Bash if... then... syntax error

Question

I am trying to remove something based on user input using bash/sh, here is my code :

echo "remove ? [Y/n]"
read REMOVE
if [ $REMOVE != "n" ] || [ $REMOVE !="N" ];then
  # ... do something ...
  echo "done"
fi

the error I am getting is something like:

./run.sh: line 8: syntax error near unexpected symbol « then »

./run.sh: line 8: `if [ $REMOVE != "n" ] || [ $REMOVE !="N" ];then'

I tried to add/remove spaces many times, and I still don't understand what happens. I also don't really understand all the differences between [[ statement ]] [statement] or ((statement)). If someone can help...

@fvu $SHELL is zsh but the script is actually running in bash — n00dl3
– n00dl3, Commented May 23, 2015 at 12:31

hymie · Accepted Answer · 2015-05-23 12:32:34Z

2

In your original problem, you need a space here

$REMOVE != "N"

In your comment response to shruti1810, it sounds like your $REMOVE variable doesn't contain what you think it contains.

Try adding

echo $REMOVE

to your script.

I typically use this construct

if [ "x$REMOVE" != "xn" ] || [ "x$REMOVE" != "xN" ]
then
  # ... do something ...
  echo "done"
fi

to ensure that my arguments are both valid.

answered May 23, 2015 at 12:32

hymie

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Comments

shruti1810 · Accepted Answer · 2015-05-23 12:26:33Z

1

Please try this:

echo "remove ? [Y/n]"
read REMOVE
if [ $REMOVE != "n" ] && [ $REMOVE != "N" ]
then
  # ... do something ...
  echo "done"
fi

edited May 23, 2015 at 12:26

answered May 23, 2015 at 12:14

shruti1810

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4 Comments

n00dl3 Over a year ago

./run.sh: line 8 : [: n : expected unary operator (maybe it's not the english output, I translated it myself)

shruti1810 Over a year ago

As far as the logic which you are trying to achieve, I feel you should be using && and not ||

lcd047 Over a year ago

Better yet: if [ x"$REMOVE" != xn -a x"$REMOVE" != xN ]; then ....

Soren Over a year ago

The x is not needed as long as you quote the $ variable -- although the answer is accepted it is erroneous and a timebomb waiting to blow

Soren · Accepted Answer · 2015-05-23 12:32:07Z

1

Quote the "$REMOVE" and insert space around the "!=" -- like this;

if [ "$REMOVE" != "n" ] || [ "$REMOVE" != "N" ];then
  # ... do something ...
  echo "done"
fi

The problem is that is REMOVE is not set or if it is set to an empty string (like if you just press return on the 'read') you will get $REMOVE substituted to nothing and your expression would look like [ != "N" ] which will produce an unary operator expected error.

answered May 23, 2015 at 12:32

Soren

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