Re: Funky behavior...
Re: Funky behavior...
- Subject: Re: Funky behavior...
- From: "Thomas Peters" <email@hidden>
- Date: Tue, 12 Aug 2003 23:42:57 -0400
- Seal-send-time: Tue, 12 Aug 2003 23:42:58 -0400
Thank you Art, thank you Chuck. My problem is solved.
Tom
----- Original Message -----
From: Chuck Hill
Cc: email@hidden
Sent: Tuesday, August 12, 2003 5:27 PM
Subject: Re: Funky behavior...
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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