在jsp中，用ajax发送复杂参数，用new object()组装发送，并接受

最新推荐文章于 2024-06-16 19:39:54 发布

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最新推荐文章于 2024-06-16 19:39:54 发布

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文章标签： java ajax js jquery

本文链接：https://blog.youkuaiyun.com/id_ip/article/details/106917557

本文介绍了如何在JSP中利用Ajax发送复杂参数到后台。通过JavaScript组装参数，包括数组piclist和objanswerList，然后使用JSON.stringify()转换为字符串。在后台，通过定义对应的Bean接收并解析请求参数。

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用ajax发送请求时，有复杂参数，如下java要接受的参数

	private List<String> piclist;
    private List<QuestionAnswer> objanswerList;
    private Long homeworkid;

第一步 js进行组装
ps:piclist和objanswerList是数组

var submitParam = new Object();
 submitParam.piclist = piclist;
 submitParam.objanswerList = objanswerList;
 submitParam.homeworkid = '<%=homeworkid%>';

第二步发送ajax请求
重点：submitParam: JSON.stringify(submitParam)

$.ajax({
   type:"post",
      url:"url",
      data:{
          submitParam: JSON.stringify(submitParam)
      },
      dataType:"json",
      async:false,
      // headers:{"Content-Type":"application/json"},
      success:function(data){
          if(data.status=="1"){
             
      }
  });

第三步后台接受
1> 先定义接受参数的bean

public class SubmitParam {
    private List<String> piclist;
    private List<QuestionAnswer> objanswerList;
    private Long homeworkid;

    public List<String> getPiclist() {
        return piclist;
    }

    public void setPiclist(List<String> piclist) {
        this.piclist = piclist;
    }

    public List<QuestionAnswer> getObjanswerList() {
        return objanswerList;
    }

    public void setObjanswerList(List<QuestionAnswer> objanswerList) {
        this.objanswerList = objanswerList;
    }

    public Long getHomeworkid() {
        return homeworkid;
    }

    public void setHomeworkid(Long homeworkid) {
        this.homeworkid = homeworkid;
    }

    @Override
    public String toString() {
        return "SubmitParam{" +
                "piclist=" + piclist +
                ", objanswerList=" + objanswerList +
                ", homeworkid=" + homeworkid +
                '}';
    }
}

2> 使用String submitParamStr = request.getParameter(“submitParam”);接受参数

 String submitParamStr = request.getParameter("submitParam");
 if(submitParamStr!=null&&!"".equals(submitParamStr)){
   SubmitParam submitParam = JsonConvert.parseObject(submitParamStr, SubmitParam.class);
 }

然后就可以使用submitParam 这个参数对象了

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