Struts2 + jQuery 实现ajax (二) 回调函数
情景描述：
???? 利用Struts2 + jQuery 实现ajax时候，如何获取返回值的信息来正确的提示用户。例如给某个电影打分rating，每个用户只能对一个电影打一次分数，如果打第二次的时候就提示用户"已经打过分了"。
1.js代码：function ratingMovieClick(userId,num,movieId){if (userId == 0) {alert("You have to be logged in to rating !")return;}else{var url = "../RatingMovie.action";var params = {userId : userId,num : num,movieId : movieId};jQuery.post(url, params, callbackFun, 'json');}}function callbackFun(data) {alert(data);}?
2.配置文件：注意这里的<param name="root">result</param>这句将请求处理返回的结果带回到页面中。<!DOCTYPE struts PUBLIC "-//Apache Software Foundation//DTD Struts Configuration 2.0//EN" "http://struts.apache.org/dtds/struts-2.0.dtd"><struts><package name="Struts2_AJAX" extends="json-default"><action name="RatingMovie" method="addRatingMovie"><result type="json"><param name="root">result</param></result></action></package></struts>
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3.java代码：这里定义的result就是在配置文件里面定义的，这2个要一致。public class RatingAction extends AbstractAction{private double avg_rating ;private int count;private String result;private static String CALL_BACK_1 = "You have ratinged !";private static String CALL_BACK_2 = "Thanks for rating !";private IRatingService ratingService;private IMovieService movieService;/** * check the records. * @param obj1 * @param obj2 * @return */public int checkRelatedTableRecord(long obj1, long obj2){if(ratingService == null){count = getRatingService().getRecordRating(obj1, obj2);}else{count = ratingService.getRecordRating(obj1, obj2);}return count;}public String addRatingMovie(){int record = this.checkRelatedTableRecord(userId, movieId);if(record > 0){this.result = CALL_BACK_1;return "success";}else{Related related = new Related();related.setType(Related.TYPE_RATING_MOVIE);related.setObj1(userId);related.setObj2(movieId);related.setDegree(num);related.setCreated(new Date());related.setModified(new Date());ratingService.addRating(related);avg_rating = ratingService.getAvgRating(Related.TYPE_RATING_MOVIE, movieId);Movie movie = movieService.getMovieById(movieId);movie.setRating(new Float(avg_rating));movieService.updateMovie(movie);this.result = CALL_BACK_2;return "success";}}public void setRatingService(IRatingService ratingService) {this.ratingService = ratingService;}public void setMovieService(IMovieService movieService) {this.movieService = movieService;}public String getResult() {return result;}public void setResult(String result) {this.result = result;}}?
??? 这样当record > 0时，数据库里面已经有记录的时候，就this.result = CALL_BACK_1;return "success"; 这里的result就会赋值成"You have ratinged !",这样在回调函数中就可以拿到这个字符串了。这里的返回值只能是"success",我试过好几个都会报错！
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下面是某个参考某论坛上的信息：http://www.iteye.com/topic/560638
一、准备一个JSP页面用于提交ajax请求，这里我使用了JQuery的$.getJSON(url,params,function callback(data))函数提交ajax请求到指定url，并且携带参数params，最后用一个回调函数callback处理请求返回结果data；二、一个处理请求的Action类，并在struts.xml文件中做相应配置：写一个action类处理ajax请求数据，并将返回结果封装成一个JSONObject对象返回给请求页面。同时在struts.xml中配置对应action，指明其返回类型为json并使其package的extends为json-default，并将要返回请求页面的数据放在名为root的param中，如<param name="root">result</param>。三、接受请求返回结果：使用JS的eval方法将返回结果data转换成JSON对象，并处理返回结果。$.getJSON(url,params,function callback(data){ // convert to json object var user = eval("("+data+")");// $("#result").each(function(){ $(this).html('welcome ,' + user.name); }); }); } public String login() throws Exception {?? ?????????? ??????? Map map = new HashMap();?? ??????? map.put("name", user.getLogName());?? ??????? map.put("password",user.getPassword());?? ??????? JSONObject obj = JSONObject.fromObject(map);?? ?????????? ?????????? ??????? result = obj.toString();?? ?????????? ??????? return SUCCESS;?? ??? }? ?
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