Shell script: if multiple conditions

Question

I want to create a shell script that echos something if 3 directories don't exist.

Here is the code I wrote: test.sh:

if [ ! -d "/home/unix/POSTagger2" ] || [! -d "/home/unix/POSTagger2/stanford-parser-full-2015-12-09"] || [! -d "/home/unix/POSTagger2/stanford-corenlp-full-2015-12-09"]; then
echo "Nope"
fi

When I run it, I get this error:

./test.sh: line 1: [!: command not found

What's wrong with my syntax?

Answer 1

You are missing some spaces, for example [! must be [ ! and "] must be " ] look to the corrected code:

#!/bin/bash
if 
[ ! -d "/home/unix/POSTagger2" ] || 
[ ! -d "/home/unix/POSTagger2/stanford-parser-full-2015-12-09" ] || 
[ ! -d "/home/unix/POSTagger2/stanford-corenlp-full-2015-12-09" ] 
then
      echo "Nope"
fi

Another way for your code:

#!/bin/bash
for dir in "/home/unix/POSTagger2" "/home/unix/POSTagger2/stanford-parser-full-2015-12-09" "/home/unix/POSTagger2/stanford-corenlp-full-2015-12-09"; do
     if [ ! -d "$dir" ]; then echo nope ; break; fi 
done  

Answer 2

You need a space between the [ and the ! for things to work correctly. This is because [ is implemented as shell-builtin command (it even used to be a separate exectuable /usr/bin/[).

You can also use:

if [ ! -d "/home/unix/POSTagger2" -o ! -d "/home/unix/POSTagger2/stanford-parser-full-2015-12-09" -o ! -d "/home/unix/POSTagger2/stanford-corenlp-full-2015-12-09") ] ; then
    echo "Nope"
fi

Bash offers an alternative [[ that is implemented as en expression. [[ uses &&, ||, etc. instead of -a, -o as operators.

if [[ ! (-d "/home/unix/POSTagger2" && -d "/home/unix/POSTagger2/stanford-parser-full-2015-12-09" && -d "/home/unix/POSTagger2/stanford-corenlp-full-2015-12-09") ]] ; then
    echo yes
fi

Edit: Thanks to comments from @LucianoAndressMartini and @pabouk for important corrections to my understanding.