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04-Idea模板创建Servlet(快速创建servlet及快速创建servlet的代码模板)

发布时间：2024-11-28 18:24:48编辑：雪饮阅读（）

servlet可以在IntelliJ IDEA中快速创建，一般在你包名鼠标右键New=>Servlet

输入好servlet的名称，就能快速创建一个servlet如：
package com;

import javax.servlet.*;
import javax.servlet.http.*;
import javax.servlet.annotation.*;
import java.io.IOException;

@WebServlet(name = "Servlet11", value = "/Servlet11")
public class Servlet11 extends HttpServlet {
    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    }

    @Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    }
}
但是这个servlet中竟然还有name，我们一般的servlet如果是没有name的，就是最好直接传入一个相对patterns的单值的WebServlet注解即可。
那么这个固定的生成模板是可以修改的，位于Settings=>Editos=>File and Code Templates=>Other=>Web=>Java code templates=>Servlet Annotated Class.java
这里的模板如
#if (${PACKAGE_NAME} && ${PACKAGE_NAME} != "")package ${PACKAGE_NAME};#end
#parse("File Header.java")

#if ($JAVAEE_TYPE == "jakarta")
import jakarta.servlet.*;
import jakarta.servlet.http.*;
import jakarta.servlet.annotation.*;
#else
import javax.servlet.*;
import javax.servlet.http.*;
import javax.servlet.annotation.*;
#end
import java.io.IOException;

@WebServlet(name = "${Entity_Name}", value = "/${Entity_Name}")
public class ${Class_Name} extends HttpServlet {
    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    }

    @Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    }
}
那么所以我可以修改成如这样：
#if (${PACKAGE_NAME} && ${PACKAGE_NAME} != "")package ${PACKAGE_NAME};#end
#parse("File Header.java")

#if ($JAVAEE_TYPE == "jakarta")
import jakarta.servlet.*;
import jakarta.servlet.http.*;
import jakarta.servlet.annotation.*;
#else
import javax.servlet.*;
import javax.servlet.http.*;
import javax.servlet.annotation.*;
#end
import java.io.IOException;

@WebServlet("/${Entity_Name}")
public class ${Class_Name} extends HttpServlet {
    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    }

    @Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    }
}
那么在下一次创建servlet就只有一个url属性的webservlet的注解了。
package com;

import javax.servlet.*;
import javax.servlet.http.*;
import javax.servlet.annotation.*;
import java.io.IOException;

@WebServlet("/Servlet12")
public class Servlet12 extends HttpServlet {
    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    }

    @Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    }
}