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Re: Funky behavior...
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Re: Funky behavior...


  • Subject: Re: Funky behavior...
  • From: Chuck Hill <email@hidden>
  • Date: Tue, 12 Aug 2003 14:27:02 -0700
  • Organization: Global Village Consulting, Inc.

The object bound to the item of a WORepetition will be set to null after the last item in the repetition is processed.

If your submit button is below the list then I would expect the exception you are getting. If you have one submit button per guest (with a multiple submit form) then it should be OK.

What are you trying to do, just select an item from a list? If so, a WOHyperlink might be more appropriate.

>>((Application)application()).addGuest(currentGuest);
Setting page/session level stuff into the application seems like a very bad idea. You are going to end up with a list of selections from different users which are in different editing contexts.



Chuck

Art Isbell wrote:
On Tuesday, August 12, 2003, at 07:35  AM, THOMAS PETERS wrote:

I said NullPointerException but am really getting java.lang.IllegalArgumentException: Attempt to insert null. 'currentGuest' is a Key on the Main component and is assigned to the WORepetition.item.


// here is the constructor for my Main WOComponent public Main(WOContext context) { super(context); currentGuest = new Guest(); }

// here is the action method for the submit button in my form.
   public WOComponent submitGuestForm() {
       ((Application)application()).addGuest(currentGuest);
       currentGuest = new Guest();
       return null;
   }


If currentGuest is bound to your repetition's "item" key, then I don't understand why your code is assigning a new Guest object to it. WORepetition will set the currentGuest value as it iterates through the objects in the object bound to its "list" key. But this isn't likely related to the exception you're experiencing. Maybe if the object bound to "list" is an empty array, currentGuest might not be assigned, so maybe assigning a new Guest object to it would prevent a null currentGuest from being added in the addGuest() method. But would merely testing for currentGuest being null in addGuest() be a better approach? If it's null at that point, you could create a new Guest object if that's what you need.

java.lang.IllegalArgumentException: Attempt to insert null into an com.webobjects.foundation.NSMutableArray.
[2003-08-12 13:30:08 EDT] <WorkerThread2> java.lang.IllegalArgumentException: Attempt to insert null into an com.webobjects.foundation.NSMutableArray.
at com.webobjects.foundation.NSMutableArray.addObject(NSMutableArray.java: 211)
at Application.addGuest(Application.java:28)


It appears that currentGuest is null at line 28 in addGuest(). You could avoid adding a null object by testing for currentGuest being null prior to adding it to the array. Or if currentGuest should never be null, you would need to debug your app to find out why it is null.

Aloha,
Art
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--

Chuck Hill                                 email@hidden
Global Village Consulting Inc.             http://www.global-village.net

Progress is the mother of all problems.
- G. K. Chesterton
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  • Follow-Ups:
    • Re: Funky behavior...
      • From: "Thomas Peters" <email@hidden>
References: 
 >Re: Funky behavior... (From: Art Isbell <email@hidden>)

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