我需要在控制器中手动呈现ModelAndView,以便将其放入JSON object中。如果我将整个ModelAndView对象传递给JSON,我会得到“找不到javassistlazyinitializer类的序列化器”异常,因为jackson不能正确处理惰性对象。谢谢
发布于 2011-06-30 16:15:39
public class JSONView implements View {
/**
* Logger for this class
*/
private static final Logger logger = Logger.getLogger(JSONView.class);
private String contentType = "application/json";
public void render(Map map, HttpServletRequest request, HttpServletResponse response)
throws Exception {
if(logger.isDebugEnabled()) {
logger.debug("render(Map, HttpServletRequest, HttpServletResponse) - start");
}
JSONObject jsonObject = new JSONObject(map);
PrintWriter writer = response.getWriter();
writer.write(jsonObject.toString());
if(logger.isDebugEnabled()) {
logger.debug("render(Map, HttpServletRequest, HttpServletResponse) - end");
}
}
public String getContentType() {
return contentType;
}
}ModelAndView returnModelAndView = new ModelAndView(new JSONView(), model);
https://stackoverflow.com/questions/6531764
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