Using AWK to get part of filename

[starlite79]

Hi,

First question: can I use awk in a Fortran program?
Second: I'd like to use substr to get the part of a file itself (not the contents of the file). For example, I have many files in a directory with the same prefix and a YYMMDD.dat suffix. I want to pick out the YY part of the filename and assign it to a variable, say yr.

Would it look something like (if I'm in the proper directory):

yr = "awk ' { print substr(FILENAME, 18, 2) } '" ?

There are many filenames I'd like to do for this process.

Am I asking too much of awk and Fortran?

 

[Annihilannic]

I'm no FORTRAN wizard, so I don't know what facilities it has for calling external programmes, if any.

There are various ways you can obtain the year from the filename in shell though:

filename=080916.dat
# with awk
yr=$(echo $filename | awk '{print substr($0,1,2)}')
# with cut
yr=$(echo $filename | cut -c1-2
# with sed
yr=$(echo $filename | sed 's/\(..\).*/\1'/)
# with shell builtin suffix removal
yr=${filename%????.dat}

Note that $( ... ) is interchangeable with ` ... ` on most shells, but I prefer to former due to its "nestability".



Annihilannic.

 

[Annihilannic]

NB I missed a closing bracket on the cut solution...

Annihilannic.