(Spring) jsonview (ModelAndView를 이용한 ajax 사용)

spring 프레임워크에서 간단하게 ajax(json data) 사용법

pom.xml

<!-- Json -->           

<groupId>net.sf.json-lib</groupId>       

    <artifactId>json-lib</artifactId>       

    <version>2.4</version>       

    <classifier>jdk15</classifier>   

</dependency>

        

<dependency>

    <groupId>org.codehaus.jackson</groupId>

    <artifactId>jackson-mapper-asl</artifactId>

    <version>1.9.2</version>

</dependency>

////////////////////////////////////////////////////////////

DispatcherServlet XML 설정파일

<bean id="jsonView" class="org.springframework.web.servlet.view.json.MappingJacksonJsonView">

<property name="contentType" value="application/json;charset=UTF-8"> </property>

</bean>

////////////////////////////////////////////////////////////

web.xml

<servlet-mapping>

<servlet-name>action</servlet-name>

<url-pattern>*.do</url-pattern>

</servlet-mapping>

<servlet-mapping>

<servlet-name>action</servlet-name>

<url-pattern>*.ajax</url-pattern>

</servlet-mapping>

////////////////////////////////////////////////////////////

Controller.java

@RequestMapping(value = "/login/selectLogin.ajax")

public ModelAndView login(CommandMap commandMap) throws Exception {

Map resultMap = new HashMap();

resultMap.put("result", 1);

ModelAndView modelAndView = new ModelAndView("jsonView", resultMap);

return modelAndView;

}

////////////////////////////////////////////////////////////

javascript

function login() {

jQuery.ajax(

{async : true

,type : "post"

,url : "/login/selectLogin.ajax"

,dataType : "json"

,data: $("#loginFrm").serialize()

,success : function(jsonData) {

var _login_result = jsonData.result;

if(_login_result == 1){

alert("로그인 성공!\n환영합니다.");

$("#loginFrm").attr("action", "/login/loginSuccess.do").submit();

} else {

alert("로그인 실패하였습니다. 아이디와 비밀번호를 재확인해 주세요.");

}

}

,error: function(request,status,error){

alert("code:"+request.status+"\n"+"message:"+request.responseText+"\n"+"error:"+error);

}

});

}