I have a text file (file_list.txt) that contains a list of filenames (e.g. contour_0.6_266_midpoints_2_0_0.dat).
I have a second file ($iw_fitted_midpoints_file.dat) which contains these filenames followed by a series of parameters and x, y information. The first 10 lines of $iw_fitted_midpoints_file.dat are:
> -Z contour_0.6_266_midpoints_2_0_0.dat 0.125512 96 0.009539 0.000017 1524.160000 1.000000 97.000000 44.000000 4 6 0.670000 0.214029 0.214030 -1.180430 -0.636157 1.307050 -0.647329 -1.101411 36.000000 -0.647530 0.145132 -1.400605 0.219184 30.949074 0.670000 48.602367 0.200000 -4.915458 -1.573416 0.428059 3376.000000
3333.000000 0.105569
3334.000000 0.106486
3335.000000 0.107501
3336.000000 0.109010
3337.000000 0.110718
3338.000000 0.107739
3339.000000 0.107382
3340.000000 0.106499
3341.000000 0.106408
I am trying to search the second file ($iw_fitted_midpoints_file.dat) for each filename and extract column 32 of the relevant line. My problem is that I cannot get the syntax for escaping the "." in the filenames correct.
This is my code
for f in `cat file_list.txt`; do
name=$(echo "$f" | cut -f 1,2,3 -d '.')
K=awk '{if ($1 == ">" && $2 ~ '\${name}') print $0}'$iw_fitted_midpoints_file.dat
echo $K
done
And this is my error
awk: {if ($1 == ">" && $2 ~ contour_0.6_96_midpoints_16_0_0.dat) print $0}
awk: ^ syntax error
For example, if I was searching for contour_0.6_266_midpoints_2_0_0.dat the above code would give me K=-1.573416 (i.e. the 32nd column from the first line in $iw_fitted_midpoints_file.dat)
instead of escaping special chars, just do an exact match, pass the value as an awk variable
$ awk -v name="$f" '$1==">" && $2==name {print $32}' datafile