bash if statement expected unary operator

Question

I'm trying to do something along lines of

if (!regular file || symbolic link)
    continue

What I have so far is

st1=$( -f "${ARRAY[$i]}" )                                                 
 if [ "$st1" -eq 0 ]

but I'm getting "expected unary operator error"

st1=$( -f "${ARRAY[$i]}" ) doesn't make sense. The $( ... ) construct requires a command between the parentheses. -f is not a command; it's an argument to the test, [, or [[ command. You should have gotten an error like bash: -f: command not found, not expected unary operator; did you copy-and-paste your exact code? — Keith Thompson
– Keith Thompson, Commented Oct 17, 2014 at 18:24

anubhava · Accepted Answer · 2014-10-17 18:15:37Z

1

You don't need to create intermediate st1 variable. Just use:

if [[ ! -f "${ARRAY[$i]}" || -h "${ARRAY[$i]}" ]]; then
    echo "${ARRAY[$i]} link exists"
fi

Your use of st1=$( -f "${ARRAY[$i]}" ) is incorrect and will cause syntax error:

-f: command not found

since shell will consider f as a command name.

answered Oct 17, 2014 at 11:33

anubhava

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5 Comments

Okay but what was the problem?

I think I mentioned about intermediate variable creation. Let me add more explanation.

Please, don't wrap non-code in code formatting. :P (Answer is better now, assuming it's correct (I don't know))

It should be an OR operator instead of AND, but the rest of it helped, thank you

Oh right, it is updated now. Glad it all worked out.