syntax error near unexpected token `fi'

thomas2004ch · 08-04-2011, 04:58 AM

I wrote a small shell script as follow:

Code:

#!/bin/bash

echo "this is my script"

RESULT=${RESULT:-"ll /var/log/jboss/mtext/"}

if [ -z $RESULT ] then
   ls /var/log/jboss/mtext/
fi
;;

But I got error "syntax error near unexpected token `fi'" when I start this script.

Thomas

acid_kewpie · 08-04-2011, 05:08 AM

yep, you will. "then" is a bash builtin command, not mere syntax. Here it's passed as an option to the if, which it's not. just put the "then" on a new line.

Your code has plenty of other issues too btw... what's the ;; for? Did you copy that from a "case" example? delete it. Also you should not use "[", use "[[" instead for various boring reasons. Basically [ is the /bin/test command whilst [[ is a genuine builtin bash operation, so handles strings more logically. in your example, if $RESULT doesn't contain anything then it will cause an error, using [[ it will know that the variable is empty.

thomas2004ch · 08-04-2011, 05:39 AM

Hi,

Many thanks. It works now.

Now I extend the script as follow:

Code:

#!/bin/bash

RESULT=${RESULT:-"grep Started /var/log/jboss/mtext/server.log"}
LOG=${LOG:-"/var/log/jboss/mtext/server.log"}

if [[ -e $LOG ]]
then
  echo "server.log exist"
  if [[ -z $RESULT ]]
  then
    echo "Jboss started"
  else
    echo "JBoss not started"
  fi
else
   echo "server.log doesn't exist"
fi

I think the syntactic should be correct.

The server.log exist and the $RESULT is not empty but it shows even though

Code:

server.log exist  (this is correct)
JBoss not started (this is not correct)

Anything wrong?

Thomas

grail · 08-04-2011, 05:48 AM

Quote:

Anything wrong?

Well I would have to say yes if you are not getting the results you expect!

Maybe you should echo the variable RESULT and see exactly what is in it??

thomas2004ch · 08-04-2011, 06:26 AM

I add the echo and the script ölooks as follow now:

Code:

#!/bin/bash

RESULT=${RESULT:-"grep Started /var/log/jboss/mtext/server.log"}
LOG=${LOG:-"/var/log/jboss/mtext/server.log"}

if [[ -e $LOG ]]
then
  echo "server.log exist"
  echo "RESULT = " $RESULT
  if [[ -z $RESULT ]]
  then
    echo "Jboss started"
  else
    echo "JBoss not started"
  fi
else
   echo "server.log doesn't exist"
fi

As I run this script I got:

Code:

server.log exist
RESULT =  grep Started /var/log/jboss/mtext/server.log
JBoss not started

1.
One can see the RESULT is not empty.

2.
I wonder why the "grep Started /var/log/jboss/mtext/server.log" doesn't work.

acid_kewpie · 08-04-2011, 06:45 AM

is does "work" you're just setting that as a string.

just use this:

RESULT=$(grep whatever)

much simpler. No idea where you've picked up all this odd syntax from.

In terms of what that command is trying to do, I would suggest a different approach though:

RESULT=$(grep -c whatever)
if [[ $RESULT -gt 0 ]]
then
...
fi

i.e. count the number of times that grep finds a hit. more elegant that way. another common alternative would be to use the exit code of the grep, not the output:

grep -q whatever
if [[ $? -eq 0 ]]
then
... exit code was zero, so grep "worked"
fi

thomas2004ch · 08-04-2011, 06:51 AM

Now I change the script as follow:

Code:

#!/bin/bash

function grepLog(){
  grep "Started in" /var/log/jboss/mtext/server.log
}

LOG=${LOG:-"/var/log/jboss/mtext/server.log"}

if [[ -e $LOG ]]
then
  echo "server.log exist"
#  echo $(grepLog)
  result=$(grepLog)
  echo $result
  if [[ -z  "$result" ]]
  then
    echo "JBoss started"
  else
    echo "JBoss not started"
  fi
else
   echo "server.log doesn't exist"
fi

The ran result is:

Code:

server.log exist
2011-08-04 12:32:54,098 INFO [org.jboss.bootstrap.microcontainer.ServerImpl] (main) JBoss (Microcontainer) [5.0.1 (build: SVNTag=JBPAPP_5_0_1 date=201003301050)] Started in 2m:19s:741ms
JBoss not started

One can see the result is not empty. But it show always "JBoss not started".

Thomas

acid_kewpie · 08-04-2011, 06:54 AM

no no, you're now trying to execute a program called grepLog, that makes no sense. You don't need a function for it.

thomas2004ch · 08-04-2011, 06:58 AM

Sorry, I make mistake. The -z means empty. I have to change the script as follow:

Code:

#!/bin/bash

function grepLog(){
  grep "Started in" /var/log/jboss/mtext/server.log
}

LOG=${LOG:-"/var/log/jboss/mtext/server.log"}

if [[ -e $LOG ]]
then
  echo "server.log exist"
#  echo $(grepLog)
  result=$(grepLog)
  echo $result
  if [[ -z  "$result" ]]
  then
    echo "JBoss not started"
  else
    echo "JBoss started"
  fi
else
   echo "server.log doesn't exist"
fi

thomas2004ch · 08-04-2011, 07:00 AM

Quote:

Originally Posted by acid_kewpie

no no, you're now trying to execute a program called grepLog, that makes no sense. You don't need a function for it.

Ok, I change the script as you recommended as follow. It print correctly.

Code:

#!/bin/bash

function grepLog(){
  grep "Started in" /var/log/jboss/mtext/server.log
}

LOG=${LOG:-"/var/log/jboss/mtext/server.log"}

if [[ -e $LOG ]]
then
  echo "server.log exist"
#  echo $(grepLog)
#  result=$(grepLog)
  result=$(grep "Started in" /var/log/jboss/mtext/server.log)
  echo $result
  if [[ -z  "$result" ]]
  then
    echo "JBoss not started"
  else
    echo "JBoss started"
  fi
else
   echo "server.log doesn't exist"
fi

Many thanks.

Thoams

grail · 08-04-2011, 07:14 AM

Your other option is to do away with the variable altogether:

Code:

if grep -q "Started in" $LOG
then
    echo "JBoss started"
else
    echo "JBoss not started"
fi

catkin · 08-04-2011, 07:53 AM

Quote:

Originally Posted by thomas2004ch

Now I extend the script as follow:

Code:

if [[ -e $LOG ]]
then
   <stuff>
fi

If you like your if-fi to line up vertically you can use

Code:

if [[ -e $LOG ]]; then
   <stuff>
fi