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Re: matching interface builder object to source
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Re: matching interface builder object to source


  • Subject: Re: matching interface builder object to source
  • From: Amul Goswamy <email@hidden>
  • Date: Wed, 21 Jan 2004 10:43:26 -0800

You can get a pointer to the NSClipView by calling superview on NSTextView.
NSClipView * clipView = [textView superview];

More details on superview are here:
http://developer.apple.com/documentation/Cocoa/Reference/ ApplicationKit/ObjC_classic/Classes/NSView.html

Amul Goswamy
Interface Builder
Apple

On Jan 21, 2004, at 10:14 AM, Matthew Cook wrote:

I've just recently moved into Cocoa after working in java Swing for a while, and it's a bit of an adjustment, but I really like this ease of laying out interface components with Interface Builder. I've been through a few cocoa/objective-c books but I'm a little fuzzy on how best to access user interface objects programatically.

For example, in one interface I have a large NSTextView that will act something like a scrolling console with messages added to the bottom from time to time. I've had no issue setting up an IBOutlet object in my controller class and correctly displaying the strings, but I need to access the NSScrollView (more specifically the NSClipView) of the NSTextView so that I can tell it to scroll down when new lines are added to the console.

If I just put an instance of an NSScrollView in my controller class, can I somehow reference the NSScrollView from the IBOutlet NSTextView object? Can I instantiate an NSScrollView in Interface Builder and somehow connect it to the NSScrollView that comes with the NSTextView? Should I just try and create, place and draw the whole NSScrollView and TextView programatically? Should I create a spicific NSScrollView in Interface Builder and place a NSTextView inside it, or will this make an NSTextView in an NSScrollView in an other NSScrollView?

Any clarification would be appreciated. Thanks in advance.
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References: 
 >matching interface builder object to source (From: Matthew Cook <email@hidden>)

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