相关疑难解决方法(0)

我在节点中有一个字符串,我想将字符串拆分为'？' 并返回数组中的最后一项.

例如,在下面的块中:

<a>
    <xsl:attribute name="href">
        /newpage.aspx?<xsl:value-of select="someNode"/>
    </xsl:attribute>
    Link text
</a>

我想分割someNode价值.

编辑:这是我用来为我的Asp.Net页面加载Xsl的VB.Net:

Dim xslDocPath As String = HttpContext.Current.Server.MapPath("~/App_Data/someXslt.xsl")
Dim myXsltSettings As New XsltSettings()
Dim myXMLResolver As New XmlUrlResolver()

myXsltSettings.EnableScript = True
myXsltSettings.EnableDocumentFunction = True

myXslDoc = New XslCompiledTransform(False)
myXslDoc.Load(xslDocPath, myXsltSettings, myXMLResolver)

Dim myStringBuilder As New StringBuilder()
Dim myXmlWriter As XmlWriter = Nothing

Dim myXmlWriterSettings As New XmlWriterSettings()
myXmlWriterSettings.ConformanceLevel = ConformanceLevel.Auto
myXmlWriterSettings.Indent = True
myXmlWriterSettings.OmitXmlDeclaration = True

myXmlWriter = XmlWriter.Create(myStringBuilder, myXmlWriterSettings)

myXslDoc.Transform(xmlDoc, argumentList, myXmlWriter)

Return myStringBuilder.ToString()

更新: …

对于家庭作业,我试图将一个XML文件转换为R中的数据框.我尝试了很多不同的东西,我在互联网上搜索了一些想法,但都没有成功.到目前为止,这是我的代码:

library(XML)
url <- 'http://www.ggobi.org/book/data/olive.xml'
doc <- xmlParse(myUrl)
root <- xmlRoot(doc)

dataFrame <- xmlSApply(xmltop, function(x) xmlSApply(x, xmlValue))
data.frame(t(dataFrame),row.names=NULL)

我得到的输出就像一个巨大的数字向量.我试图将数据组织到数据框中,但我不知道如何正确调整我的代码以获得它.