java - 错误 - 错误的请求；检查错误消息以获取详细信息

Question

我用 spring 创建了 API。当我调用该 API 时显示错误

这是 Controller 类。

@RequestMapping(value={"/dPIUsagePackageInfo"},method=RequestMethod.POST)
  public ResponseEntity<DPIUsagePackageInfoRs> dPIUsagePackageInfo(@RequestBody List<DPIUsagePackageInfoRq> dPIUsagePackageInfoRq){

      //
      DPIUsagePackageInfoRs response = this.ccbsBusiness.dPIUsagePackageInfo(dPIUsagePackageInfoRq);
      return new ResponseEntity(response, response.getStatus());
  }

这是请求类

public class DPIUsagePackageInfoRq {    

    private List<String> srvName;


    public List<String> getSrvName() {
        return srvName;
    }

    public void setSrvName(List<String> srvName) {
        this.srvName = srvName;
    }


}

我传递了这个json主体

{
    "dPIUsagePackageInfoRq" : {
        "srvName": ["xxx","rrr","rrrrr"]
    }
}

但是这样的回应

{
    "resultCode": "000400",
    "resultDesc": "ERROR - Bad request; check the error message for details."
}

我的代码哪里有问题。在此先感谢。

Answer 1

您只发送一个 DPIUsagePackageInfoRq 实例，您应该在列表中发送它。请尝试在正文的开头和结尾添加“[”和“]”，使其成为一个列表。

您发送的请求应如下所示:

[
    {
        "srvName": ["xxx","rrr","rrrrr"]
    }
]

如果您想发送多个 DPIUsagePackageInfoRq 对象，您可以像这样使用递增对象:

[
    {
        "srvName": ["xxx","rrr","rrrrr"]
    },
    {
        "srvName": ["xxx","rrr","rrrtrr"]
    }
]