Display all fields except the last

Both these sed and awk solutions work independent of the number of fields.

Using sed:

$ sed -r 's/(.*)\..*/\1/' file1.2.3.4sanma.namc.d.b

Note: -r is the flag for extended regexp, it could be -E so check with man sed. If your version of sed doesn't have a flag for this then just escape the brackets:

sed 's/\(.*\)\..*/\1/' file1.2.3.4sanma.namc.d.b

The sed solution is doing a greedy match up to the last . and capturing everything before it, it replaces the whole line with only the matched part (n-1 fields). Use the -i option if you want the changes to be stored back to the files.

Using awk:

$ awk 'BEGIN{FS=OFS="."}{NF--; print}' file1.2.3.4sanma.namc.d.b

The awk solution just simply prints n-1 fields, to store the changes back to the file use redirection:

$ awk 'BEGIN{FS=OFS="."}{NF--; print}' file > tmp && mv tmp file

Reverse, cut, reverse back.

rev file | cut -d. -f2- | rev >newfile

Or, replace from last dot to end with nothing:

sed 's/\.[^.]*$//' file >newfile

The regex [^.] matches one character which is not dot (or newline). You need to exclude the dot because the repetition operator * is "greedy"; it will select the leftmost, longest possible match.

With cut on the reversed string

cat youFile | rev |cut -d "." -f 2- | rev