SPRINGMVC Study Notes

First, SPRINGMVC environment construction

1. Create a new file and select Java→pack Explore view

2, the Requai under the Libs under the jar package into the

3. Web. XML start Configuration Springmvc

<servlet>      <Servlet-name>Springmvc</Servlet-name>      <Servlet-class>Org.springframework.web.servlet.DispatcherServlet</Servlet-class>"It's supposed to be under the Spring-webmvc-3.2.0.release.jar under this jar.org.springframework.web.servlet root directory】
         <Init-param>              <Param-name>Contextconfiglocation</Param-name>              <Param-value>Classpath*:config/spring-servlet.xml</Param-value>          </Init-param>      <Load-on-startup>1</Load-on-startup>  </servlet>    <servlet-mapping>      <Servlet-name>Springmvc</Servlet-name>      <Url-pattern>/</Url-pattern>  </servlet-mapping>

4. Configuring the SRIMGMVC configuration file

<?XML version= "1.0" encoding= "UTF-8"?><Beansxmlns= "Http://www.springframework.org/schema/beans"Xmlns:context= "Http://www.springframework.org/schema/context"xmlns:p= "http://www.springframework.org/schema/p"Xmlns:mvc= "Http://www.springframework.org/schema/mvc"Xmlns:xsi= "Http://www.w3.org/2001/XMLSchema-instance"xsi:schemalocation= "Http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/ Spring-beans-3.0.xsd Http://www.springframework.org/schema/context Http://www.springframework.org/schema/con Text/spring-context.xsd Http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/ Spring-mvc-3.0.xsd ">                  <Beanname= "/test1/helloworld"class= "Com.tgb.web.controller.HelloWorldController" />        <BeanID= "Viewresolver"class= "Org.springframework.web.servlet.view.InternalResourceViewResolver">        < Propertyname= "prefix"value="/"></ Property>        < Propertyname= "suffix"value= ". jsp"></ Property>    </Bean> </Beans>  

The above class= "Org.springframework.web.servlet.view.InternalResourceViewResolver"

Under the Spring-webmvc-3.2.0.release.jar.

Org.springframework.web.servlet.view under the

Parser for Internalresourceviewresolver.class View decomposition

Second, section II

1, create a new controller.

 PackageCom.tgb.web.controller;Importjavax.servlet.http.HttpServletRequest;ImportJavax.servlet.http.HttpServletResponse;ImportOrg.springframework.web.servlet.ModelAndView;ImportOrg.springframework.web.servlet.mvc.Controller; Public classHelloworldcontrollerImplementscontroller{@Override PublicModelandview handlerequest (httpservletrequest arg0, httpservletresponse arg1)throwsException {System.out.println ("HelloWorld, video of the first Springmvc"); return NewModelandview ("/welcome"); }}

How does spring find this controller?

In spring, the configuration file Spring-servlet.xml "defaults to its path and to the Web. XML sibling. Web.xml→spring-servlet.xml "

Could not find the resource file Spring-servlet.xml "solution: Spring-servlet.xml modified to Springmvc-servlet.xml. Modified to the default filename."

Fill in <bean name= "/test1/helloworld" class= "Com.tgb.web.controller.HelloWorldController"/>

Finally: Browser input http://localhost:8080/springMVC1/test1/helloworld can be accessed

Iii. modifying configuration files and passing parameters to the interface

1. The configuration file is not in the default path

Create a new config directory under SRC,

View the source code for the Distributor. Can know how to find the directory. To Contextconfiglocation

Modifying the initialization parameters of Web. XML SPRINGMVC

    < Init-param >              < Param-name >contextconfiglocation</param-name>              <  Param-value>classpath*:config/spring-servlet.xml</param-value >          </ Init-param >

2. Passing parameters

Inside control//return string number and//the following is the return map data

String Hello = "lsh Hello accelerated";
map<string,object> map = new hashmap<string,object> ();
Map.put ("Map1", "Accelerated 1");
Map.put ("Map2", "Accelerated 2");
Map.put ("Map3", "Accelerated 3");
Returns a String data
return new Modelandview ("/welcome", "result", "hello");
Here is the return map data
return new Modelandview ("/welcome", "map", map);

To use the ESTL expression, introduce the following two packages.

Jstl's jar Package

standard-1.1.2 's jar Package

Iv. SPRINGMVC implement a controller to write multiple methods

Create a new controller

1, public class Multicontroller inherit a Multiactioncontroller

Package Com.tgb.web.controller;Import Javax.servlet.http.HttpServletRequest;Import Javax.servlet.http.HttpServletResponse;Import Org.springframework.web.servlet.ModelAndView;Import Org.springframework.web.servlet.mvc.multiaction.MultiActionController;Public class Multicontroller extends Multiactioncontroller {public Modelandview Add (httpservletrequest request,httpservletresponse Response) {System.out.println ("----Add method OK----");return new Modelandview ("/multi", "Method", "add");} public Modelandview update (HttpServletRequest request,httpservletresponse Response) {System.out.println ("-----Update---method");return new Modelandview ("/multi", "Method", "Update");    }            }

2, SPRINGMVC configuration inside how to write, is under the Springmvc-servlet.xml

 <  bean  id  = "Parammethodresolver"   class  = "Org.springframework.web.servlet.mvc.multiaction.ParameterMethodNameResolver " >  <  property  name  = "ParamName"  Span style= "color: #ff0000;" > value  = "action"  ></ property  >  </ bean  >  

Under the Spring WEBMVC Rack package under MVC multiaction. Parametermethodnameresolver

Value= "Action"> when requested, the parameter also changes with this.

A parser for the request token

Continue to configure

 <Beanname= "/test1/multi"class= "Com.tgb.web.controller.MultiController">        < Propertyname= "Methodnameresolver">            <refBean= "Parammethodresolver"/>        </ Property>    </Bean>    

Then browser input: Http://localhost:8080/springMVC1/test1/multi?action=add

Not requested, solution public Modelandview Add (httpservletrequest request,httpservletresponse response) requires two parameters in parentheses

Section V, 5th, access to static files

1. Create a new controller

 PackageCom.tgb.web.controller;Importjavax.servlet.http.HttpServletRequest;ImportJavax.servlet.http.HttpServletResponse;ImportOrg.springframework.web.servlet.ModelAndView;ImportOrg.springframework.web.servlet.mvc.multiaction.MultiActionController; Public classStaticcontrollerextendsMultiactioncontroller { Publicmodelandview img (httpservletrequest request,httpservletresponse response) {return NewModelandview ("/staticfile"); }         Publicmodelandview jquery (httpservletrequest request,httpservletresponse response) {return NewModelandview ("/jquery"); }    }

2. Create a new JSP

<%@ page language= "java" contenttype= "text/html; Charset=utf-8 "    pageencoding=" UTF-8 "%><% @taglib prefix=" C "uri=" http://java.sun.com/jsp/jstl/ Core "%><! DOCTYPE HTML PUBLIC "-//w3c//dtd HTML 4.01 transitional//en" "Http://www.w3.org/TR/html4/loose.dtd" >
<Beanname= "/test1/img"class= "Com.tgb.web.controller.StaticController">        < Propertyname= "Methodnameresolver">            <refBean= "Parammethodresolver"/>        </ Property>    </Bean>    
The address is correct, the picture cannot be taken, the static file cannot get
Continue to write an MVC tag in the SPRINGMVC configuration file.
Do not intercept, direct access
<!--static resource access-     <mvc:resources location= "/img/" mapping= "/img/**"/> <mvc:resources location=       "/js/" mapping= "/js/**"/>       
SPRINGMVC Study Notes