Closed andreykrivoshapov closed 3 years ago
Hi @andreykrivoshapov , your modification is correct, but I suggest to use another solution that provides the best compatibility with different systems.
Instead of hard coding the directory separator char /, we can use Path.DirectorySeparatorChar as below:
/
Path.DirectorySeparatorChar
return $"{_options.ResourcesPath}{Path.DirectorySeparatorChar}{typeName}.{{0}}.xml";
fix provided with vNext update under 9647936e
Thanks, Ziya
Hi @andreykrivoshapov , your modification is correct, but I suggest to use another solution that provides the best compatibility with different systems.
Instead of hard coding the directory separator char
/
, we can usePath.DirectorySeparatorChar
as below: