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# 1  
Old 01-05-2011
echo
hi,

i am facing an issuse in echo with double quates " " .

Code:
 
while read i
do
uid=`echo $i | awk '{print $3}'`
login=`echo $i |awk '{print $2}'`
name=`echo $i | awk '{print $1}'`
echo "/usr/sbin/useradd -u $uid -g 200 -d /local/home/$login -m -s /bin/ksh -c "/"$name_CRQ"/" $login "
done <eg1

eg1 file contains user info

i am expecting an o/p like this

Code:
/usr/sbin/useradd -u 12345 -g 200 -d /local/home/e1234 -m -s /bin/ksh -c "john_CRQ" e1234

but i am getting

Code:
/usr/sbin/useradd -u 12345 -g 200 -d /local/home/e1234 -m -s /bin/ksh -c " " e1234

please help me with this.
Last edited by radoulov; 01-05-2011 at 07:15 PM.. Reason: Additional code tags added.
# 2  
Old 01-05-2011
Escape your double-quotes, or just avoid wrapping double-quotes around your "echo" command ... they technically aren't required.

Shell sees you opening your quote after the
Code:
echo

, then assumes you closed it after the
Code:
-c

.

Option 1: Escape double-quotes

Code:
echo "/usr/sbin/useradd -u $uid -g 200 -d /local/home/$login -m -s /bin/ksh -c \"/\"$name_CRQ\"/\" $login"

Option 2: No quotes
Code:
echo /usr/sbin/useradd -u $uid -g 200 -d /local/home/$login -m -s /bin/ksh -c "/"$name_CRQ"/" $login

-dufftime
This User Gave Thanks to dufftime For This Post:
sudharson
# 3  
Old 01-06-2011
In addition, try this:
Code:
echo "/usr/sbin/useradd -u $uid -g 200 -d /local/home/$login -m -s /bin/ksh -c \"${name}_CRQ\" $login"

The curly brackets are necessary here, because an underscore (_) is a valid character in a variable name, so without them it the value of the variable $name_CRQ will be used which is empty.
This User Gave Thanks to Scrutinizer For This Post:
sudharson
# 4  
Old 01-11-2011
sorry for late reply. now its working fine. Thanks guys
# 5  
Old 01-11-2011
Code:
while read i
do
uid=`echo $i | awk '{print $3}'`
login=`echo $i |awk '{print $2}'`
name=`echo $i | awk '{print $1}'`

could be coded more succinctly:
Code:
while read name login uid rest
do

Or
Code:
awk '{printf("/usr/sbin/useradd -u %s -g 200 -d /local/home/%s -m -s /bin/ksh -c \"%s_CRQ\" %s\n",$3,$2,$1,$2)}' eg1

This User Gave Thanks to binlib For This Post:
sudharson
# 6  
Old 01-19-2011
Quote:
awk '{printf("/usr/sbin/useradd -u %s -g 200 -d /local/home/%s -m -s /bin/ksh -c \"%s_CRQ\" %s\n",$3,$2,$1,$2)}' eg1
this works fine, thanks for u help binlib SmilieSmilieSmilie
# 7  
Old 01-20-2011
In sh version you can parse line using read. Why last is xxx ? If line include extra data, xxx include it.
Code:
while read name login uid xxx
do
   echo "/usr/sbin/useradd -u $uid -g 200 -d /local/home/$login -m -s /bin/ksh -c \"${name}_CRQ\" $login "
done < eg1

If you need print <> " ; ... then HERE is nice solution - "printout template"
Code:
while read name login uid xxx
do
   cat <<-EOT
   /usr/sbin/useradd -u $uid -g 200 -d /local/home/$login -m -s /bin/ksh -c "${name}_CRQ" $login 
EOT
done < eg1

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