reference:http://json-lib.sourceforge.net/apidocs/jdk15/index.html
This tutorial will walk through how to configure Spring MVC to return a JSON object to client browser.
One of the main decisions to be taken while developing AJAX applications is the format of messages passed by the server to the client browser. There are many options to choose from including plain text, XML, CSV etc. One of the more popular choices today is the JavaScript Object Notation (JSON). JSON provides a nice name-value pair data format that is easy to generate and parse.
How to do it?
You can use json-lib-ext-spring. There are other libs, this is the one I found. If you know or use another one, please leave a comment with the library name.
Do not forget to download Json-lib and its dependencies.
pom.xml file:
<!-- Jackson for AJAX -->
<dependency>
<groupId>net.sf.json-lib</groupId>
<artifactId>json-lib-ext-spring</artifactId>
<version>1.0.2</version>
<exclusions>
<exclusion>
<artifactId>log4j</artifactId>
<groupId>log4j</groupId>
</exclusion>
</exclusions>
</dependency>
Now you have to configure your XML files:
Create a views.xml file under WEB-INF folder and paste the following code into it:
<?
xml
version
=
"1.0"
encoding
=
"UTF-8"
?>
<
beans
xmlns
=
"http://www.springframework.org/schema/beans"
xmlns:xsi
=
"http://www.w3.org/2001/XMLSchema-instance"
xmlns:util
=
"http://www.springframework.org/schema/util"
xsi:schemaLocation= "http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-2.5.xsd
http://www.springframework.org/schema/util
http://www.springframework.org/schema/util/spring-util-2.5.xsd">
<
bean
name
=
"jsonView"
class
=
"net.sf.json.spring.web.servlet.view.JsonView"
/>
</
beans
>
or
<bean id="jsonView" class="net.sf.json.spring.web.servlet.view.JsonView">
<property name="contentType" value="application/json; charset=UTF-8" />
</bean>
Add this config to you spring configuration file:
<!-- json -->
<
bean
id
=
"xmlFileViewResolver"
class
=
"org.springframework.web.servlet.view.XmlViewResolver"
>
<
property
name
=
"location"
>
<
value
>/WEB-INF/views.xml</
value
>
</
property
>
<
property
name
=
"order"
>
<
value
>1</
value
>
</
property
>
</
bean
>
Make sure to set the order if you are using any other view resolvers.
or
<bean class="org.springframework.web.servlet.view.XmlViewResolver" p:order="0"
p:location="/WEB-INF/ajaxviewresolver.xml" />
Now you just have to use “jsonView” as the viewname and the model will be converted to JSONbefore being sent back to the client:return
new
ModelAndView(
"jsonView"
, modelMap);
Here is an example:
public
ModelAndView getColumnsJson(HttpServletRequest request,
HttpServletResponse response)
throws
Exception {
Map<String,Object> modelMap =
new
HashMap<String,Object>(
2
);
modelMap.put(
"rows"
, service.generateColumns());
return
new
ModelAndView(
"jsonView"
, modelMap);
}
project sample1:
@RequestMapping(value = "/ajax/getCustomerDetailsById", method = RequestMethod.POST)
public String getCustomerDetailsById(@RequestParam("id") Integer customerInternalKey, final ModelMap model) {
CustomerOrganizationSite customerSite = customerOrganizationService
.retrieveSingleCustomerOrganizationSiteByCustomerOrganizationSiteId(1,
assessmentAuthInfo.getApplicationId(), customerInternalKey);
model.addAttribute("customerSite", customerSite);
return JSON_VIEW;
}
project sample2:
try {
JSONObject jsonObject = new JSONObject();
jsonObject.put("validateResult", validateResult);
jsonObject.put("errorMessage", errorMessage);
response.getWriter().println(jsonObject.toString());
} catch (IOException e) {
logger.error("IO exception happened");
}
Happy coding!