1

I have created a small script to move video files from a helmet mounted camera to my mythbuntu machine and there is a variable that for some reason will just not be passed onto the script. The variable in question is 

d=`echo $file | awk '{ print $6 }'`

which is captured OK, however in the statement

scp -l 5000 $name $SavePath$d_$hour.AVI

will just not insert a value.

#!/bin/bash
# This script does copy files from /media/disk/DCIM/100DSCIM to 
# 192.20.0.200:/media/Elements/mythtv/videos/BikeRiding and renames the files using the time stamp
set -x
SavePath=root@198.20.0.200:/media/Elements/mythtv/videos/BikeRiding/
SourcePath=/media/disk/DCIM/100DSCIM/
ls $SourcePath*.AVI --full-time > file_list
cat file_list | while read file
do
    d=`echo $file | awk '{ print $6 }'`
    hour=`echo $file | awk '{ print $7 }'`
    name=`echo $file | awk '{ print $9 }'`
    scp -l 5000 $name $SavePath$d_$hour.AVI
done
Mat
  • 52,586
Inspector Gadget
  • 11
  • 1

1 Answers1

13

This is because bash interprets variable name as d_. Use curly braces to border variable name in that case:

scp -l 5000 $name $SavePath${d}_$hour.AVI

or even 

scp -l 5000 ${name} ${SavePath}${d}_${hour}.AVI

and it's much better to double quote variable to prevent error with name with strange symbols like space:

scp -l 5000 "${name}" "${SavePath}${d}_${hour}.AVI"
rush
  • 27,403
  • This answer is exactly right. But I wish I had a nickel for every time I've seen scripts using ${...} where it wasn't needed, most often in the apparent belief that ${foo} worked like "$foo". – dubiousjim Oct 19 '12 at 14:30