>Ok. Does it seem like mitigating this might be a reasonable enhancement for
>JXPath? I don't know yet how it would work, but I see it as a flag that when
>on, would simply deal with this by extending the list automatically.
I am not sure. You can file an issue for that if you'd like. From
> -Original Message-
> From: Bruno P. Kinoshita [mailto:brunodepau...@yahoo.com.br.INVALID]
> Sent: Thursday, June 15, 2017 2:51 AM
> To: Commons Users List
> Subject: Re: [jxpath] How to add an entry to a list?
>
> Thanks for the link David. Posted a complete
Thanks for the link David. Posted a complete reply there, but here's the TL;DR.
The only way I found of doing that, was by pre-populating your list with an
empty element. That way, /numbers[1] calls List#set(0, "123") with no
exceptions.
Hope that helps
Bruno
Hi David,
Do you have some code you could share? Maybe looking at your code others (I
would try as well, but can't promise will know how to help) might be able to
help.
Cheers
Bruno
From: "KARR, DAVID"
To: Commons Users List
If a property in a bean I'm trying to manipulate with jxpath is a List, how do
I set a value in an entry of that list?
I read the info at "Modifying Object Graphs", but it's still not clear from
this how I would do this.
I've figured out how to get the list created, either with an explicit