Using an AjaxSubmitLink outside of a Form does not set the form property
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Key: WICKET-1855
URL: https://issues.apache.org/jira/browse/WICKET-1855
Project: Wicket
Issue Type: Bug
Components: wicket
Affects Versions: 1.3.4
Reporter: Pieter van Prooijen
Priority: Minor
When constructing an AjaxSubmitLink outside of a form using the constructor:
public AjaxSubmitLink(String id, final Form form)
The form argument is not set in the link. This prevents the form from seeing
the link as its submitting component, which in turn means that using thing like
enableDefaultFormProcessing() on the link don't work.
Changing the constructor of AjaxSubmitLink to:
public AjaxSubmitLink(String id, final Form form)
{
super(id, form) /* was: super(id) */
...
}
will probably fix the problem.
A temporary workaround is to implement a getForm() method in the subclass of
the submit link which answers the form of the link
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