делаю это:
| Код |
<c:import var="xmlfile" url="/temp.xml"/>
<c:import var="xslfile" url="/temp.xsl"/>
<x:parse var="doc" xml="${xmlfile}"/>
<x:transform xml="${xmlfile}" xslt="${xslfile}"/>
|
и вываливается ексепшн
| Код |
javax.servlet.ServletException: javax.servlet.jsp.JspException: javax.xml.transform.TransformerConfigurationException: javax.xml.transform.TransformerException: java.lang.IllegalStateException: can't declare any more prefixes in this context
org.apache.jasper.runtime.PageContextImpl.doHandlePageException(PageContextImpl.java:821)
org.apache.jasper.runtime.PageContextImpl.handlePageException(PageContextImpl.java:758)
org.apache.jsp.WEB_002dINF.jsp.ProjectPage_jsp._jspService(ProjectPage_jsp.java:142)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:94)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:324)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:292)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:236)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
org.springframework.web.servlet.view.InternalResourceView.renderMergedOutputModel(InternalResourceView.java:97)
org.springframework.web.servlet.view.AbstractView.render(AbstractView.java:250)
org.springframework.web.servlet.DispatcherServlet.render(DispatcherServlet.java:961)
org.springframework.web.servlet.DispatcherServlet.doDispatch(DispatcherServlet.java:738)
org.springframework.web.servlet.DispatcherServlet.doService(DispatcherServlet.java:658)
org.springframework.web.servlet.FrameworkServlet.processRequest(FrameworkServlet.java:392)
org.springframework.web.servlet.FrameworkServlet.doGet(FrameworkServlet.java:347)
javax.servlet.http.HttpServlet.service(HttpServlet.java:689)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
|
никак не могу понять, в чем проблема... |