I am using save state service -> $scope.gridApi.saveState.save() to save the current state of the grid. But the save state object doesn't contain the property 'field' defined in column Def. It contains only name, visible, width, sort and filter properties.
Is there any way to get the 'field' property of the column from saveStateService?
If you want to get the value of a field defined in columnDef, you can do it in the following way:
example of a definition of columnDef
let gridColumnDef = {
enableFiltering: true,
columnDefs: [
{ field: "Types", enableCellEdit: false, wordWrap: false },
{ field: "FileName", displayName: "File Name", enableCellEdit: false, width: "30%", wordWrap: true },
{ field: "NameState", displayName: "Name State", enableCellEdit: false, width: "9%" },
{ field: "Requirement", displayName: "Prop./Insu.", enableCellEdit: false, width: "7%" },
{ field: "Version", displayName: "Ver.", enableCellEdit: false, width: "5%" },
{
field: 'Actions', displayName: 'Action', width: "25%",
cellTemplate: `<div class="ui-grid-cell-contents">
<span ng-click="grid.appScope.SelectedRowGridConfig(row.entity)" style="margin-top: -7px !important;font-size:15px !important"
data-ng-if="(row.entity.Requirement === 'Approach') && row.entity.NameState !== 'uncharged'">
<i class="fa fa-cloud-download color-black"></i>
</div>`
}]
};
you can also access methods from ui-grid: grid.appScope.nameMethod(params), in my case I call the method SelectedRowGridConfig and I pass you row.entity as a parameter:
grid.appScope.SelectedRowGridConfig(row.entity)
I hope it helps you.
