小编Vin*_*eng的帖子

Spring Web 客户端返回异常“bodyType=[responseObject] 不支持内容类型‘application/json’”

我使用Web客户端发布第三方API,第三方API返回响应如下:

\n
{\n    "RetailTransactionGenericResponse": {\n        "authID": 1146185,\n        "product": {\n            "amount": 20,\n            "balance": 0,\n            "status": "D",\n            "metafields": {\n                "metafield": [\n                    {\n                        "name": "serialNum",\n                        "value": 8095843490\n                    }\n                ]\n            }\n        },\n        "responseCode": 0,\n        "responseText": "Success",\n        "transactionID": "b0924cca-f2a9-477f-915b-9153e74ebce0"\n    }\n}\n
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所以在我的payload包中,它有RetailTransactionGenericResponseWrapper类,其中包含RetailTransactionGenericResponse类。

\n

RetailTransactionGenericResponse这是类的代码

\n
import com.fasterxml.jackson.annotation.JsonCreator;\nimport com.fasterxml.jackson.annotation.JsonInclude;\nimport com.fasterxml.jackson.annotation.JsonProperty;\nimport lombok.AllArgsConstructor;\nimport lombok.Builder;\nimport lombok.Data;\nimport lombok.ToString;\n\n@Data\n@ToString\n@Builder\n@AllArgsConstructor(onConstructor_ = @JsonCreator)\n@JsonInclude(JsonInclude.Include.NON_NULL)\npublic class RetailTransactionGenericResponse {\n\n    @JsonProperty\n    private String authID;\n\n    @JsonProperty\n    private Product product;\n\n    @JsonProperty\n    private String responseCode;\n\n    @JsonProperty\n    private String responseText;\n\n    @JsonProperty\n    private String transactionID;\n}\n …
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java spring-boot spring-webclient

8
推荐指数
1
解决办法
8087
查看次数

打字稿获取当前位置并传递给后端以获取结果并传递给材料表数据源

我有我的 Angular 2 项目试图获取用户的当前位置,因此在我的班级中我有以下代码:

export class CurrentLocation {
  constructor(private http: Http) { }
  private lat : any;
  private lon : any;
  private params  = new URLSearchParams();
  private url = 'api/search/location';


  getPosition = (lat, lon) => {
    navigator.geolocation.getCurrentPosition((position) => { 
      this.lat = position.coords.latitude; 
      this.lon = position.coords.longitude;
     });
  }
  getCurrentLocation(): Observable<any> {
    this.getPosition(this.lat, this.lon);
    console.log(this.lat+ "," + this.lon);
    //console.log(this.lat + ":" + this.lon);
    this.params.set('lat', this.lat);
    this.params.set('lon', this.lon);
    //console.log(this.params.toString());
    var result = this.http.get(this.url, { search: this.params });
    result.toPromise();
    return result;
  }
}
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但是 …

typescript angular

5
推荐指数
1
解决办法
1万
查看次数