[C.S.PuteraParticipant]

Thank you for the reply. Yes, I had seen the example before and already follow the guidelines :

1. Set the id of the row. In my case id : ‘delivery_confirmation_id’
2. Set the selectionmode. In my case selectionmode : ‘checkbox’

But the checkbox does not show up. Is there another setting I should do ?

I set the rows as the root => root : ‘rows’ ( instead of “entry” like the example ) and did not set the record: “content”. Should I set the record to “content” ? And If I should, what is that for ?

Thank you

[C.S.PuteraParticipant]

Thank you for the reply. Now the foreign key is shown, but I still not able to set the filter. e.g, I have 2 tables : table vehicle and table vehicle type. Now I’d like to set the filter of vehicle grid, as the drop down, which values taken from all data in table vehicle type. How could I do that ? I am using json, not xml. Please help, thank you

[C.S.PuteraParticipant]

Problem solved, thank you so much Peter

[C.S.PuteraParticipant]

I found the reason. There is conflict between jquery used by jqwidgets with jquery used by Yii form ( which is generated by Yii automatically to handle validation-related functions ). Is there a way to handle this conflict ? Thank you

[Author]

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