Some more information ...
in my program I am able to upload the file content of an input field (class="easyui-filebox") with the code below.
The problem is how to use this code with a filebox field in a datagrid.
In other words:
if in case of an input field (class="easyui-filebox") the variable "f" is:
var f = $('#Pdf').next().find('.textbox-value');
what would it be in case the filebox is a field in a datagrid ?
Thanks for any help.
Miche
var form_data = new FormData();
var f = $('#Pdf').next().find('.textbox-value');
file_data = f[0].files[0];
form_data.append('Pdf', file_data);
$.ajax({
url: 'upload.cgi',
data: form_data,
...
});