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Is there a way to simulate the click event of a custom UITableViewCell from the UITableViewController class?
I have a custom TableViewCell that has an UIImageView as its accessory. I need to be able to simulate a click on the imageView from the table view controller.
A:
There are lots of ways of doing that but the way you look at it is that as the table view cell is not the UIViewController, it will not have any reference to the source of the selector.
You could simply store in the cell a reference to a class or object which will call the selector but that sounds a little bit strange.
I suspect you might not get the “clicking” response you want using what you have tried so far.
Perhaps you could add the click event to the UIViewController using something like this:
[self.tableView.superview addGestureRecognizer:recognizer];
The superview will point to the cell. It might be more sensible though to add the gesture recognizer to the cell and use the UIViewController as the delegate.
Also it seems like it would make sense to respond to the click in the cell by updating the view model of the cell. You might then be able to respond to it from the UI to say “I’ve been clicked” or something like that.
Chelsea’s Cole Ploughs His Own Way To Fame
As the popularity of Rugby grows in the UK, we see rising stars of Rugby League emerging across the nation. Today I will be talking about one of our premier players, Liam Cole, the fullback for the Championship-winning London Broncos!
Liam Cole has represented St. Albans, Hertfordshire at football since he was eleven years old and has been turning heads with a very promising career. He has already been capped by the England U17 squad and has been training in the NRL for Australia over the summer break. Liam Cole
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What does it mean in physics when one speaks of diverging geometric or other “coordinates”?
I’m reading a paper on the non-relativistic limit of the radial Schrödinger equation, ( and I’m trying to understand a piece of mathematics.
We have a parameter $a$ such that $a \rightarrow 0$ and we have the equation
$$\frac{d^2 R}{dr^2} + \left( \frac{l(l+1)}{r^2} – \frac{2M\alpha}{r} \right) R = 0$$
on some interval of $r$ such that $l > 0$, and $M
e 0$. The author makes the following move:
In the limit $\alpha \rightarrow 0$, we may let $R = r^l f(\theta)$, where $\theta$ is the angle between the vector from the origin to
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