ArcEngine的分级显示问题(ClassbreakRenderer)

<P>请问各位大侠,在VB中如何用ArcGIS Engine对图层进行分组(分级)显示啊(ClassbreakRenderer)? 帮助文件中有如下代码:</P>
<P>Private Sub Button3_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button3.Click
        Dim pTable As ITable
        Dim pClassify As IClassify
        Dim pTableHistogram As ITableHistogram
        Dim pHistogram As IHistogram
        Dim dataFrequency As Object
        Dim dataValues As Object
        Dim strOutput As String
        ' We're going to retrieve frequency data from a population field
        ' and then classify this data
        '
        pTable = m_pGeoFeatureLayer
        pTableHistogram = New TableHistogram
        pHistogram = pTableHistogram</P>
<P>        ' Get values and frequencies for the population field
        ' into a table histogram object
        pTableHistogram.Field = strPopField1
        pTableHistogram.Table = pTable
        pHistogram.GetHistogram(dataValues, dataFrequency)</P>
<P>        ' Put the values and frequencies into an Equal Interval classify object
        '
        pClassify = New EqualInterval
        pClassify.SetHistogramData(dataValues, dataFrequency)</P>
<P>        ' Now a generate the classes
        ' Note:
        ' 1/ The number of classes returned may be different from requested
        '    (depends on classification algorithm)
        ' 2/ The classes array starts at index 0 and has datavalues starting
        '    from the minumum value, going to maximum
        '
        Dim Classes() As Double
        Dim ClassesCount As Long</P>
<P>        pClassify.Classify(numDesiredClasses)
        Classes = pClassify.ClassBreaks
        ClassesCount = UBound(Classes)</P>
<P>        ' Initialise a new class breaks renderer and supply the number of
        ' class breaks and the field to perform the class breaks on.
        '
        Dim pClassBreaksRenderer As IClassBreaksRenderer</P>
<P>        pClassBreaksRenderer = New ClassBreaksRenderer
        pClassBreaksRenderer.Field = strPopField1
        pClassBreaksRenderer.BreakCount = ClassesCount
        pClassBreaksRenderer.SortClassesAscending = True</P>
<P>        ' Use an algorithmic color ramp to generate an range of colors between
        ' yellow to red (taken from ArcMaps colorramp properties)
        '</P>
<P>        ' Set the initial color to yellow
        '
        Dim pFromColor As IHsvColor</P>
<P>        pFromColor = New HsvColor
        pFromColor.Hue = 60         ' Yellow
        pFromColor.Saturation = 100
        pFromColor.Value = 96</P>
<P>        ' Set the final color to be red
        '
        Dim pToColor As IHsvColor</P>
<P>        pToColor = New HsvColor
        pToColor.Hue = 0         ' Red
        pToColor.Saturation = 100
        pToColor.Value = 96</P>
<P>        ' Set up the HSV colour ramp to span from yellow to red
        '
        Dim pRamp As IAlgorithmicColorRamp
        Dim pEnumColors As IEnumColors
        Dim ok As Boolean</P>
<P>        pRamp = New AlgorithmicColorRamp
        pRamp.Algorithm = esriColorRampAlgorithm.esriHSVAlgorithm
        pRamp.FromColor = pFromColor
        pRamp.ToColor = pToColor
        pRamp.Size = ClassesCount
        pRamp.CreateRamp(ok)
        pEnumColors = pRamp.Colors</P>
<P>
        ' Iterate through each class brake, setting values and corresponding
        ' fill symbols for each polygon, note we skip the minimum value (classes(0))
        '
        Dim pColor As IColor
        Dim pFillSymbol As ISimpleFillSymbol
        Dim breakIndex As Long</P>
<P>        For breakIndex = 0 To ClassesCount - 1</P>
<P>            ' Retrieve a color and set up a fill symbol,
            ' put this in the symbol array corresponding to the class value
            '
            pColor = pEnumColors.Next
            pFillSymbol = New SimpleFillSymbol
            pFillSymbol.Color = pColor
            pFillSymbol.Style = esriSimpleFillStyle.esriSFSSolid
            pClassBreaksRenderer.Symbol(breakIndex) = pFillSymbol
            pClassBreaksRenderer.Break(breakIndex) = Classes(breakIndex + 1)</P>
<P>            ' Store each break value for user output
            strOutput = strOutput ; "- " ; Classes(breakIndex + 1) ; vbNewLine</P>
<P>        Next breakIndex</P>
<P>        ' Assign the renderer to the layer and update the display
        '
        m_pGeoFeatureLayer.Renderer = pClassBreaksRenderer</P>
<P>        'MapControl1.Refresh(esriViewGeography)</P>
<P>        'txbOutput.Text = "Class Breaks Renderer" ; vbNewLine ; vbNewLine ; _
        '"Fills the States polygons with a color according to population from field " ; strPopField1 ; "." ; vbNewLine ; _
        '"Colors range from yellow (low) to red (high)." ; vbNewLine ; _
        '"The " ; pClassify.MethodName ; " classification is used to generate " ; ClassesCount ; " classes." ; _
        '"Class breaks are:" ; vbNewLine ; _
        'strOutput
    End Sub</P>
<P>可是实际上这段代码无法工作,因为找不到"Dim pHistogram As IHistogram"和"pTableHistogram = New TableHistogram"这两行代码所定义的类型. 我把所有Engine的类库都添加引用,但还是没发现有这两个类, 请大侠指点!!  <a href="mailttfsgis2@zaas.org" target="_blank" >tfsgis2@zaas.org</A></P>

<P>你找不到这两行代码的定义类型是因为你没有引入必要的类库，同时，这段代码移植到VB.NET中是无法使用的，这是ESRI的一个BUG。呵呵</P>

帮助里没有IHistogram这个接口说明，但是engine是有这个接口的，我这里是有的，我只引用了所有engine的类型库

我把所有的Engine类库都引用了还是找不到, 在VB里和VS.NET里都不能用啊,

IHistogram和TableHistogram是desktop库中的接口和对象，用engine中相应的接口和对象替换就行了，分别是IBasicHistogram和BasicTableHistogram

谢谢这位大哥! 我原先就按这样改了,还是不行.

<H1>IHistogram Interface (<a href="mk:@MSITStore:C:\Program%20Files\ArcGIS\DeveloperKit\Help\Com\VB\esriCartoUI.chm::/CartoUI_library.htm" target="_blank" >esriCartoUI</A>)</H1>
<br>
<P></P>
<P>Provides access to members that control histogram objects created from different data sources. </P>
<P><B>Product Availability</B></P>
<DIV>Available with ArcGIS Desktop.</DIV>

GIS说的对，IHistogram在 esriCartoUI中，以后遇到这种问题，用Develop Tools里的Esri Library Locator工具。

大哥,能具体点说说吗? 谢谢!

不错