01-30-2007 08:52 AM
01-30-2007 09:07 AM
From the FP pop-up on the ring (just the ring!) and select create property node.
Ben
01-30-2007 09:16 AM
And for the more generic solution, you can get a reference to the controls inside a cluster by using the cluster's Controls[] property.
The order of the references is the tabbing order and you can use the To More Specific Class function to cast the reference to your target class.
01-30-2007 09:57 AM
10-03-2011 10:57 AM
But is there a way to directly get ring item text for rings in an array without converting to cluster per below code:
10-03-2011 11:10 AM - edited 10-03-2011 11:13 AM
@GerdW wrote:
And you can check the Label.Text property to be sure you got the correct item in the cluster!
But you do have to be careful with this. If the label changes, the code no longer finds the reference. I prefer Ben's method (not the one listed above
) of bundling all references into a cluster in an action engine, or a gui manager. I then create a new action for each different gui management action I want to take. The looping through references is definitely a quicker solution, but I like having a cluster with the explicit names of the references, to force any other developers (or myself) to explicitly unbundle the reference they want to use.
10-03-2011 11:24 AM
Note that when I created the VI snippet, the reference name changed from 'cluster' to 'cluster 2'. VI snippet must not have been able to determine that the reference was from the cluster.
10-04-2011 06:41 PM
If you right click on your array and click Create»Property Node»Ring Text»Text then you shouldn't have convert to a cluster. If you read a string from this property node, it will give you string value of whatever array index is selected.
10-05-2011 11:01 AM - edited 10-05-2011 11:01 AM
If you right-click on the array ELEMENT, you can create a property node to get ring text. However, you can create a similar property node for the ARRAY of menu rings. As you can see below, there is no way to get ring text in this case.
10-06-2011 12:09 PM
Oh yeah -- sorry about that DonRoth. I understand better what you're trying to do now. I can see how you would prefer not to create a property node for each element in the array.