Script to set display name in iCS Calendar view to the LDAP CN

Question

Script to set the display name in the Calendar view for iPlanet Calendar
Server(iCS) to the LDAP common name(CN)
By default, iCS uses a user ID(uid) based on an employee number, rather than on
an employee's first and last name, as the calendar ID(calid).
The current release of iCS (5.0 P2) does not create a display name for a
calendar when a user enables a calendar by logging in; by default, it will
list the calid again in the Display Name field of the Calendar view.
For example, if an employee has a calid of "12345," when you click the
Calendar tab to view the calendar, the Display Name will appear as follows:
<P>
12345 (12345)
<P>
A problem arises when a user tries to subscribe to another user's calendar.
Although the search criteria are based on the calid and the Display
Name, the only information currently stored in the calendar database is the
calid. Therefore, users will be unable to subscribe to another
user's calendar unless they know the calid of that person. The next
patch release for iCS will remedy this problem by using common names(CNs) as
the Display Names. That is, the database will store the CN values
from LDAP for users, and the Calendar view will appear something as follows:
<P>
12345 (John Doe)
<P>
Until this next release of iCS, there are two options to work around this
problem.
<P>
<OL>
<LI>You can "provision" users by running the cscal
administrative utility with the
Display Name option.
<P>
OR
<P>
<LI>If the user community already exists, you can use the sample Perl script
below to search through the calendars of users.
<P>
Note: If a default calid exists that doesn't have a Display Name, the script
will search the LDAP directory to find a CN to set as the Display Name for
that calendar.
</OL>
<P>
<HR>
<P>
<B>Sample Perl Script:</B>
#!/bin/perl5.004
sub TRUE {1}
sub FALSE {"}
$SIG{INT} = 'handler';
$SIG{QUIT} = 'handler';
$mypath = $ENV{'LD_LIBRARY_PATH'};
$savepath = $mypath;
$ENV{'LD_LIBRARY_PATH'} = $mypath.';.';
#--------------INITIALIZATION----------------
$host="ldaphost";
$base_dn="ou=People,o=iplanet.com";
$port=389;
$auth_dn="cn=Directory Manager";
$auth_pwd="password";
$found_confile = TRUE;
$default_cal = FALSE;
open(CSCAL,"./cscal -v list |");
while($cal_list = <CSCAL>)
if ($cal_list =~ m/: owner=/)
@calid = split(' ',$cal_list);
chop($calid[0]);
print "\ncalid: $calid[0] ... ";
$default_cal = TRUE if ($calid[0] !~ m/:/);
} elsif (($default_cal) && ($cal_list =~ m/^ name=([a-zA-Z ]*)/)) {
chomp($1);
print "cal name: $1";
if (($1 EQ ") || ($1 EQ $calid[0]))
open(LDAPSEA,"./ldapsearch -h $host -p $port -b \"$base_dn\" -D
\"$auth_dn\" -w \"$auth_pwd\" uid=$calid[0] |");
while($ldap_list = <LDAPSEA> )
if ($ldap_list =~ m/^cn: ([a-zA-Z ]*)/)
chomp($1);
`./cscal -n "$1" modify $calid[0]`;
print "The display name for $calid[0] is being modified to be: $1\n";
sleep(1);
close(LDAPSEA);
$default_cal = FALSE;
close(CSCAL);
sub handler
local($sig) = @_;
print "... Caught a SIG$sig--closing down shop\n";
close(CSCAL);
close(LDAPSEA);
exit(0);
}

Answer 1

anne wrote:
Hi David,
About your confuse about"case when 1=2 then "product_d"."name" else "calendar_d"."year" end".
You can try in your locale.
You will find they are different.
BASED ON THAT you have two table product_d and calendar_d AND they are related by one Fact table.
THEN When you type in
"case when 1=2 then "product_d"."name" else "calendar_d"."year" end"
AND
"calendar_d"."year",
IT WILL SHOW U TWO different RESULTS.
I need to show year which its related product is not null. but I cannot use SHOW->SQL RESULTS->TYPE SOME "WHERE..." because I also need to use "constrain"..
That why I choose to use a case when function..
So, do you have any idea about this?
Regards,
AnneWhy not use two filters in your request? Have something like this:
product_d.name IS NOT NULL
AND
calendar.year IS PROMPTED?
...instead of using a CASE statement? This way you can have both filters show the way they should in a meaningful way.