How to grep on a line that contains a semicolon; ?

Question

linux

Hello, I need to "grab" from such a line

ID=PMZ_0036570;Dbxref=InterPro:IPR000175,PANTHER:PTHR11616,PANTHER:PTHR11616:SF125,Pfam:PF00209,ProSiteProfiles:PS50267,SUPERFAMILY:SSF161070;Name=Slc6a18;desc=Sodium-dependent

a section called Name=blablabla (in this particular case - Name=Slc6a18)
I get the above line using and then I try to do
awk '{print $9}' PPP_genes.gff3.txt | head -3

awk '{print $9}' PPP_genes.gff3.txt | head -3 | grep "Name=" PPP_genes.gff3.txt

and nothing comes out .. I
tried to do it cut -d; -f 3(this is the 3rd column) - I get an error, they say -d does not understand the semicolon ...
In general, I may have done something, but I just need to "grab" this from each line here is the very Name.
Tell me, please, how to do it?
Thanks

Answer 1

• awk '{print $9}' PPP_genes.gff3.txt | head -3 | cut -d";" -f 3

  
  

•  &&
      echo ${BASH_REMATCH[1]}

  
  

Answer 2

Compose a regexp starting with Name=and ending with a semicolon. What's in the middle - and pull out.
https://regex101.com