Sharing violation when loading mime message from file

[alex-becker-startp]

Hi,

the method MimeMessage.Load(options, filePath) fails when the file is already opened by another process.
This is because you do not specify the FileShare attribute within the File.Open method.
To prevent the issue, modify the code as following:

using (FileStream fileStream = File.Open(fileName, FileMode.Open, FileAccess.Read))
using (FileStream fileStream = File.Open(fileName, FileMode.Open, FileAccess.Read, FileShare.Read))

Alex.

[jstedfast]

Thanks for catching this!