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1.é¦å ä¸è¿ä¸¤ä¸ªå commons-fileupload-1.2.1.jarï¼commons-io-1.3.2.jar
2.ç¼ååå°é¡µé¢
<%@ page language="java" pageEncoding="gbk"%>
<%@ taglib uri="/struts-tags" prefix="s" %>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4. Transitional//EN">
<html>
<body>
<form action="<%=request.getContextPath()%>/UploadServlet" method="post" enctype="multipart/form-data">
username:<input type="text" name="username"><br>
password:<input type="password" name="password"><br>
file:<input type="file" name="file"><br>
<input type="submit" value="submit"><br>
</form>
</body>
</html>
3.ç¼åservlet
package cn.jci.upload.servlet;
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.util.List;
import javax.servlet.ServletException;
import javax.servlet..jci.upload.servlet.UploadServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>UploadServlet</servlet-name>
<url-pattern>/UploadServlet</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
</web-app>
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不能访问web-inf下JSP页面
因此,可以让servlet进行访问,如web-inf下有a.jsp则可以用request.getrequestdispatcher("/web-inf/a.jsp").forward(request,response);进行派遣访问.但如果web-inf下有a.htm,则用request.getrequestdispatcher("/web-inf/a.htm").forward(request,response);就不能访问.至于原理的话,可以去看看Tomcat的稳操胜券源码源码。多多学习。暗黑战神 源码
小黄鸡源码2024-11-20 11:48
2024-11-20 11:47
2024-11-20 11:25
2024-11-20 10:38
2024-11-20 09:58