PivotGridControl.GetFieldValue(PivotGridField, Int32) Method

Returns the specified column or row field’s value for the cell addressed by its zero-based index in the Data Area.

Namespace: DevExpress.XtraPivotGrid

Assembly: DevExpress.XtraPivotGrid.v23.2.dll

NuGet Package: DevExpress.Win.PivotGrid

Declaration

C#

public object GetFieldValue(
    PivotGridField field,
    int lastLevelIndex
)

VB.NET

Public Function GetFieldValue(
    field As PivotGridField,
    lastLevelIndex As Integer
) As Object

Parameters

Name	Type	Description
field	PivotGridField	A PivotGridField object representing a column or row field whose value is to be obtained.
lastLevelIndex	Int32	A zero-based index of a cell in the Data Area that identifies the required field value. Indexes are numbered starting from the left edge for column fields, and from the top edge for row fields.

Returns

Type	Description
Object	An object representing the field’s value.

Remarks

The following image shows cell indexes numerating rows (columns are numerated in the same manner):

In this example, the GetFieldValue method will return:

• Argentina for the Country field and cell’s index=0;
• Buenos Aires for the City field and cell’s index=0;
• Austria for the Country field and cell’s index=1;
• null for the City field and cell’s index=1;

Related GitHub Examples

The following code snippet (auto-collected from DevExpress Examples) contains a reference to the GetFieldValue(PivotGridField, Int32) method.

Note

The algorithm used to collect these code examples remains a work in progress. Accordingly, the links and snippets below may produce inaccurate results. If you encounter an issue with code examples below, please use the feedback form on this page to report the issue.

C#

winforms-pivot-customize-appearance-of-the-even-odd-groups-of-cells/CS/WindowsApplication53/Form1.cs#L71

for (int i = 0; i < fields.Length ; i++) {
    current[i] = pivot.GetFieldValue(fields[i], index);
}

VB.NET

winforms-pivot-customize-appearance-of-the-even-odd-groups-of-cells/VB/WindowsApplication53/Form1.vb#L74

For i As Integer = 0 To fields.Length - 1
    current(i) = pivot.GetFieldValue(fields(i), index)
Next