Print lines that match regex on xth string

Hello,

I need an awk command to print only the lines that match regex on xth field from file.
For example if I use this command

awk -F"|" ' $22 == "20130117090000.*" '

It wont work, I think, because single quotes wont allow the usage of the metacharacter star * . On the other hand I dont know what other syntax should I use to avoid the usage of the single quotes.

awk -F"|" '$22 ~ /20130117090000/' ...

Hello, thank you, yes this was what I was looking for. Didn't think to use "~"