I am setting up a software package to work with our LDAP server. Since we have limited licenses, I need to limit what group a user can be a member of to be included. I have a field called LDAP Search Base where I can specify something like this:
(sAMAccountName={0})
The {0} is filled in with the login name. I would like to limit the users to be part of our Development LDAP group, but the memberOf group can have multiple values, Doing something like this:
(&(sAMAccountName={0})(memberOf=*Development*))
doesn't work.
Is there a way to query LDAP to specify that I am looking through all of the values of memberOf looking for a groups that matches *Development*?
Looks like memberOf is a Distinguished name. Substring matches using wildcards are not supported on DNs
Related
I understand you can't simply filter on dn, but I have something like this:
dc=lvl3,dc=lvl2,dc=lvl1, and someone could have a dn like this:
CN=Last, First,OU=ou1,OU=retired,OU=ou1,DC=lvl3,DC=lvl2,DC=lvl1.
Is there a way I can filter results for those who do not have 'retired' in any ou?
No. Not as an LDAP filter.
You could of course do a search and then sort on the CLient-side.
Generally organization of "characteristics" of users should be done using Attributes and not by the Directory Structure.
Something like:
employeeType=Active
employeeType=Retired
employeeType=etc
To avoid these conditions.
Maybe, but it depends on the server's LDAP implementation.
There is a filter syntax which allows matching against DN attributes (in addition to entry attributes):
(ou:dn:=Retired)
(Within the LDAP specification this is known as the 'dnAttributes' field, part of the 'extensibleMatch' filter type. See RFC 4511 section 4.5.1.7.7)
Not all directory servers support this. For example, OpenLDAP handles it correctly, but Active Directory (MS AD and Samba) will ignore it, behaving as if you used (ou=Retired) instead.
If your server supports this, then a negative match is simply done by wrapping the filter in (!(...)) as you would normally do.
However, in other cases you should either a) use a custom attribute, or b) perhaps check for membership in a global "Active employees" group or something similar.
We have an LDAP with a number of groups that follow this pattern:
Acme-MyApp-ABC-Admin
Acme-MyApp-ABC-Bottlewasher
Acme-MyApp-ABC-Cook
Acme-MyApp-DEF-Admin
Acme-MyApp-DEF-Bottlewasher
Acme-MyApp-DEF-Cook
etc repeated many times.
(&(objectClass=person)(memberOf=cn=Acme-MyApp-ABC-Admin,ou=Groups,dc=acme,dc=com))
correctly returns members of the Acme-MyApp-ABC-Admin group. We'd like to find members of all of the Admin groups.
(&(objectClass=person)(memberOf=cn=*-Admin,ou=Groups,dc=acme,dc=com))
Is it possible to put a wildcard within a DN?
Generally, Wildcard searches on DN's syntax attributes are not supported.
Some LDAP server implementation may support them. You question is tagged as OpenLDAP but the search filter appears to be more like an AD implementation.
I did find "Question about using an LDAP filter to get memberOf from an AD Group" on TechNet stating, ".. that wildcards are no allowed." (I am assuming he met NOT vs no)
-jim
While I do not think that this can be done with the ldap filter directly. I have faced similar requirements many times and my go to method is as following:
Create a group that encapsles all relevant groups, in your case the admin groups.
Acme-MyApp-Admins
Acme-MyApp-ABC-Admin
Acme-MyApp-DEF-Admin
Then setup a filter based on the recursive membership of that group.
(&(objectClass=person)(memberOf:1.2.840.113556.1.4.1941:=CN=Acme-MyApp-Admins,ou=Groups,dc=acme,dc=com))
This way you only need to take care that every new admin group is added as a member of the access group, but you do not need to modify the ldap filter.
I have a series of users and groups. Users' DNs can be added to the group's uniquemember attribute. How can I query LDAP such that I send a user dn and it returns to me the OU of the group that user belongs to?
I am doing this based on the LDAP at ldap.forumsys.com.
See the screenshot below:
Thanks.
What you want is a pretty common equality match filter, for example:
(uniquemember=uid=einstein,dc=example,dc=com)
Note, that the value is not quoted. You only need to escape / replace special characters (e.g. parenthesis). More information about LDAP filter syntax can be found in RFC 4515.
Also keep in mind that LDAP filters are very simple and provide only simple value matching (this can be confusing to people used to SQL and complex queries with joins, subselects, functions, etc.).
The user records in my company's LDAP data contain a field indicating
the manager-subordinate relationship, which contains the
distinguishedName for that employee's manager, like this:
CN=Some\\, Name,OU=WO1,OU=IT,OU=StandardUser,OU=ITUser,DC=emea,DC=foobar,DC=biz
I'd like to be able to search all employee that report to the same
person, by doing something like this in the query:
(manager=CN=Some\\, Name,OU=WO1,OU=IT,OU=StandardUser,OU=ITUser,DC=emea,DC=foobar,DC=biz)
but this doesn't seem to work (nor does it look right), nor do wildcard searches on the manager
field, which I was ready to use as a workaround.
I am using perl, with the Net::LDAP module, which is working perfectly
for other searches.
Can anyone help?
Strictly speaking, a backslash \ must be followed by two hexadecimal characters that define the byte that has been encoded.
This LDAP query successfully enumerates all users within a group:
memberOf=CN=MySubGroup1,OU=MyGroup1,OU=Global Groups,DC=mycompany,DC=com
The group MyGroup1 has two subgroups: MySubGroup1, MySubGroup2.
In order to get all the users of MyGroup1, I could make a query to get the users of MySubGroup1, another query to get the users of MySubGroup1, and then make the union.
However, I am asking how I can achieve the same results with only one LDAP query,
asking for all the users within MyGroup1 and sub-groups.
Any idea?
There is no such thing as a subgroup, just groups. The correct term is subordinate,
i.e., cn=mysubgroup1 is subordinate to ou=mygroup1, and so forth.
Use the following parameters in an LDAP search request:
base object: OU=MyGroup1,OU=Global Groups,DC=mycompany,DC=com
search scope: sub if there is more than one 'level' beneath ou=mygroup1, one otherwise
filter: (|(cn=mysubgroup1)(cn=mysubgroup2))
requested attribute: whichever multi-valued attribute whose value is the distinguished name
of each member of the group
These search request parameters should result in a search result with two entries, the distinguished
of each entry, and the attributes whose values are the distinguished names of the members of each group.
see also
LDAP: Seach best practices
LDAP: Programming practices
If your server is Microsoft Active Directory then you can use some extended rules. One of those rules does basically what you are looking for. Look at this answer.
Try this:
memberof:1.2.840.113556.1.4.1941:=CN=Some Group,OU=My Organization Unit,DC=company,DC=com
Quoting from that answer:
[...] it is possible, when using Microsoft AD LDAP, to do authorization using nested groups by using LDAP_MATCHING_RULE_IN_CHAIN matching rule. This is much faster than searching subgroups on the client, because it is done on the DC server with less queries over network.
1.2.840.113556.1.4.1941 LDAP_MATCHING_RULE_IN_CHAIN This rule is limited to filters that apply to the DN. This is a special "extended match operator that walks the chain of ancestry in objects all the way to the root until it finds a match.