I am trying to write a bash/shell script to zip up a specific folder and ignore certain sub-dirs in that folder.
This is the folder I am trying to zip "sync_test5":
My bash script generates an ignore list (based on) and calls the zip function like this:
#!/bin/bash
SYNC_WEB_ROOT_BASE_DIR="/home/www-data/public_html"
SYNC_WEB_ROOT_BACKUP_DIR="sync_test5"
SYNC_WEB_ROOT_IGNORE_DIR="dir_to_ignore dir2_to_ignore"
ignorelist=""
if [ "$SYNC_WEB_ROOT_IGNORE_DIR" != "" ];
then
for ignoredir in $SYNC_WEB_ROOT_IGNORE_DIR
do
ignorelist="$ignorelist $SYNC_WEB_ROOT_BACKUP_DIR/$ignoredir/**\*"
done
fi
FILE="$SYNC_BACKUP_DIR/$DATETIMENOW.website.zip"
cd $SYNC_WEB_ROOT_BASE_DIR;
zip -r $FILE $SYNC_WEB_ROOT_BACKUP_DIR -x $ignorelist >/dev/null
echo "Done"
Now this script runs without error, however it is not ignoring/excluding the dirs I've specified.
So, I had the shell script output the command it tried to run, which was:
zip -r 12-08-2014_072810.website.zip sync_test5 -x sync_test5/dir_to_ignore/**\* sync_test5/dir2_to_ignore/**\*
Now If I run the above command directly in putty like this, it works:
So, why doesn't my shell script exclude working as intended? the command that is being executed is identical (in shell and putty directly).
This is what I ended up with:
After a few trial and error, I have managed to fix this problem by changing this line:
to:
Not sure why this worked, but it does :)
Because backslash quotings in a variable after word splitting are not evaluated.
If you have
a='123\4'
,echo $a
would giveBut if you do it directly like
echo 123\4
, you'd getClearly the arguments you pass with the variable and without the variables are different.
You probably just meant to not quote your argument with backslash:
Btw, what actual works is a non-evaluated glob pattern:
You can verify this with
And this is my suggestion: