Springboot入门到精通-fastJson返回结果敏感字段剔除

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2022-05-27

百度发现有个filter 方法 SimplePropertyPreFilter

 


SimplePropertyPreFilter filter = new SimplePropertyPreFilter(); filter.getExcludes().add("name"); filter.getExcludes().add("weight"); filter.getExcludes().add("sex"); return ApiResult.success(JSONObject.parse(JSONObject.toJSONString(poJo , filter, SerializerFeature.WriteMapNullValue)));

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