I have a lot of images (hundreds) and I need to put some of them as a resource of ImageView
, but if I would create If Then set Image to Image name, I die from these tons of code. I want to set Image resource which name is inside a variable, but I can't find out how. If I have
img.setImageResource(R.drawable.my_image);
How can I assign variable name at my_image
instead of true image name?
Please help.
Thanks.
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Phantômaxx
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Black Dragon
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Possible duplicate of [Android getResources().getDrawable() deprecated API 22](https://stackoverflow.com/questions/29041027/android-getresources-getdrawable-deprecated-api-22) – shizhen Apr 02 '19 at 15:10
4 Answers
1
Try this:
String resourceName = "clouds";
int resourceIdentifier = getResources().getIdentifier(resourceName, "drawable", Constants.CURRENT_CONTEXT.getPackageName());
Then
imgView.setImageResource(resourceIdentifier);

S-Sh
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I don't know if I clearly understood your request but why don't you try to do that instead :
int my_image = R.id.my_image;
img.setImageResource(ContextCompat.getDrawable(this, my_image));
https://stackoverflow.com/a/29146895/8526518 for more informations about ContextCompat

Adama Traore
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if you can prefer assets folder rather than resource, then you can try this code
create sub-folder in assets directory. use getAssets().list() for getting all file names from assets:
String[] images =getAssets().list("images");
ArrayList<String> listImages = new ArrayList<String>(Arrays.asList(images));
Now to set image in imageview you first need to get bitmap using image name from assets :
InputStream inputstream=mContext.getAssets().open("images/"
+listImages.get(position));
Drawable drawable = Drawable.createFromStream(inputstream, null);
imageView.setImageDrawable(drawable);

Amar Sharma
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just a convenient refactoring of the S-Sh answer if you use kotlin and like using its extensions functions:
fun ImageView.loadDrawableFromName(name:String, ctx: Context, visible:Int=View.VISIBLE){
val resId = ctx.resources.getIdentifier(name, "drawable", ctx.packageName)
this.setImageResource(resId)
this.visibility = visible
}
you can call it in both ways:
imgname="IMGNAME"
myimageview.loadDrawableFromName(imgname, ctx)
and
imgname="IMGNAME"
myimageview.loadDrawableFromName("@drawable/$imgname", ctx)

albaspazio
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