Create a new color drawable
JavaAndroidAndroid DrawableColordrawableJava Problem Overview
I am trying to convert a hex value to an int so I can create a new color drawable. I'm not sure if this is possible, but according to the documentation, it should. It plainly asks for
> public ColorDrawable (int color) > > Added in API level 1 Creates a new ColorDrawable with the specified > color. > > Parameters color The color to draw.
So, my code isn't working because I'm getting an Invalid int: "FF6666" error. Any ideas?
int decode = Integer.decode("FF6666");
ColorDrawable colorDrawable = new ColorDrawable(decode);
Java Solutions
Solution 1 - Java
Since you're talking about hex you have to start with 0x
and don't forget the opacity.
So basically: 0xFFFF6666
ColorDrawable cd = new ColorDrawable(0xFFFF6666);
You can also create a new colors.xml file into /res and define the colors like:
<?xml version="1.0" encoding="utf-8"?>
<resources>
<color name="mycolor">#FF6666</color>
</resources>
and simply get the color defined in R.color.mycolor
getResources().getColor(R.color.mycolor)
Solution 2 - Java
For using with ContextCompat and rehuse the color you can do something like this:
ColorDrawable colorDrawable = new ColorDrawable(ContextCompat.getColor(this, R.color.white));
Solution 3 - Java
It should be like this...
ColorDrawable cd = new ColorDrawable(0xffff6666);
Note I used 8 hex digits, not 6 hex digit . which add to transparency
Solution 4 - Java
By followingthe above advice,to be a summary of this question:
-
ColorDrawable colorDrawable = new ColorDrawable(Color.parseColor("#ce9b2c"));`
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ColorDrawable colorDrawable = new ColorDrawable(0xFFCE9B2C); Note there is 8 hex digits, not 6 hex digit,which no work. Case all
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ColorDrawable colorDrawable = new ColorDrawable(ContextCompat.getColor(mContext,R.color.default_color));
Selecting up to you!
Solution 5 - Java
I think you have to use :
> public static int parseColor (String colorString) > > Added in API level 1 Parse the color string, and return the > corresponding color-int. If the string cannot be parsed, throws an > IllegalArgumentException exception. Supported formats are: #RRGGBB #AARRGGBB red, blue, green, black, white, gray, cyan, magenta, yellow, lightgray, darkgray, grey, lightgrey, darkgrey, aqua, fuschia, lime, > maroon, navy, olive, purple, silver, teal
Solution 6 - Java
> This is how I converted a Hex color to int and applied to a Background > of a View
Let's say that we have a color #8080000.
1) Hex to int conversion
int myColor = Color.parseColor("#808000");
###2) Set background
view.setBackgroundColor(context.getColor(myColor));