syntax error near unexpected token `then'

voleg · 11-11-2013, 02:43 AM

man test:

Quote:

! EXPRESSION
EXPRESSION is false

You use

Quote:

if [ ! "$NAME" ]

what can be translated as "NOT variable".

Unbreakable test should be:

Code:

if [ 'x'"$NAME" = 'x' ] ; then
..
fi

catkin · 11-11-2013, 03:03 AM

Quote:

Originally Posted by voleg

Unbreakable test should be:

Code:

if [ 'x'"$NAME" = 'x' ] ; then
..
fi

This line is not easy to understand ...

Code:

if [ ! "$NAME" ]

... but it is perfectly valid bash so is not the cause of the problem which is a syntax problem.

Another version of the same test, robust and transparent is

Code:

if [[ $NAME != '' ]]; then

or (less transparent IMHO)

Code:

if [[ -z $NAME ]]; then

tuananhk53 · 11-11-2013, 03:03 AM

Quote:

Originally Posted by voleg

man test:

You use

what can be translated as "NOT variable".

Unbreakable test should be:

Code:

if [ 'x'"$NAME" = 'x' ] ; then
..
fi

hi voleg, i'm an amater system anh linux
can you fix this script helps me?
thanhks voleg

voleg · 11-11-2013, 03:58 AM

No, you have to do this yourself. This is a "dao" way to learn.
You will like scripting, I promisse.

voleg · 11-11-2013, 04:03 AM

Quote:

Unbreakable test should be:

Code:

if [ 'x'"$NAME" = 'x' ] ; then
..
fi

This line is not easy to understand ...

This construction works for me for /bin/sh and /bin/ksh before bash was invented.
Old dog will not teach new trick.

tuananhk53 · 11-11-2013, 04:36 AM

Quote:

Originally Posted by voleg

This construction works for me for /bin/sh and /bin/ksh before bash was invented.
Old dog will not teach new trick.

hi voleg,
i understand what you say.
Thanks for your help!

catkin · 11-12-2013, 08:45 PM

Sometimes new tricks make life easier (as long as portability is not required): http://mywiki.wooledge.org/BashFAQ/031

tuananhk53 · 11-13-2013, 03:02 AM

Quote:

Originally Posted by catkin

Sometimes new tricks make life easier (as long as portability is not required): http://mywiki.wooledge.org/BashFAQ/031

hi catkin
i don't understand what you say
hmm, maybe that is a censure for me......

Firerat · 11-13-2013, 03:38 AM

@tuananhk53

don't worry about it, it is not worth it.

if in any doubt use voleg's 'form'

Code:

if [ 'isitnull'"$VAR" = 'isitnull' ]
then
   echo "yes , VAR is \'null\' "
else
   echo "no , VAR = $VAR"
fi

the simple fact is , it works when [ "$VAR" ] doesn't AND will still work when it does

you should use it in place of [ ! "$NAME" ] in your script, it might 'fix it'

example

Code:

   UID1=`echo $line | cut -d: -f1`
   NAME=`echo $line | cut -d: -f5 | cut -d: -f1`
   if [ ! "$NAME" ]
   then
      NAME=$UID1

Code:

   UID1=`echo $line | cut -d: -f1`
   NAME=`echo $line | cut -d: -f5 | cut -d: -f1`
   if [ 'x'"$NAME" = 'x' ]
   then
      NAME=$UID1

catkin · 11-23-2013, 03:14 AM

Quote:

Originally Posted by tuananhk53

hi catkin
i don't understand what you say
hmm, maybe that is a censure for me......

Was it "as long as portability is not required" or the linked BASH FAQ page that you did not understand?