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Mar 9, 2016 at 10:00 comment added user1181378 Yeah. Know how to use $this out of scope of the widget class without getting an error?
Mar 9, 2016 at 9:13 answer added Emin Özlem timeline score: 0
Mar 9, 2016 at 8:28 comment added user1181378 My widget is fine. I just grabbed the part of function form($instance) from it. I wanted to show that I'm trying to add a form input through a class. The issue is just with using $this (of the widget) outside it.
Mar 9, 2016 at 8:06 history asked user1181378 CC BY-SA 3.0