Bash to store result in variable for other lines in script to use
I can not figure out how to capture the $filename variable store by the bash.
Code:
#!/bin/bash
# oldest folder stored as variable for analysis, version log created, and quality indicators matched to run
dir=/home/cmccabe/Desktop/NGS/test
find "$dir" -maxdepth 1 -mindepth 1 -type d -printf '%T+\t%P\0' | sort -rz |
while read -r -d $'\t' time && read -r -d '' filename
do
printf "The oldest folder is $filename, created on $time and analysis done using v1.4 by $USER at $(date "+%D %r")\n" >> /home/cmccabe/medex.logs/folder.log
awk -v FL="$filename" '
FNR == 1 {filenum++}
filenum==1 && index($0, FL) {
match($0, "_0*([0-9]+)/")
FNUM=substr($0,RSTART+1,RLENGTH-2)
gsub(/^0+/,"", FNUM)
}
filenum==2 && $0 ~ FNUM".pdf$"' /home/cmccabe/s5_files/downloads/list /home/cmccabe/s5_files/pdf/pdf > /home/cmccabe/s5_files/pdf/output
break
done
echo $filename
So after running the bash, $filename is 12345R12.
So, the result of the echo is 12345R12. Currently after the bash runs the $filename is not stored (maybe because of the done).
The result stored in $filename needs to be available for my script to execute. Currently, I manually copy it in but I thought that this might be a better way, but I can not seem to get the syntax correct. Thank you .
Last edited by cmccabe; 01-17-2017 at 02:00 PM..
Reason: fixed format
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Discussion started by: cmccabe
2 Replies
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01-17-2017
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