2016-02-12 97 views
0

我想用XSL和groovy轉換一些XML。我爲它使用javax.xml.transform.TransformerFactory。詞法比較的問題

但是在我的XLS文件中比較並不像我想象的那樣工作。

它不能告訴我,2.0.1大於2.0。爲什麼?我認爲這應該是因爲xsl:stylesheet version =「2.0」。我做錯了什麼?

這裏是我的文件:

XML

<?xml version="1.0" encoding="UTF-8"?> 
<apis> 
    <api version="2.0.1"> 
     <resource> 
      <description>doc for API 2.0.1</description> 
     </resource> 
    </api> 
</apis> 

XSL

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> 

<xsl:param name="api-version" select="2.0"/> 
<xsl:template match="/apis/api[@version &gt; $api-version]/resource"> 
    <resource> 
     <xsl:value-of select="description"></xsl:value-of> 
    </resource> 
</xsl:template> 

</xsl:stylesheet> 

和Groovy腳本

import javax.xml.transform.TransformerFactory 
import javax.xml.transform.stream.StreamResult 
import javax.xml.transform.stream.StreamSource 

def workspacePath 
def xslPath 
def xslFileName 
def xmlPath 
def xmlFileName 
def outputPath 
def outputFileName 
def xslt 
def transformer 
def xml 
def output 
def apiVersions 

workspacePath = "C:/test/" 
xslPath = "transformations/" 
xslFileName = "test5.xsl" 
xmlPath = "pendingFeature/" 
xmlFileName = "test5.xml" 
outputPath = "outputs/" 

xslt = new File(workspacePath + xslPath + xslFileName).getText() 
transformer = TransformerFactory.newInstance().newTransformer(new StreamSource(new StringReader(xslt))) 
xml = new File(workspacePath + xmlPath + xmlFileName).getText() 
outputFileName = "doc.html" 
output = new FileOutputStream(workspacePath + outputPath + outputFileName) 
transformer.transform(new StreamSource(new StringReader(xml)), new StreamResult(output)) 
output.close() 

回答

1

你需要把一個XSLT 2.0處理器一樣撒克遜9在課程路徑中有XSLT 2.0支持。如果你想比較字符串而不是數字,那麼做

<xsl:param name="api-version" select="'2.0'"/> 
<xsl:template match="/apis/api[@version &gt; $api-version]/resource">