Help with script

[Junior C.]

Why am I getting this error when I run this script. script and output error attach

Thanks,
Junior

[Rick Garland]

Put a backslash in front of the *. The shell is trying to interpret

[Andreas Voss]

Hi,

could it be that whithin your subirs there are files that begin with -* ?
Check this with a find command.
If so change your ll command to:
ll ./$FILE
This will prevent interpreting the filename as an option for ll.

Regards

Andrew

[Alan Riggs]

Might I suggest replacing the entire structure with:

ll -d $RETAILRE/*/* $RETAILRE/*/data.*/* | grep -v "^d" >> $LOGFILE

[Jeremiah Campbell]

What is it you are trying to do? Perhaps that would help to figure out the best way to get this script to work.

[Douglas Nakamoto]

I would think that the problem is with the syntax of your 'for' statement.

The following logic (utilizing your structure, excerpted from your script - it will have to be inserted, replacing the old logic, to work properly) will do what I think you want:

cd $RETAILRE
for DIR in `ll | grep ^d | awk '{print $9}'`
do
echo $DIR >> $LOGFILE
cd $DIR
for FILE in `ls $DATA`
do
if [[ -f $FILE ]]
then
ll $FILE >> $LOGFILE
fi
done
cd $RETAILRE
done

[Jeremy Dean_3]

Alan and Douglas are on to a winner.

for file in *

will work fine when there are files to match, however when there are none it will return "*" as file once.
This could have been explicitly tested in the original coding loop as an exception case.
Alternatively Alan and Douglas's suggestions are (better) alternatives.

for file in $DATA/*; do
if [ "$file" = "$DATA/*" ]; then
break
fi
...
done

Note that the shell expands $DATA, but not * within the quotes!