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# 1  
Old 04-05-2012
Help Linux Shell Group exists

I am having some problems when writing shell as follows:
shell runs but returns no results
Code:
echo "enter group name: "
            dir="/home"
            read group
            if id -g $group > /dev/null 2>&1
            then
            echo "group exits"
            else
            echo "group not exits"
            fi;;

can you help me!
thank very....
# 2  
Old 04-05-2012
You have a syntax error. The double semicolons are not needed at the end of your fi

I also don't know about your version of id, but the one installed on my version of linux accepts a -g option, but not with an additional parameter. If the return is always false, you might want to read the man page for the id command.
This User Gave Thanks to agama For This Post:
# 3  
Old 04-06-2012
Are you trying to see if a group already exists? If so, try this:
Code:
read -p "enter group name: " group
if grep -q $group /etc/group
then
   echo "group exits"
else
   echo "group not exits"
fi

These 2 Users Gave Thanks to fpmurphy For This Post:
# 4  
Old 04-06-2012
hey guys, i can not run this script, as it says, Line 4: [: too many arguments, i am new in scripting
question is write a script in which user type the number and if its divisible by 15 and get remainder 0, then it is a unique number otherwise not a unique number.
Code:
#/bin/bash
echo "write the number"
read a
if [ $a % 15 -eq 0 ];
then
  echo "$a is special"
else
  echo "$a is not special"
fi

please help in solving this, i am having very hard time with this.

Last edited by Franklin52; 04-06-2012 at 07:51 AM.. Reason: Please use code tags for data and code samples, thank you
# 5  
Old 04-06-2012
Quote:
Originally Posted by ping2
hey guys, i can not run this script, as it says, Line 4: [: too many arguments, i am new in scripting
question is write a script in which user type the number and if its divisible by 15 and get remainder 0, then it is a unique number otherwise not a unique number.
Code:
#/bin/bash
echo "write the number"
read a
if [ $a % 15 -eq 0 ];
then
  echo "$a is special"
else
  echo "$a is not special"
fi

please help in solving this, i am having very hard time with this.
@ping2, please start a new thread for your question next time rather than post it in an existing thread! Thanks.

Regarding your question you can try this:
Code:
if (( $a % 5 == 0 ))
then
  echo "$a is special"
else
  echo "$a is not special"
fi

This User Gave Thanks to Franklin52 For This Post:
# 6  
Old 04-06-2012
ok thanks franklin and sorry to write in the middle of the thread, as that was my first question in this form, i have never used this before, thanks for your help.
# 7  
Old 04-10-2012
Quote:
Originally Posted by fpmurphy
Are you trying to see if a group already exists? If so, try this:
Code:
read -p "enter group name: " group
if grep -q $group /etc/group
then
   echo "group exits"
else
   echo "group not exits"
fi

thank sir. this's answers that I need

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