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The sort flag is set to true. When sorting from 'a' to 'z' is applied, the results are sorted according to an unknown algorithm.
Only sorting by row number works fine.
All data (except row number) are fetched from API via future methods.
Filtering also works only for row number:
For the other columns they do not work..
Hi Damian,
We have reviewed the information you provided, but we
were not able to reproduce the reported issue from our end. As the sorting and
filtering are working properly on our end, we suspect that the issue may be
specific to the sample level. To assist you better, could you please provide us
with more detailed information about how you fetch the data from the API and
assign it to the DataGrid? It would also be helpful if you can confirm that
you’re setting the same column name on the GridColumn.columnName and DataGridCell.columnName
in DataGridRow. We have attached a sample that we tested, along with an
output video reference. If possible, could you please provide the sample you
are having trouble with? Alternatively, you can modify the attached sample to
be reproducible, so we can investigate and provide a solution as soon as
possible.
List<GridColumn> get getColumns { return [ GridColumn( columnName: 'id', label: Container(alignment: Alignment.center, child: const Text('ID'))), GridColumn( columnName: 'name', label: Container( alignment: Alignment.center, child: const Text('Name'))), GridColumn( columnName: 'designation', label: Container( alignment: Alignment.center, child: const Text('Designation'))), GridColumn( columnName: 'salary', label: Container( alignment: Alignment.center, child: const Text('Salary'))), ]; }
class EmployeeDataSource extends DataGridSource { EmployeeDataSource(List<Employee> employees) { buildDataGridRow(employees); }
void buildDataGridRow(List<Employee> employeeData) { dataGridRow = employeeData.map<DataGridRow>((employee) { return DataGridRow(cells: [ DataGridCell<int>(columnName: 'id', value: employee.id), DataGridCell<String>(columnName: 'name', value: employee.name), DataGridCell<String>( columnName: 'designation', value: employee.designation), DataGridCell<int>(columnName: 'salary', value: employee.salary), ]); }).toList(); } } |
Regards,
Tamilarasan