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Full Discussion: Error"syntax error at line 15: `end of file' unexpected"
Top Forums Shell Programming and Scripting Error"syntax error at line 15: `end of file' unexpected" Post 302815137 by millan on Friday 31st of May 2013 04:14:56 AM
Old 05-31-2013
Yeah, it worked.

Code:
dbSID="SWQE"
usrname="apps"
password="Wrgthrk3"
count=0
while [ $count -lt 3 ]
do
    sqlplus $usrname/$password@$dbSID <<-EOF 
    WHENEVER OSERROR EXIT 9;
    WHENEVER SQLERROR EXIT SQL.SQLCODE;
    prompt Connected to the database;
    quit;
EOF======================> i kept EOF without any space before i,e at the starting of the line and it worked.
  count=`expr $count + 1`
done

Last edited by Franklin52; 05-31-2013 at 09:22 AM.. Reason: Please use code tags
 

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LEARN ABOUT SUSE

gpasswd

gpasswd(1)						      General Commands Manual							gpasswd(1)

NAME

       gpasswd - change group password

SYNOPSIS

       gpasswd [-r|-l|-u] [group]

DESCRIPTION

       gpasswd	changes  passwords  for  group accounts. Only an administrator may change the password for any group. The used hash algorithmus is
       defined by the value of GROUP_CRYPT or, if not specified, CRYPT from /etc/default/passwd.  If not configured, the traditinal des  algorith-
       mus is used.

OPTIONS

       -r     Remove group password.

       -l     A system administrator can lock the password of the specified group.

       -u     A  system  administrator	can  unlock  the password of the specified group, if the group is not passwordless afterwards (it will not
	      unlock a group that has only  "!" as a password).

FILES

       /etc/group - group account information /etc/default/passwd - default values for password hash

SEE ALSO

       group(5), groupadd(8), groupdel(8), groupmod(8)

AUTHOR

       Thorsten Kukuk <kukuk@suse.de>

pwdutils							     July 2006								gpasswd(1)
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