如果我有这样的ExpandoObject:
dynamic d = new ExpandoObject();
d.x = "a";
d.y = "b";然后使用Serilog将其记录到RollingFile,使用如下JsonFormatter:
_logger.Debug("{@d}", d);它将像这样被序列化为json:
[{"_typeTag":"KeyValuePair`2","Key":"x","Value":"a"},{"_typeTag":"KeyValuePair`2","Key":"y","Value":"b"}]如果我使用Newtonsoft.Json来像这样序列化相同的ExpandoObject:
JsonConvert.SerializeObject(d)我去拿这个:
{"x":"a","y":"b"}如何使Serilog产生与Newtonsoft.Json相同的json?
发布于 2018-02-25 22:51:43
增加:
.Destructure.ByTransforming<ExpandoObject>(e => new Dictionary<string,object>(e))你的LoggerConfiguration应该这么做。
https://stackoverflow.com/questions/48958444
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