我有使用嵌入式Jetty实例的spring应用程序。
project | src | controller | webapps | jsp | WEB-INF | web.xml | applicationContext.xml | spring-servlet.xml
我的罐子有相同的树状结构,但我不断
d:\test>java -jar springtest.jar 2011-11-22 15:37:02.576:INFO::jetty-7.x.y-SNAPSHOT 2011-11-22 15:37:02.686:WARN::Failed startup of context o.e.j.w.WebAppContext{/,[file:/C:/Users/me/AppData/Local/Temp/jetty-0.0.0.0-8080-webapps-_-any-/webinf /, jar:file:/d:/test/springtest.jar!/org/jcvi/webapps/]} java.io.FileNotFoundException: d:\test\org\eclipse\jetty\webapp\webdefault.xml (The system cannot find the path specified) at java.io.FileInputStream.open(Native Method) at java.io.FileInputStream.<init>(FileInputStream.java:106) at java.io.FileInputStream.<init>(FileInputStream.java:66) at sun.net.www.protocol.file.FileURLConnection.connect(FileURLConnection.java:70) at sun.net.www.protocol.file.FileURLConnection.getInputStream(FileURLConnection.java:161) at com.sun.org.apache.xerces.internal.impl.XMLEntityManager.setupCurrentEntity(XMLEntityManager.java:653) at com.sun.org.apache.xerces.internal.impl.XMLVersionDetector.determineDocVersion(XMLVersionDetector.java:186) at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:772) at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:737) at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(XMLParser.java:119) at com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.parse(AbstractSAXParser.java:1205) at com.sun.org.apache.xerces.internal.jaxp.SAXParserImpl$JAXPSAXParser.parse(SAXParserImpl.java:522) at javax.xml.parsers.SAXParser.parse(SAXParser.java:395) at org.eclipse.jetty.xml.XmlParser.parse(XmlParser.java:188) at org.eclipse.jetty.xml.XmlParser.parse(XmlParser.java:204) at org.eclipse.jetty.webapp.Descriptor.parse(Descriptor.java:60) at org.eclipse.jetty.webapp.WebDescriptor.parse(WebDescriptor.java:140) at org.eclipse.jetty.webapp.MetaData.setDefaults(MetaData.java:141) at org.eclipse.jetty.webapp.WebXmlConfiguration.preConfigure(WebXmlConfiguration.java:46) at org.eclipse.jetty.webapp.WebAppContext.preConfigure(WebAppContext.java:412) at org.eclipse.jetty.webapp.WebAppContext.doStart(WebAppContext.java:448) at org.eclipse.jetty.util.component.AbstractLifeCycle.start(AbstractLifeCycle.java:58) at org.eclipse.jetty.server.handler.HandlerWrapper.doStart(HandlerWrapper.java:89) at org.eclipse.jetty.server.Server.doStart(Server.java:258) at org.eclipse.jetty.util.component.AbstractLifeCycle.start(AbstractLifeCycle.java:58) at org.jcvi.ServerRunner.startServer(ServerRunner.java:83) at org.jcvi.MainServer.main(MainServer.java:18) 2011-11-22 15:37:02.748:INFO::Started SelectChannelConnector@0.0.0.0:8080 STARTING
我有以下运行码头服务器实例的Java类
String webDir = this.getClass().getClassLoader().getResource("webapps").toExternalForm(); Server server = new Server(8080); WebAppContext context = new WebAppContext(); context.setContextPath("/"); context.setResourceBase(webDir); context.setParentLoaderPriority(true); HandlerList handlers = new HandlerList(); handlers.setHandlers(new Handler[] { context, new DefaultHandler() }); server.setHandler(context); server.start();
我的web.xml看起来像
<welcome-file-list> <welcome-file>index.jsp</welcome-file> </welcome-file-list> <context-param> <param-name>contextConfigLocation</param-name> <param-value>/WEB-INF/applicationContext.xml</param-value> </context-param> <listener> <listener-class> org.springframework.web.context.ContextLoaderListener </listener-class> </listener> <servlet> <servlet-name>spring</servlet-name> <servlet-class> org.springframework.web.servlet.DispatcherServlet </servlet-class> <load-on-startup>2</load-on-startup> </servlet> <servlet-mapping> <servlet-name>spring</servlet-name> <url-pattern>/</url-pattern> </servlet-mapping>
如果我在IDE中运行,此应用程序运行良好,但JAR失败。我该如何解决此问题,这样我才能拥有其中包含Web应用程序的单个jar文件?
我遇到了类似的问题,并通过以下主类实现解决了这个问题:
private static final int PORT = 8080; private static final String WAR_LOCATION = "src/webapps"; //in your case I guess private static final String CONTEXT_PATH = "/movence"; //change it if you want public static void main(String[] args) throws Exception { Server server = new Server(); WebAppContext context = new WebAppContext(); SocketConnector connector = new SocketConnector(); setupConnector(connector); setupContext(server, context); setupServer(server, context, connector); startServer(server); } private static void startServer(Server server) throws Exception, InterruptedException { server.start(); server.join(); } private static void setupServer(Server server, WebAppContext context, SocketConnector connector) { server.setConnectors(new Connector[] { connector }); server.addHandler(context); } private static void setupConnector(SocketConnector connector) { connector.setPort(PORT); } private static void setupContext(Server server, WebAppContext context) { context.setServer(server); context.setContextPath(CONTEXT_PATH); context.setWar(WAR_LOCATION); }