Boolean Expressions in Shell Scripts

Question

What's the "right" way to do the following as a boolean expression?

for i in `ls $1/resources`; do
    if [ $i != "database.db" ]
    then
     if [ $i != "tiles" ]
     then
      if [ $i != "map.pdf" ]
      then
       if [ $i != "map.png" ]
       then
        svn export -q $1/resources/$i ../MyProject/Resources/$i
...

Answer 1

for i in `ls $1/resources`; do
    if [ $i != "database.db" ] && [ $i != "tiles" ] && [ $i != "map.pdf" ] && [ $i != "map.png" ]; then
        svn export -q $1/resources/$i ../MyProject/Resources/$i

Answer 2

Even shorter:

for i in `ls $1/resources`; do
  if [ $i != databse.db -a $i != titles -a $i != map.pdf ]; then
    svn export -q $1/resources/$i ../MyProject/Resources/$i
  fi
done;

The -a in the if expression is the equivalent of the boolean AND in shell-tests. For more see man test

Answer 3

The above solutions have a couple of common mistakes: http://www.pixelbeat.org/programming/shell_script_mistakes.html

1. for i in ls ... is redundant/problematic just do: for i in $1/resources*; do ...

2. [ $i != file1 -a $1 != file2 ] This actually has 2 problems. a. The "$i" is not quoted, hence names with spaces will cause issues b. -a is inefficient if stating files as it doesn't short circuit (I know the above is not stating files).

So instead try:

for i in $1/resources/*; do
    if [ "$i" != "database.db" ] &&
       [ "$i" != "tiles" ] &&
       [ "$i" != "map.pdf" ] &&
       [ "$i" != "map.png" ]; then
        svn export -q "$i" "../MyProject/Resources/$(basename $i)"
    fi
done

Answer 4

Consider using a case statement:

for i in $(ls $1/resources); do
    case $i in
        database.db|tiles|map.pdf|map.png)
           ;;
        *)
           svn export -q $1/resources/$i ../MyProject/Resources/$i;;
    esac
done