shell script to print ldap users

cmontr · 08-18-2008, 02:36 PM

Hello guys,

Could someone tell me why the code below doesn't print? I tried modifying few times but getting the same result. It is supposed to read data from a backup file, run ldapsearch, print all needed IDs defined in the ldapsearch. Somehow it prints 0, though ldapsearch in my sample ldif file works fine. It is Solaris machine.

Any help appreciated

----------------------
code:

#!/bin/ksh
while read g
do
echo ""
ldapsearch -L -h localhost -p 5555 -D "uid=sm_admin,ou=abc_admin,o=abc.com" -w sm_passwd -b o=vz.com -s sub uid=$g
done < g.ldif > g_out.txt

------------------------

input file (g.ldif):

dn: uid=testID,ou=abc,o=abc.com
objectclass: top
objectclass: someuser
uid: testID

Hko · 08-19-2008, 02:26 AM

What do expect/want the script to do?
I can't figure out from the script what it is supposed to do.

cmontr · 08-19-2008, 10:36 AM

Quote:

Originally Posted by Hko

What do expect/want the script to do?
I can't figure out from the script what it is supposed to do.

It is supposed to run ldapsearch in the loop and read data from backup ldif file and print all user ids. I will be able to modify the ldapsearch with the correct attributes. Just want to make this one working for now as sample. Thanks for the reply.

Hko · 08-19-2008, 01:43 PM

Quote:

Originally Posted by cmontr

It is supposed to run ldapsearch in the loop and read data from backup ldif file and print all user ids. I will be able to modify the ldapsearch with the correct attributes. Just want to make this one working for now as sample.

OK, something like this?
(note: this is bash, I don't know ksh, zo you may need too adapt it a little to ksh)

Code:

#!/bin/bash

sed -n 's/^uid: //p' g.ldif | while read g
do
echo ""
#ldapsearch -L -h localhost -p 5555 -D "uid=sm_admin,ou=abc_admin,o=abc.com" -w sm_passwd -b o=vz.com -s sub uid=$g >>g_out.txt
done