Servlet configuration method in tomcat environment

xiaoxiao2021-03-06  43

Write on Web.xml: voteservlet vote.voteServlet where voteServlet is the name of the file Vote is a file Package name. voteServlet / voteservlet where the VoteServlet of the first line is the name, the second line The / voteServlet is also the name of the file, just add one / stop; if implemented in servlet technology: get commands as long as you register in Web.xml. URL is: and the file add name letter in the same directory in the Tomcat directory, add the name of the servlet. For example, the folder named in EXAMPLE is: MyApp's servletServlet1 file name under the Class folder inside: TEST: URL is: http: // localhost: 8080 / myApp / test Remember not to file the name of the file Upper suffix: .jsp. If implemented in servlet technology is: post command, you must register in Web.xml. The front desk must have a JSP file call, pay attention to the writing of the action path in the form, of course, Method must write "POST" first write "/" then write and write the name of the folder in the examples, such as: myApp and then write "/" then Write the name of the servlet file. For example: servlet1. That is: action = "/ myApp / servlet1".

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