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[Solved] take name of directory and files as variables

 
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# 22  
Old 12-02-2011
I tested with /usr/bin/sh and /usr/bin/ksh but have no bash on this server, sorry...(This one is a HPUX 11.00 )

End of execution with /bin/sh:

Code:
.
.
.

test:
total 0
drwxrwxrwx   3 vbe        bin             96 Nov 23 17:38 samples
Finished!
Copied 119 files 
zra1:/home/vbe $ what copy2   
copy2:
         copy2 v1.0   :Copy all files from current directory to directory given as parameter.
zra1:/home/vbe $

Last edited by vbe; 12-02-2011 at 12:30 PM.. Reason: added output
This User Gave Thanks to vbe For This Post:
me.
# 23  
Old 12-02-2011
thank you vbe for that code, I will work on it and try to get my one working now!

pint of Guinness tastes best in Ireland (even Obama said so Smilie) so you are welcome for one or two here Smilie

---------- Post updated at 01:02 PM ---------- Previous update was at 11:52 AM ----------

ok I did it Smilie it was quite easy now, with your help vbe !

the code works and I think I am happy with it, for now any way Smilie.

Code:
!/bin/bash
dirname=/home/andre/Documents/FRIDAY
files=$(ls)
list=$(ls -m)

if [ -d $dirname ]; then
        echo " destination direcotry $dirname exists... now will copy
files "

elif
 [ ! -d $dirname ]; then
        echo " destination directory doesnt exists...creating
direcotry "
echo " ... "
sleep 3
mkdir $dirname
echo " ... "
echo " direcotry created...processing to copying files "
fi

for i in $files
do
echo " files $i is being copied... "
cp -i $i /home/andre/Documents/FRIDAY
echo " DONE "
done

echo " copied files are "
echo " $list "

thank you
# 24  
Old 12-05-2011
You welcome!
 
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