Not able to call Super

Question

When I invoke this.Super("initWidget") in derived widget, the browser is giving an exception

**TypeError: this.Super is not a function**

Can anyone help me solve this issue?

Insufficient information. Please provide some sample code. – RAS Dec 19 '12 at 06:36 — RAS, Dec 19 '12 at 06:36

score 0 · Answer 1 · answered Dec 18 '12 at 12:51

0

this and super are keywords. this refers to the current object, while super refers to the parent class. If you try to invoke super, you are invoking the super class' constructor. In your case with one parameter. You should use only

super("initWidget");

because the dot operator after the this means you want to use a method in the current object.

answered Dec 18 '12 at 12:51

mikereem

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he is not using keyword super but Super() – Bhavik Shah Dec 19 '12 at 06:51
I assumed he want to call the superclass constructor with super, because he didn't provided any sample code, but he talked about derived widget. – mikereem Dec 19 '12 at 07:43

Not able to call Super

1 Answers1