Hi, Tobias Thanks for reply.
I am newbie of customizing grid component. I guess what you mean like this?: <table t:type="grid" t:source="userSource" include="location_city"> <t:parameter name="location_cityheader"> </t:parameter> <t:parameter name="location_citycell"> </t:parameter> </table> But exception message still exists: "Bean editor model for User does not contain a property named 'location_city'". Thanks! DH ----- Original Message ----- From: "Tobias Wehrum" <[EMAIL PROTECTED]> To: "Tapestry users" <users@tapestry.apache.org> Sent: Friday, April 18, 2008 4:20 PM Subject: Re: T5: Is there a simple way to display property of embedded instance in Grid? > Hi DH, > > you can teach Location a standard way to be outputted by overwriting the > toString() function of Location. > > Now you can output the String returned by toString() simply by including > "location". > > If you want to output different properties of Location and not in one > cell, I think you will have to add location_city, location_street etc > and implement <t:parameter> blocks for it. > > Hope that helps, > Tobias > > dhning schrieb: >> Hi, All >> >> Case: A user own a location while the location is comprised of city, >> street... >> public class User { >> private Location location; >> // setter & getter >> } >> public class Location { >> private String city; >> // setter & getter >> } >> >> In the user list page, I want to display the city as one column in Grid. >> But it doesn't work like this <table t:type="grid" t:source="userSource" >> include="location.city"></table>. >> The exception message is "User does not contain a property named >> 'location.city'". >> >> Is there a simple way to implement such function? >> >> Thanks in advance. >> DH > > > --------------------------------------------------------------------- > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] > >