If I have a URL like http://localhost/Test/edit/{id} and I would like
the {id} transform to a parameter instead of URL path part.
The problem is that if I have a template named edit.html under webapp/
Test, which is the template I use to edit an item.
And I have a Menu instance like the following:
<code>
Menu (Loc("Test", List("Test") -> true, "Test"))
</code>
It would only match URL like http://localhost/Test/edit, not anything
like http://localhost/Test/edit/1
Is it best way to do it by create a menu with RewriteRequest? Because
I found it a little boilerplate if I have lot URL pattern like this.
<code>
val menu = Menu(new Loc[Unit] {
override def name = "Test"
override def text = "Test"
override def link = (List ("Test"), true)
override def params = Nil
override def defaultValue = Full(())
def isTarget (path: ParsePath) = path match {
case ParsePath (List("Test", "edit", id), _, _, _) => true
case _ => false
}
override def rewrite = Full ( NamedPF("Test") {
case RewriteRequest (path, _, _) if isTarget(path) =>
RewriteResponse(List("Test", "edit"),
Map("id" -> "1024")) -> ()
})
})
</code>
--
You received this message because you are subscribed to the Google Groups
"Lift" group.
To post to this group, send email to [email protected].
To unsubscribe from this group, send email to
[email protected].
For more options, visit this group at
http://groups.google.com/group/liftweb?hl=en.
