javascript - spring4.3.7 獲取ajax傳遞的list等復雜類型的參數呢
問題描述
spring4.3.7如何接收ajax傳遞的list等復雜類型呢?直接傳就被報
org.springframework.beans.InvalidPropertyException: Invalid property’uAnswers0’ of bean class [cn.xxx.entity.UAnswerList]:Property referenced in indexed property path ’uAnswers0’ isneither an array nor a List nor a Map; returned value was[cn.xxx.entity.UAnswer@2d3e47ed]
搜了下網上的解決方案,基本都是以前的,3.x的居多:
1.添加 jackson jar包 :
jackson-core jackson-mapper jackson-databind
2.配置xml
<bean class='org.springframework.web.servlet.mvc.annotation.AnnotationMethodHandlerAdapter'><property name='messageConverters'> <list ><bean /> </list></property> </bean>
一個是AnnotationMethodHandlerAdapter屬于被廢棄的了,再一個就是依舊報錯,被告知
org.springframework.web.util.NestedServletException: Requestprocessing failed; nested exception isorg.springframework.beans.InvalidPropertyException: Invalid property’uAnswers0’ of bean class [cn.xxx.entity.UAnswerList]:Property referenced in indexed property path ’uAnswers0’ isneither an array nor a List nor a Map; returned value was[cn.xxx.entity.UAnswer@70671304]
詳細代碼如下:1、UAnswer類
public class UAnswer { private Integer uaid; private Integer quid;//用戶id private Integer qnid;//問卷id private Integer qid;//問題id private Integer oid;//選項id private String remark;//備注 private String createtime; public Integer getUaid() {return uaid; } public void setUaid(Integer uaid) {this.uaid = uaid; } public Integer getQuid() {return quid; } public void setQuid(Integer quid) {this.quid = quid; } public Integer getQnid() {return qnid; } public void setQnid(Integer qnid) {this.qnid = qnid; } public Integer getQid() {return qid; } public void setQid(Integer qid) {this.qid = qid; } public Integer getOid() {return oid; } public void setOid(Integer oid) {this.oid = oid; } public String getRemark() {return remark; } public void setRemark(String remark) {this.remark = remark; } public String getCreatetime() {return createtime; } public void setCreatetime(String createtime) {this.createtime = createtime; }}
2、UAnswerList類
public class UAnswerList { private Integer quid;//用戶id private Integer qnid;//問卷id private List<UAnswer> uAnswers; public Integer getQuid() {return quid; } public void setQuid(Integer quid) {this.quid = quid; } public Integer getQnid() {return qnid; } public void setQnid(Integer qnid) {this.qnid = qnid; } public List<UAnswer> getuAnswers() {return uAnswers; } public void setuAnswers(List<UAnswer> uAnswers) {this.uAnswers = uAnswers; }}
3、controller
@ResponseBody @RequestMapping('addAnswer.do') public String addAnswer(HttpServletRequest request, HttpServletResponse response, UAnswerList uAnswers){ReturnResult r = new ReturnResult();r.setCode(1);r.setMsg('findAllNowCity success');if(uAnswers==null||uAnswers.getQnid()==null||uAnswers.getQuid()==null||uAnswers.getuAnswers()==null){ r.setCode(-1); r.setMsg('尚未填寫問卷'); return r.toJsonString();} //相關操作return r.toJsonString(); }
4、ajax模擬的數據
function yaya(){ $.ajax({ url: 'http://192.168.0.251:8080/jsapi/addAnswer.do', data: {quid:'odeq5uFtR835lZGTKxLpGy9jHEDU',qnid:1,uAnswers:[ {qid:1,oid:2 }, {qid:1,oid:3 }, {qid:2,oid:7 }, {qid:3,oid:14 }, {qid:6,oid:15,remark:'hahaha' },] }, type: 'post', cache : false, dataType:’json’, success: function(data) {console.log(data);document.getElementById('resurl').innerHTML= JSON.stringify(data); }, }); }
也可能是ajax這塊數據拼接格式的問題,但暫時不知該如何解決了。。
問題解答
回答1:用@RequestBody UAnswerList uAnswers
回答2:@ResponseBody @RequestMapping('addAnswer.do') public String addAnswer(HttpServletRequest request, HttpServletResponse response, @RequestBody UAnswerList uAnswers){ReturnResult r = new ReturnResult();r.setCode(1);r.setMsg('findAllNowCity success');if(uAnswers==null||uAnswers.getQnid()==null||uAnswers.getQuid()==null||uAnswers.getuAnswers()==null){ r.setCode(-1); r.setMsg('尚未填寫問卷'); return r.toJsonString();} //相關操作return r.toJsonString(); }回答3:
2個地方需要修改:
使用@RequestBody注解,表示使用http body的內容,如下:
@ResponseBody@RequestMapping('addAnswer.do')public String addAnswer(HttpServletRequest request, HttpServletResponse response, @RequestBody UAnswerList uAnswers){ 省略 }
js端要使用json傳送數據,所以需要指定application/json類型并使用JSON.stringify來將對象轉成json格式,如下:
function yaya(){ $.ajax({ url: 'http://192.168.0.251:8080/jsapi/addAnswer.do', contentType:'application/json', //新增 data: JSON.stringify({quid:'odeq5uFtR835lZGTKxLpGy9jHEDU',qnid:1,uAnswers:[ {qid:1,oid:2 }, {qid:1,oid:3 }, {qid:2,oid:7 }, {qid:3,oid:14 }, {qid:6,oid:15,remark:'hahaha' },] }), type: 'post', cache : false, dataType:’json’, success: function(data) {console.log(data);document.getElementById('resurl').innerHTML= JSON.stringify(data); }, }); }
相關文章:
1. Java反射問題:為什么android.os.Message的recycleUnchecked方法不能通過反射獲取到?2. 如何分別在Windows下用Winform項模板+C#,在MacOSX下用Cocos Application項目模板+Objective-C實現一個制作游戲的空的黑窗口?3. html5和Flash對抗是什么情況?4. php如何獲取訪問者路由器的mac地址5. javascript - 在 vue里面用import引入js文件,結果為undefined6. 前端 - node vue webpack項目文件結構7. 小程序怎么加外鏈,語句怎么寫!求救新手,開文檔沒發現8. python - linux怎么在每天的凌晨2點執行一次這個log.py文件9. javascript - vue-resource中如何設置全局的timeout?10. thinkPHP5中獲取數據庫數據后默認選中下拉框的值,傳遞到后臺消失不見。有圖有代碼,希望有人幫忙
