Change shape border color at runtime

删除回忆录丶 提交于 2019-12-03 04:27:06
rodalves

I had the same problem. In my case, I had a GridView, which the items in grid could have the border color changed by the user at runtime.

So, in the gridviewAdapter for that grid, I did the following in the getView method (the one that generates the view for the adapter)

public View getView(int position, View convertView, ViewGroup parent) {
    LayoutInflater inflater = (LayoutInflater) mContext.getSystemService(Context.LAYOUT_INFLATER_SERVICE);

    convertView = inflater.inflate(R.layout.griditem, null);
    GradientDrawable gradientDrawable = (GradientDrawable) convertView.getBackground(); 

    gradientDrawable.setStroke(2, mColor); 
    convertView.invalidate();
    return convertView;
}

mColor is a int that represents the color, much like we do in the xml files. In java code, instead of "#" we use "0x" to define it in the AARRGGBB format. For example, use 0xFF000000 for 100% opaque BLACK and 0xFF0000FF for 100% opaque BLUE. Explaining this here since the google api 'helpfully' tells that the int color is "the color of the stroke".

This solved my problem... I guess you can try something similar for your case.

Talha

Hi you can try create your background on runtime, then you can change it whenever you want.

RoundRectShape rect = new RoundRectShape(
  new float[] {30,30, 30,30, 30,30, 30,30},
  null,
  null);
ShapeDrawable bg = new ShapeDrawable(rect);
bg.getPaint().setColor(0x99FFFFFF);
view.setBackgroundDrawable(bg);
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!