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Old 08-11-2008, 03:45 AM   #1
TalkingMarble
Member
 
Registered: May 2008
Location: Deventer, The Netherlands
Distribution: CentOS 5.0
Posts: 50

Rep: Reputation: 17
Bash script : string problem


Hi all,

I'm new to bash scripting and have a little problem with the following script:

if [ $# -ne 0 ] && [ -d /vmware/$1 ]
then
vmxFile=`find /vmware/$1 -name "*.vmx"`
else
echo "Directory does not exist."
exit 1
fi

cd /scripts

./PowerOff.pl $1 $vmxFile

sleep 10

tar -cvzf $1.tgz /vmware/$1
if [ $? -eq 0 ]
then
if [ -e $1.tgz ]
then
lftp -c put $1.tgz -o ftp://10.11.71.25/VMwareBackups/
if [ $? -eq 0 ]
then
send_mail $1 "`hostname`: Backup Virtual Machine Successful"
rm $1.tgz
./PowerOn.pl $1 $vmxFile
else
send_mail $1 "`hostname`: Backup Virtual Machine Failed"
fi
fi
else
send_mail $1 "`hostname`: Backup Virtual Machine Failed"
fi

When i run the script the find command retrieves the name of the VMX file from the virtual machine to be stopped, and puts the name in the vmxFile string. The string contains spaces and when i pass string vmxFile as an argument to ./PowerOff.pl it treats the string like several small arguments instead of one big argument, causing ./PowerOff.pl to crash.

How can i accomplish that the string will be seen as one argument instead of several small ones?

Any help will be appreciated.

TIA
 
Old 08-11-2008, 04:20 AM   #2
Alien Bob
Slackware Contributor
 
Registered: Sep 2005
Location: Eindhoven, The Netherlands
Distribution: Slackware
Posts: 6,350

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Add double quotes around the argument:

./PowerOff.pl $1 "$vmxFile"

Eric
 
Old 08-11-2008, 04:25 AM   #3
matthewg42
Senior Member
 
Registered: Oct 2003
Location: UK
Distribution: Kubuntu 12.10 (using awesome wm though)
Posts: 3,530

Rep: Reputation: 63
Quote:
Originally Posted by TalkingMarble View Post
How can i accomplish that the string will be seen as one argument instead of several small ones?
Use quotes.

For example, here I write a function which prints out the arguments it receives, and then call it with quoted arguments to show how this affects commands. The same will apply to calling your script:

Code:
#!/bin/bash

arg_test () {
    n=1
    for arg in "$@"; do
        echo "arg $n is $arg"
        let n+=1
    done
    echo ""
    echo ""
}

echo "first time:"
arg_test one two three

echo "second time:"
arg_test "one two three"

echo "third time:"
arg_test one "two three"
output:
Code:
first time:
arg 1 is one
arg 2 is two
arg 3 is three


second time:
arg 1 is one two three


third time:
arg 1 is one
arg 2 is two three
 
Old 08-11-2008, 04:34 AM   #4
TalkingMarble
Member
 
Registered: May 2008
Location: Deventer, The Netherlands
Distribution: CentOS 5.0
Posts: 50

Original Poster
Rep: Reputation: 17
As suggested i surrounded the string with quotes, and the scripts now works as intended. Thank you all for the help.
 
  


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