Title: Single field with datagrid Post by: neatgadgets on December 19, 2016, 11:02:38 AM I want to have a datagrid where I go through and tick on and off items, than give this selection a name and save it to the database. How can I add a field at the topas a name, with a button to save (insert all the ticked items as rows in a table with the field name)?
Andrew e.g. Selection Field Name: MySelection A B C Selected Fred 0 1 Tick Barney 1 3 Unticked Wilma 2 1 Tick So rows Fred and Wilma are saved with the name "MySelection" into a table Title: Re: Single field with datagrid Post by: jarry on December 19, 2016, 04:59:41 PM If the ticked row is the selected row, you can call 'getSelections' method to get all the selected rows and then post them to the server.
Code: var rows = $('#dg').datagrid('getSelections'); Title: Re: Single field with datagrid Post by: neatgadgets on December 20, 2016, 12:46:31 AM Thanks, does that mean I can include a form on the datagrid and pass the name of the selected list as well?
Title: Re: Single field with datagrid Post by: jarry on December 20, 2016, 07:46:21 PM The form field values are combined by name-value pairs. But what you want is to post all the selected rows, so you just need to serialize these rows and post to server. In your server side, you should unserialize to restore the rows.
Title: Re: Single field with datagrid Post by: neatgadgets on December 22, 2016, 04:25:11 PM So what I want to do is like this
Code: var rows = $('#dg').datagrid('getSelections'); Only when I try and read it on the php page called I get Undefined index: dg Code: $dg = $_REQUEST['dg']; Yet the charter_type came through |