Question

我正在尝试将jsonObject从jsp发布到控制器，我收到400 Bad Request Error。

我的jsp代码

$.ajax({
   url : 'save.web',
   type: "post", 
   dataType: 'json',
   contentType:'application/json',
   data: "data="+JSON.stringify(jsonArray),
   success:function(data){

   }
   }); 

   });

  My json Object
var jsonArray="{"+'"'+"cds"+'"'+":"+"["; 
for(i=0;i<newRow.length;i++)
{
 jsonArray=jsonArray+"{"+'"'+"dno"  +'":'+newRow[0]+","+'"'+"CampaignCode"  
 +'":'+newRow[1]+","  +'"'+"CampaignName"+'":'+newRow[2]+","  
 +'"'+"ServiceNo"  +'":'+newRow[3]+"}";
}
jsonArray=jsonArray+"]}";

我的控制器代码

 @RequestMapping(value = "/save.web", method = RequestMethod.POST)
 @ResponseBody
 public String save(WebRequest webRequest,Model model,HttpServletRequest    
 request, HttpServletResponse response,@RequestParam CampaignDisplay[] cds)        
{

    for(CampaignDisplay inputs : cds){
        System.out.println(inputs.getId());
        }
    return "menu";

}

输入我传递的Json对象 { “CDS”：[{ “DNO”：8中， “CampaignCode”：d “CAMPAIGNNAME”：E， “服务无”：F}，
{“dno”：7，“CampaignCode”：a，“CampaignName”：b，“ServiceNo”：c}]}

我不知道为什么我会得到400个错误？我也设置了标题

Answer 1

似乎控制器无法理解请求。原因是“您正在发布JSON，而不是表单数据，但您尝试将其作为表单参数读取”。将您的方法更改为

**public String save(WebRequest webRequest,Model model,HttpServletRequest    
 request, HttpServletResponse response,@RequestBody Map<String, Object> inputParameter)        
{**

点击此链接以供参考 400 Bad request on Spring Jquery Ajax Post

发布Json对象时400 Bad Request

1 个答案: