Unix - if else EOF error

Question

I am writing the following code:

if [ $opt -ge $max -o $opt -le  0 ]
then
    echo "Bad";
else
    echo "Good";
    if [ $opt = "\" -o $opt = "/"  ]
    then
        echo "Good";
    else
        echo "Invlaid"; //Line number 21
    fi
fi //Line number 23 no Line number 24.

this shows an error:

./file.sh: line 21: unexpected EOF while looking for matching `"'
./file.sh: line 24: syntax error: unexpected end of file

If I place this code:

if [ $opt -ge $max -o $opt -le  0 ]
then
    echo "Bad";
else
    echo "Good";
fi //Line number 23 no Line number 24.

Then there is no error. I am not able to figure out the problem.

See the syntax highlighting - you have accidentily "escaped" the second quote in if [ $opt = "\" -o $opt = "/" ]. — Andreas Fester
– Andreas Fester, Commented Jul 26, 2013 at 5:55
Note that -o for logical OR is deprecated; you should use [ "$opt" -ge "$max" ] || [ "$opt" -le 0 ] instead--and quote your expansions! — chepner
– chepner, Commented Jul 26, 2013 at 13:09

Joni · Accepted Answer · 2013-07-26 05:55:41Z

3

Where you write "\" you start a string literal whose first character is a double quote. To include a back slash in a string you have to precede it with another:

"\\"

answered Jul 26, 2013 at 5:55

Joni

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Comments

anubhava · Accepted Answer · 2013-07-26 06:03:32Z

1

Backslash \ inside double quotes needs to be escaped or else you can use single quotes like this:

if [ $opt = '\' -o $opt = '/' ]; then
   echo "Good"
fi

Single quotes treat the wrapped string literally that's the precise reason single quote in shell cannot be escaped.

answered Jul 26, 2013 at 6:03

anubhava

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Comments

Mandar Pande · Accepted Answer · 2013-07-26 06:41:03Z

0

opt="\\"

echo $opt
\

if [ $opt = "/" -o $opt = "\\" ]; then echo "Hi"; else echo "bye"; fi
Hi

answered Jul 26, 2013 at 6:41

Mandar Pande

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