将servlet定义为springBean的方法
需要完成一个比较晓得web服务端提供给外部系统调用，又不想用比较复杂web层框架（struts），只是采用servlet即可，不过每定义一个servlet都需要在web.xml里面写入配置，不方便管理，并且servlet和spring的交互感觉不是那么的河蟹……
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这里自己想到了一个办法，将所有的http请求都转发到springBean里面，只需要定义一个servlet即可：
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首先需要定义一个接口叫servlet处理 器
import java.io.IOException;import javax.servlet.ServletException;import javax.servlet.http.HttpServletRequest;import javax.servlet.http.HttpServletResponse;public interface HttpHandler {/** * 处理 http请求 * * @param request * @param response * @throws ServletException * @throws IOException */public void requestHandler(HttpServletRequest request, HttpServletResponse response) throws ServletException,IOException;}?
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servlet? class
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import java.io.IOException;import java.util.HashMap;import java.util.Map;import javax.servlet.ServletConfig;import javax.servlet.ServletContext;import javax.servlet.ServletException;import javax.servlet.http.HttpServlet;import javax.servlet.http.HttpServletRequest;import javax.servlet.http.HttpServletResponse;import org.apache.log4j.Logger;import org.springframework.web.context.WebApplicationContext;import org.springframework.web.context.support.WebApplicationContextUtils;public class ServletDispacher extends HttpServlet {private static final long serialVersionUID = -4397593743538525329L;private Map<String, HttpHandler> handlers = new HashMap<String, HttpHandler>();Logger log = Logger.getLogger(this.getClass());@SuppressWarnings("unchecked")@Overridepublic void init(ServletConfig config) throws ServletException {ServletContext context = config.getServletContext();WebApplicationContext appContext = WebApplicationContextUtils.getRequiredWebApplicationContext(context);handlers = appContext.getBeansOfType(HttpHandler.class);for (HttpHandler handler : handlers.values()) {if (handler instanceof ServletBean) {ServletBean bean = (ServletBean) handler;bean.init(appContext);}}super.init(config);}@Overrideprotected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException,IOException {String beanName = request.getPathInfo();HttpHandler handler = handlers.get(beanName);if (handler == null) {String wrongInfo = "cant't find servlet bean name " + beanName;log.error(wrongInfo);throw new RuntimeException(wrongInfo);}if (handler instanceof ServletBean) {ServletBean bean = (ServletBean) handler;bean.printLog(request);}handler.requestHandler(request, response);}@Overrideprotected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {doPost(req, resp);}}?
servlet 在web.xml的配置：
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<servlet><servlet-name>remote</servlet-name><servlet-class>com.cqcis.ecsc.app.servlet.ServletDispacher</servlet-class></servlet><servlet-mapping><servlet-name>remote</servlet-name><url-pattern>/remote/*</url-pattern></servlet-mapping>
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servletBean（这个是我自己定义的名字）的java代码
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package com.cqcis.ecsc.app.servlet;import java.util.Map;import javax.servlet.http.HttpServletRequest;import org.apache.commons.logging.Log;import org.apache.commons.logging.LogFactory;import org.springframework.web.context.WebApplicationContext;public class ServletBean implements HttpHandler {@SuppressWarnings("unchecked")public void requestHandler(HttpServletRequest request, HttpServletResponse response) throws ServletException,IOException {// 这里和普通的servlet一样的写}}?
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最后把servletBean 配置为springBean
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<bean name="/testDatabase" ></bean>
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最后我们可以根据路径/remote/testDatabase 来访问我们定义的servletBean
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这样带来两点好处
1.再也不需要关心web.xml里面定义大量servlet的问题了，我们只需要在spring配置文件里面定义新的bean就可以打到新增servlet的目的
2 servlet里面引用普通springBean也变得容易了，直接按照一般规则注入就行了