Hallo zusammen,
ich versuche, eine Web-Anwendung über Tomcat zu starten. Leider bekomme ich folgende Fehlermeldung:
----------------------------
StandardWrapperValve[action]: Servlet.service() for servlet action threw exception
java.io.IOException: Stream closed
at org.apache.jasper.runtime.JspWriterImpl.ensureOpen (JspWriterImpl.java:202)
at org.apache.jasper.runtime.JspWriterImpl.clearBuffe r(JspWriterImpl.java:157)
at org.apache.jsp.webcore.layout_jsp._jspService(layo ut_jsp.java:443)
at org.apache.jasper.runtime.HttpJspBase.service(Http JspBase.java:94)
at javax.servlet.http.HttpServlet.service(HttpServlet .java:802)
at org.apache.jasper.servlet.JspServletWrapper.servic e(JspServletWrapper.java:324)
at org.apache.jasper.servlet.JspServlet.serviceJspFil e(JspServlet.java:292)
at org.apache.jasper.servlet.JspServlet.service(JspSe rvlet.java:236)
at javax.servlet.http.HttpServlet.service(HttpServlet .java:802)
---------------------------
Weiß jmnd. was ihm noch fehlt!!
Danke Euch im Voraus!
LG
Vicky
ich versuche, eine Web-Anwendung über Tomcat zu starten. Leider bekomme ich folgende Fehlermeldung:
----------------------------
StandardWrapperValve[action]: Servlet.service() for servlet action threw exception
java.io.IOException: Stream closed
at org.apache.jasper.runtime.JspWriterImpl.ensureOpen (JspWriterImpl.java:202)
at org.apache.jasper.runtime.JspWriterImpl.clearBuffe r(JspWriterImpl.java:157)
at org.apache.jsp.webcore.layout_jsp._jspService(layo ut_jsp.java:443)
at org.apache.jasper.runtime.HttpJspBase.service(Http JspBase.java:94)
at javax.servlet.http.HttpServlet.service(HttpServlet .java:802)
at org.apache.jasper.servlet.JspServletWrapper.servic e(JspServletWrapper.java:324)
at org.apache.jasper.servlet.JspServlet.serviceJspFil e(JspServlet.java:292)
at org.apache.jasper.servlet.JspServlet.service(JspSe rvlet.java:236)
at javax.servlet.http.HttpServlet.service(HttpServlet .java:802)
---------------------------
Weiß jmnd. was ihm noch fehlt!!
Danke Euch im Voraus!
LG
Vicky
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