How to extract final character of a specific string from within an awk and append it to column?

Question

I have many data files that look something like this: ,8/9/2015 Timezone,-6 , Serial No.,19000000395CCE41 Location:,LS_trap_9u High temperature limit (�C),20.12 Low temperature limit (�C),0.05 Date - Time,Temperature (�C) 5/28/2015 6:00,20 5/28/2015 8:00,22.6 5/28/2015 10:00,27.1 5/28/2015 12:00,26.1 5/28/2015 14:00,27.1 5/28/2015 16:00,26.1 5/28/2015 18:00,24.6 5/28/2015 20:00,23.6 5/28/2015 22:00,22.6 5/29/2015 0:00,22.1 I parse these files with this script:

awk -vFS=, -vOFS=, \
   '{gsub("\"","")}
    FNR==4{s=$2}
    FNR==5{l=$2}
    FNR>8{gsub(" ",OFS);print l,s,FILENAME,$0}' \
   *.csv > formatted_log.csv
printf "\nDone\n"

I want to extract the last character from the 'loc' string (in this case "u") and append it to the another column.

The final file should look something like this:

LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,5:59,20.1
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,7:59,27.6
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,9:59,30.1
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,11:59,29.6
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,13:59,29.6
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,15:59,28.1
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,17:59,26.1
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,u,5/28/2015,19:59,23.6

My attempt thus far looks like this:

awk -vFS=, -vOFS=, \
   '{gsub("\"","")}
    FNR==4{ser=$2}
    FNR==5{loc=$2}
    FNR>8{gsub(" ",OFS);print loc,ser,FILENAME,${loc:(-1)},$0}' \
   *.csv > formatted_log.csv

I get the following errors:

awk: cmd. line:4:     FNR>8{gsub(" ",OFS);print loc,ser,FILENAME,${loc:(-1)},$0}
awk: cmd. line:4:                                                 ^ syntax error
awk: cmd. line:4:     FNR>8{gsub(" ",OFS);print loc,ser,FILENAME,${loc:(-1)},$0}
awk: cmd. line:4:                                                           ^ syntax error
awk: cmd. line:4:     FNR>8{gsub(" ",OFS);print loc,ser,FILENAME,${loc:(-1)},$0}
awk: cmd. line:4:                                                              ^ syntax error

Changing the script to this:

    awk -vFS=, -vOFS=, \
       awk -vFS=, -vOFS=, \
   '{gsub("\"","")}
    FNR==4{ser=$2}
    FNR==5{loc=$2}
  my_loc="${loc:(-1)}"
    FNR>8{gsub(" ",OFS);print loc,ser,FILENAME,my_loc,$0}' \
   *.csv > formatted_log.CSV
printf "\nDone1\n"
awk -vFS=, -vOFS=, \
   '{gsub("\"","")}
    FNR==4{ser=$2}
    FNR==5{loc=$2}
  my_loc="${loc:(-1)}"
    FNR>8{gsub(" ",OFS);print loc,ser,FILENAME,my_loc,$0}' \
   *.csv > formatted_log.CSV
printf "\nDone1\n"

adds unwanted extra lines to the formattted_log.csv file. Which look something like this:

LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,5/28/2015,5:59,20.1
5/28/2015 7:59,27.6
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,5/28/2015,7:59,27.6
5/28/2015 9:59,30.1
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,5/28/2015,9:59,30.1
5/28/2015 11:59,29.6
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,5/28/2015,11:59,29.6
5/28/2015 13:59,29.6
LS_trap_9c,3.6E+15,trap9c_3600000039654841_150809.csv,5/28/2015,13:59,29.6
5/28/2015 15:59,28.1

How can I extract final character of a specific string from within an awk?

Answer 1

To extract the last character in AWk, you could use:

substr(var,length(var),1)

The script will be:

awk -vFS=, -vOFS=, \
   '{gsub("\"","")}
   FNR==4{ser=$2}
   FNR==5{loc=$2}
   FNR>8{gsub(" ",OFS);print loc,ser,FILENAME,substr(loc,length(loc),1),$0}' \
   *.csv > formatted_log.csv

From man awk:

substr(s, i [, n])
Return the at most n-character substring of s starting at i. If n is omitted, use the rest of s.