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UNIX for Dummies Questions & Answers
[Solved] take name of directory and files as variables
Post 302575372 by me. on Monday 21st of November 2011 12:06:25 PM
11-21-2011
sorry, I was just all over the place with that script and missed your responses...
tried what you wrote but it didnt work so I was messing with it but without success...
i tried
dont work as it should be
these dont work either,
its like
I am getting nowhere so far, to bad I missed your replies because would be helpful to chat with someone like you while doing....
tried what you wrote but it didnt work so I was messing with it but without success...
i tried
PHP Code:
#!/bin/bash
dir="/home/andre/Documents/Scripts/Copy/"
FILE=(ls)
for i in $FILE
do
cp -i "$1" "${dir}${1}"
echo " file $1 has been copied to $dir "
cp -i $3 $dir $3
echo " file $2 has been copied to $dir "
cp -i $4 $dir $4
echo " file $3 has been copied to $dir "
done
PHP Code:
done
for i in $(pwd) $(ls)
do
echo "$i"
if [ -d $dir ]; then
cp -a * ~/Documents/Scripts/Copy #these would copy all files to Copy dir
echo " files copied to directory $dir "
else
mkdir Copy && cp $i $dir
echo " $2 copied to directory $dir "
end
its like
I am getting nowhere so far, to bad I missed your replies because would be helpful to chat with someone like you while doing....|
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