2 references to RunFileApiJsonSerializerContext
dotnet (2)
Commands\Run\Api\RunApiCommand.cs (2)
31
RunApiInput input = JsonSerializer.Deserialize(line,
RunFileApiJsonSerializerContext
.Default.RunApiInput)!;
45
string json = JsonSerializer.Serialize(message,
RunFileApiJsonSerializerContext
.Default.RunApiOutput);