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
> 
> 
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