Question

What is the easiest way for a Word macro to execute XPath expressions such as:

"string(/alpha/beta)" 

"not(string(/alpha/beta)= true )"

谁应该分别返回地貌和ool? (而不是xml node或delist)

我想避免已经赢得的DLL在2007年或2010年机器运行台上出现。

功能选择Single Node (queryString As String) 返回了IXMLDOMNode,从而赢得了一定地位。

换言之,类似于......。 NET sxpathnavigator.evaluate ?

http://msdn.microsoft.com/en-us/library/2c16b7x8.aspx”rel=“nofollow”http://msdn.microsoft.com/en-us/library/2c16b7x8.aspx

Answer 1

您可使用XSL转变来评价XPath的表述,特别是 xsl: Value-of 。

我写了<代码>Evaluate功能,该功能与这一原则有关。它创造了一种XSL风格的记忆,其中包含一个XSL模板,该模板将采用XPath的表述方式,评价它,并退回一份新的XML文件,其中载有<代码><result> node的结果。它对确保<条码>价值-of退回了部分内容(如果没有,就留下了一个错误)进行核对,如果是,将XPath的表述结果转换为以下几种数据之一:<条码>:Long、<条码>Double<>>>>>>>、<条码>>Boolean>、或<条码>。

这里是我用来实施该守则的一些测试。 I used the books.xml 如你想要进行这些测试,则从MSDN网页上存档(如果你想进行这些测试,那么你必须改变通往books.xml的道路)。

Public Sub Test_Evaluate()

    Dim doc As New DOMDocument
    Dim value As Variant

    doc.async = False
    doc.Load "C:DevelopmentStackOverflowXPath Evaluationooks.xml"

    Debug.Assert (doc.parseError.errorCode = 0)

      Sum of book prices should be a Double and equal to 30.97
     
    value = Evaluate(doc, "sum(descendant::price)")
    Debug.Assert TypeName(value) = "Double"
    Debug.Assert value = 30.97

      Title of second book using text() selector should be "The Confidence Man"
     
    value = Evaluate(doc, "descendant::book[2]/title/text()")
    Debug.Assert TypeName(value) = "String"
    Debug.Assert value = "The Confidence Man"

      Title of second book using string() function should be "The Confidence Man"
     
    value = Evaluate(doc, "string(/bookstore/book[2]/title)")
    Debug.Assert TypeName(value) = "String"
    Debug.Assert value = "The Confidence Man"

      Total number of books should be 3
     
    value = Evaluate(doc, "count(descendant::book)")
    Debug.Assert TypeName(value) = "Long"
    Debug.Assert value = 3

      Title of first book should not be "The Great Gatsby"
     
    value = Evaluate(doc, "not(string(/bookstore/book[1]/title))= The Great Gatsby ")
    Debug.Assert TypeName(value) = "Boolean"
    Debug.Assert value = False

      Genre of second book should be "novel"
     
    value = Evaluate(doc, "string(/bookstore/book[2]/attribute::genre)= novel ")
    Debug.Assert TypeName(value) = "Boolean"
    Debug.Assert value = True

      Selecting a non-existent node should generate an error
     
    On Error Resume Next

    value = Evaluate(doc, "string(/bookstore/paperback[1])")
    Debug.Assert Err.Number = vbObjectError

    On Error GoTo 0

End Sub

此处为<代码>Evaluate功能(IsLong功能,是使数据类型转换代码变得更可读的辅助功能):

<Note:As barrowc 在评论中,你可以明确你想要用哪一种版本的MSXML,将/code>替换成具体版本的类别名称,例如。 DOMDocument30 (MSXML3) or DOMDocument60 (MSXML6). 书面守则将不使用MSXML3,目前这一系统被更广泛地利用,但MSXML6的性能较好,也是最新版本,是目前推荐的Microsoft。

参看问题我应使用何种版本的MSXML? 关于MSXML不同版本的更多信息。

Public Function Evaluate(ByVal doc As DOMDocument, ByVal xpath As String) As Variant

    Static styleDoc As DOMDocument
    Dim valueOf As IXMLDOMElement
    Dim resultDoc As DOMDocument
    Dim result As Variant

    If styleDoc Is Nothing Then

        Set styleDoc = New DOMDocument

        styleDoc.loadXML _
            "<xsl:stylesheet version= 1.0  xmlns:xsl= http://www.w3.org/1999/XSL/Transform >" & _
                "<xsl:template match= / >" & _
                    "<result>" & _
                        "<xsl:value-of />" & _
                    "</result>" & _
                "</xsl:template>" & _
            "</xsl:stylesheet>"

    End If

    Set valueOf = styleDoc.selectSingleNode("//xsl:value-of")
    valueOf.setAttribute "select", xpath

    Set resultDoc = New DOMDocument
    doc.transformNodeToObject styleDoc, resultDoc

    If resultDoc.documentElement.childNodes.length = 0 Then
        Err.Raise vbObjectError, , "Expression  " & xpath & "  returned no results."
    End If

    result = resultDoc.documentElement.Text

    If IsLong(result) Then
        result = CLng(result)
    ElseIf IsNumeric(result) Then
        result = CDbl(result)
    ElseIf result = "true" Or result = "false" Then
        result = CBool(result)
    End If

    Evaluate = result

End Function

Private Function IsLong(ByVal value As Variant) As Boolean

    Dim temp As Long

    If Not IsNumeric(value) Then
        Exit Function
    End If

    On Error Resume Next

    temp = CLng(value)

    If Not Err.Number Then
        IsLong = (temp = CDbl(value))
    End If

End Function

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