Question

如何在spring-boot中自定义dispatchServlet的访问路径

Answer 1

在程序的主入口中定义dispatchRegistration
<span style="font-size:18px;">package com.lzl;

import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.boot.web.servlet.ServletComponentScan;
import org.springframework.boot.web.servlet.ServletRegistrationBean;
import org.springframework.context.annotation.Bean;
import org.springframework.web.servlet.DispatcherServlet;

import com.lzl.servlet.MyServlet;

@SpringBootApplication
public class SpringBootDemoApplication {

public static void main(String[] args) {
SpringApplication.run(SpringBootDemoApplication.class, args);
}

@Bean
public ServletRegistrationBean dispatcherRegistration(DispatcherServlet dispatcherServlet) {
ServletRegistrationBean reg = new ServletRegistrationBean(dispatcherServlet);
reg.getUrlMappings().clear();
reg.addUrlMappings("*.html");
reg.addUrlMappings("*.do");
return reg;
}
}</span>

Answer 2

添加mvc配置，新增application.yml

server:

port: 8081 #端口

contextPath: /PojectName #项目名 (不配置默认 "/" 根)

spring:

mvc:

view:

prefix: /page/ #前缀

suffix: .jsp #后缀