Hello,
EDIT: No longer a question I need an answer to!
I have a controller to handle a login form. It extends Spring's SimpleFormController. According to the documentation for this controller (see
http://static.springsource.org/spring/docs/2.5.x/api/org/springframework/web/servlet/mvc/SimpleFormController.html):
The submit behavior can be customized by overriding one of the onSubmit methods. Submit actions can also perform custom validation if necessary (typically database-driven checks), calling showForm in case of validation errors to show the form view again.
This is what I want to do. Please note: I have a validator linked to the form which I want to deal with "plain data" validation only. I don't want to do the database-driven validation there.
I'm now looking at the documentation for the showForm() methods (
http://static.springsource.org/spring/docs/2.5.x/api/org/springframework/web/servlet/mvc/SimpleFormController.html#showForm(javax.servlet.http.HttpServletRequest,%20javax.servlet.http.HttpServletResponse,%20org.springframework.validation.BindException)). They take an object of BindException so figure I have to create one to pass to showForm().
Looking at the documentation for BindException: (
http://static.springsource.org/spring/docs/2.5.x/api/org/springframework/validation/BindException.html) and I get informed that:
As of Spring 2.0, this is a special-purpose class. Normally, application code will work with the BindingResult interface, or with a DataBinder that in turn exposes a BindingResult via DataBinder.getBindingResult().
So, can someone please tell me what I need to do, or how I can perform my database-backed validation?
Thanks,
Ed
EDIT: the onSubmit receives the a BindException object called errors! I really apologise for the post!!