Open rixed opened 2 years ago
This does not parse: avg [s1; s2; s3] as s but this does: avg([s1; s2; s3]) as s.
avg [s1; s2; s3] as s
avg([s1; s2; s3]) as s
This does not parse:
avg [s1; s2; s3] as s
but this does:avg([s1; s2; s3]) as s
.