XSLT使用命名空间转换XML？ [英] XSLT Transform XML with Namespaces ?

问题描述

我有一个关于XSLT的问题。

i有一个xml文件：

 <？  xml-stylesheet     type   =  text / xsl    href   =  XSLTFile1.xslt ？  >  
 <   otx     xmlns   =  http://iso.org/OTX/1.0.0    xmlns：i18n   =  http：// i so.org/OTX/1.0.0/i18n\"   xmlns：diag   =  http://iso.org/OTX/1.0.0/DiagCom    xmlns：measure   =  http://iso.org/OTX/1.0.0/Measure\"   xmlns：string   =  http://iso.org/OTX/1.0.0/StringUtil    xmlns：fileXml   =  http://vwag.de/OTX/1.0.0/XmlFile    xmlns：日志  =  http://iso.org/OTX/1.0.0/Logging    xmlns：file   =  http：// vwag.de/OTX/1.0.0/File\"   xmlns：dataPlus   =  http://iso.org/OTX/1.0.0/DiagDataBrowsingPlus    xmlns：event   =  http://iso.org/OTX/1.0.0/Event\"   xmlns：quant   =  http://iso.org/OTX/1.0.0/Quantities    xmlns：hmi   =  http://iso.org/OTX/1.0.0/HMI    xmlns：数学 <跨度class =code-keyword> =  http://iso.org/OTX/1.0.0/Math    xmlns：flash   =  http://iso.org/OTX/1.0.0/Flash    xmlns：data   =  http://iso.org/OTX/1.0.0/DiagDataBrowsing    xmlns：xsi   =  http://www.w3.org/2001/XMLSchema-instance    xmlns ：dt   =  http://iso.org/OTX/1.0.0/DateTime    xmlns：eventPlus   = < span class =code-keywo rd> http://iso.org/OTX/1.0.0/EventPlus    xmlns：corePlus   =  http://iso.org/OTX/1.0.0/CorePlus    xmlns：xmime   =  http://www.w3.org/2005/05/xmlmime    xmlns：job   =  http://iso.org/OTX/1.0.0/Job     id   = < span class =code-keyword> id_e895af47737740b1b0aca75a4f43dd86    name   =  NewDocument1   包  =  NewOtxProject1Package1    version   =  1.0.0.0     timestamp   =  2014-04-03T15：11：12.3634145 + 07:00 >  
 <  程序 >  
 < span class =code-keyword><   procedure     id   =  id_25f45459817744f083d360ff75c0ea5b   名称  =  main    visibility   =  PUBLIC >  
 <   / procedure  >  
 <   / procedures  >  
 <   / otx  >  

这是我的XSLT

< pre lang =xml> <？ xml version = 1.0 encoding = utf-8 ？ >
< xsl：stylesheet 版本 = 1.0 xmlns：xsl = http：/ /www.w3.org/1999/XSL/Transform\" xmlns = http://iso.org/OTX/1.0.0

xmlns：msxsl = urn：schemas-microsoft-com：xslt exclude-result-prefixes = msxsl

>
< xsl：output 方法 = html 缩进 = 是 / >

<！ - < xsl：template match =@ * | node（）>
< xsl：copy>
< xsl：apply-templates select =@ * | node（）/>
< / xsl：copy> - >

< xsl：template < span class =code-attribute> match = / >
< html lang = en >
< body >
< 表格 class = form-horizo​​ntal 角色 = 表单 >
< div class = form-group >
< div class = col-sm-10 >
< label > 程序：< / label >
< div 类 = 表格响应 >
< table >
< xsl：for-each 选择 = otx / procedures / procedure >
< tr >
< td < span class =code-keyword>>
< xsl：value-of 选择 = < span class =code-keyword> @ name / >
< / td >
< / tr >
< / xsl：for-each >
< / table >
< / div >
< / div >
< / div >
< / form >
< / body >
< / html >

< / xsl：template >
< / xsl：stylesheet >

当我点击xml文件时，它在浏览器上显示时无法获取数据。

请帮助我，XSLT有什么问题？

解决方案

I have a problem about XSLT.
i have a xml file :

<?xml-stylesheet type="text/xsl" href="XSLTFile1.xslt"?>
<otx xmlns="http://iso.org/OTX/1.0.0" xmlns:i18n="http://iso.org/OTX/1.0.0/i18n" xmlns:diag="http://iso.org/OTX/1.0.0/DiagCom" xmlns:measure="http://iso.org/OTX/1.0.0/Measure" xmlns:string="http://iso.org/OTX/1.0.0/StringUtil" xmlns:fileXml="http://vwag.de/OTX/1.0.0/XmlFile" xmlns:log="http://iso.org/OTX/1.0.0/Logging" xmlns:file="http://vwag.de/OTX/1.0.0/File" xmlns:dataPlus="http://iso.org/OTX/1.0.0/DiagDataBrowsingPlus" xmlns:event="http://iso.org/OTX/1.0.0/Event" xmlns:quant="http://iso.org/OTX/1.0.0/Quantities" xmlns:hmi="http://iso.org/OTX/1.0.0/HMI" xmlns:math="http://iso.org/OTX/1.0.0/Math" xmlns:flash="http://iso.org/OTX/1.0.0/Flash" xmlns:data="http://iso.org/OTX/1.0.0/DiagDataBrowsing" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:dt="http://iso.org/OTX/1.0.0/DateTime" xmlns:eventPlus="http://iso.org/OTX/1.0.0/EventPlus" xmlns:corePlus="http://iso.org/OTX/1.0.0/CorePlus" xmlns:xmime="http://www.w3.org/2005/05/xmlmime" xmlns:job="http://iso.org/OTX/1.0.0/Job" id="id_e895af47737740b1b0aca75a4f43dd86" name="NewDocument1" package="NewOtxProject1Package1" version="1.0.0.0" timestamp="2014-04-03T15:11:12.3634145+07:00">
  <procedures>
    <procedure id="id_25f45459817744f083d360ff75c0ea5b" name="main" visibility="PUBLIC">
    </procedure>
  </procedures>
</otx>

and this is my XSLT

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns="http://iso.org/OTX/1.0.0"

    xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl"

>
  <xsl:output method="html" indent="yes"/>

  <!--<xsl:template match="@* | node()">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()"/>
        </xsl:copy>-->

  <xsl:template match="/">
    <html lang="en">
      <body>
        <form class="form-horizontal" role="form">
          <div class="form-group">
            <div class="col-sm-10">
              <label>Procedures:</label>
              <div class="table-responsive">
                <table>
                  <xsl:for-each select="otx/procedures/procedure">
                    <tr>
                      <td>
                        <xsl:value-of select="@name"/>
                      </td>
                    </tr>
                  </xsl:for-each>
                </table>
              </div>
            </div>
          </div>
        </form>
      </body>
    </html>

  </xsl:template>
</xsl:stylesheet>

When i clicked on xml file, it cannot get data when showing on browser.
please help me, what is wrong in the XSLT?

解决方案

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