"
+ String.valueOf(event.getY()),
Toast.LENGTH_SHORT).show();*/
ImageView imageView = (ImageView) v;
Bitmap bitmap = ((BitmapDrawable)imageView.getDrawable()).getBitmap();
int pixel = bitmap.getPixel(x,y);
int redValue = Color.red(pixel);
int blueValue = Color.blue(pixel);
int greenValue = Color.green(pixel);
Toast.makeText(getBaseContext(), "Color Code :"+
redValue+blueValue+greenValue,Toast.LENGTH_SHORT).show();
return true;
}
});
}
else
{
Toast.makeText(getBaseContext(),
"No Image Found",
Toast.LENGTH_LONG).show();
}
}
使用选择器怎么样 在res/drawable文件夹中，生成如下代码所示的xml文件
<?xml version="1.0" encoding="UTF-8"?>
<selector xmlns:android="http://schemas.android.com/apk/res/android">
<item android:state_selected="true" android:color="@color/white"/> <!-- selected -->
<item android:state_pressed="true" android:color="@color/white" /> <!-- pressed -->
<item android:color="@color/black" />
</selector>
让我们从第二个问题开始：为了将
颜色整数#RGB, #RRGGBB, #RRRGGGBBB, #RRRRGGGGBBBB
/**
* Converts the passed integer (color) into it's hexadecimal representation
* @param pixel the Color to convert
* @return a String: the <code>pixel</code> parameter in #RRGGBB format
*/
private String rgbToHex(int pixel)
{
int redValue = Color.red(pixel);
int blueValue = Color.blue(pixel);
int greenValue = Color.green(pixel);
return "#" + getBaseColorAsHex(redValue) +
getBaseColorAsHex(greenValue) +
getBaseColorAsHex(blueValue);
}
/**
* Returns the hex representation of the <code>baseColor</code>
* parameter, padding up to 2 characters if necessary.
* @param baseColor the color whose hex representation is needed
* @return the hex code for the passed parameter
*/
private String getBaseColorAsHex(int baseColor)
{
final String hex = Integer.toHexString(baseColor);
return (hex.length() == 1 ? '0' : "") + hex;
}
同样，你应该考虑把问题分成两个部分，因为在这个表单中它不适合堆栈溢出的指导方针！ 编辑：关于提出的第一个问题：一个严重的问题可能是，您正在分析的图像比屏幕小，并且被拉伸，因此当您点击屏幕并检索
xy除非您使用
try catch