jQuery UI Widgets › Forums › Grid › How to create new column after init of grid?
Tagged: #jqwidgets-grid, create new column, grid, javascript grid, jquery grid, not show
This topic contains 1 reply, has 2 voices, and was last updated by Hristo 5 years, 9 months ago.
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Hello.
I need your help.
How to create new column at grid?1
My code:$scope.loadGrid = function() { $scope.settings = getGridDispositionSettings(); $scope.settings.source = new $.jqx.dataAdapter({ localdata: [] }); var s = $scope.settings.source._source.localdata; $scope.settings.rowclick = function(event) { $scope.selectedRow = event.args.row.bounddata; $scope.selectedRowIndex = event.args.rowindex; } $scope.settings.source = new $.jqx.dataAdapter({ localdata: s }); $scope.settings.editable = true; $scope.settings.groupable = false; $scope.settings.editmode = 'click'; $scope.settings.columns = [ { text: "Напр", datafield: "_direction", width: 25, filtertype: 'checkedlist', resizable: false, cellsrenderer: asColumnRender, editable: false, type: "string", pinned: true, rendered: tooltiprenderer }, { text: "Рег.Номер", datafield: "_RegistrationNumbers", width: 90, resizable: false, cellsrenderer: asColumnRender, editable: false, type: "string", pinned: true, rendered: tooltiprenderer }, { text: "Караван", datafield: "_caravansForTractor", width: 90, type: "string", filtertype: 'checkedlist', cellsrenderer: caravanRender, editable: false, rendered: tooltiprenderer }, { text: "Блок", datafield: "_blocksForTractor", width: 80, type: "string", filtertype: 'checkedlist', cellsrenderer: blockRender, editable: false, rendered: tooltiprenderer }, { text: "Комм. инженера", datafield: "EngineerComment", cellsrenderer: asColumnRender, cellendedit: rowEditEngineerComment, width: 100, type: "string", rendered: tooltiprenderer } ]; $scope.settings.columngroups = [ { text: 'Календарь', align: 'center', name: 'Calendar' } ]; if (typeof $scope.calendarStart == 'undefined' || $scope.calendarStart !== new Date() && !$scope.calendarFlag) { $scope.changeCalendarForGrid(); } var firstDate = $scope.calendarStart; var secondDate = $scope.calendarEnd; var diffDays = Math.round(Math.abs((firstDate.getTime() - secondDate.getTime()) / (oneDay))); $scope.datafields = []; for (var k = 0; k <= diffDays; k++) { var date = new Date(); date = firstDate.addDays(k); var day = date.getDate(); var month = date.getMonth() + 1; var datafield = '_Date_' + month + '_' + day; $scope.settings.columns.push({ text: day + '-' + getDayOfWeek(date.getDay()), datafield: datafield, columnsreorder: false, width: 50, draggable: false, type: "string", columngroup: 'Calendar', cellsrenderer: calendarRender }); $scope.datafields.push(datafield); } }
In current time I have columngroup ‘Calendar’ with next column “day + ‘-‘ + getDayOfWeek(date.getDay())”
but in one moment i need to change this period on the other
How to correctly add new columns to columngroup and to remove unnecessary from columngroup ?
Thx for attention!With best regards,
Alexey Zabelsky- This topic was modified 5 years, 9 months ago by Zabelsky.
Hello Alexey Zabelsky,
You could change initial object for the columngroups with a new one.
Please, take a look at this topic, it demonstrates that approach:
https://www.jqwidgets.com/community/topic/jqxgrid-columngroups-coumns-text-change-dynamcially/
On the other side, you could change for each one particular column its ‘collumngroup’ bysetcolumnproperty
method of the Grid.Best Regards,
Hristo HristovjQWidgets team
http://www.jqwidgets.com -
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