How can I login KM with windows AD?
PostPosted:Mon Sep 16, 2013 5:16 am
Open.xml
Pls help, thanks.
Code: Select all
Configuration
<beans:bean id="ldapAuthProvider" class="org.springframework.security.ldap.authentication.LdapAuthenticationProvider">
<beans:constructor-arg ref="ldapBindAuthenticator"/>
<beans:constructor-arg ref="ldapAuthoritiesPopulator"/>
</beans:bean>
<beans:bean id="contextSource" class="org.springframework.security.ldap.DefaultSpringSecurityContextSource">
<!-- MS Active Directory -->
<beans:constructor-arg value="ldap://AD IP:389"/>
<beans:property name="userDn" value="cn=AD account,cn=Users,dc=domain"/>
<beans:property name="password" value="AD PWD"/>
<beans:property name="baseEnvironmentProperties">
<beans:map>
<beans:entry>
<beans:key>
<beans:value>java.naming.referral</beans:value>
</beans:key>
<beans:value>follow</beans:value>
</beans:entry>
</beans:map>
</beans:property>
</beans:bean>
<beans:bean id="ldapBindAuthenticator" class="org.springframework.security.ldap.authentication.BindAuthenticator">
<beans:constructor-arg ref="contextSource"/>
<beans:property name="userSearch" ref="userSearch"/>
</beans:bean>
<beans:bean id="userSearch" class="org.springframework.security.ldap.search.FilterBasedLdapUserSearch">
<!-- MS Active Directory -->
<!-- user-search-base; relative to base of configured context source -->
<beans:constructor-arg index="0" value="dc=domain"/>
<!-- user-search-filter -->
<beans:constructor-arg index="1" value="(sAMAccountName={0})"/>
<beans:constructor-arg index="2" ref="contextSource"/>
</beans:bean>
<beans:bean id="ldapAuthoritiesPopulator" class="org.springframework.security.ldap.userdetails.DefaultLdapAuthoritiesPopulator">
<beans:constructor-arg ref="contextSource" />
<beans:constructor-arg value="dc=domain" />
<beans:property name="groupSearchFilter" value="(member={0})"/>
<beans:property name="groupRoleAttribute" value="CN" />
<beans:property name="rolePrefix" value=""/>
<beans:property name="searchSubtree" value="true"/>
<beans:property name="convertToUpperCase" value="false"/>
<beans:property name="ignorePartialResultException" value="true"/>
</beans:bean>Code: Select all
An other qns:In KM, should I have to create an account which is used in AD?system.login.lowercase=true
principal.adapter=com.openkm.principal.LdapPrincipalAdapter
principal.database.filter.inactive.users=true
principal.hide.connection.roles=false
principal.ldap.mail.attribute=mail
principal.ldap.mail.search.base=dc=domain
principal.ldap.mail.search.filter=(&(objectclass=person)(sAMAccountName={0}))
principal.ldap.referral=follow
principal.ldap.role.attribute=cn
principal.ldap.role.search.base=dc=domain
principal.ldap.role.search.filter=(cn=*)
principal.ldap.roles.by.user.attribute=memberOf
principal.ldap.roles.by.user.search.base=dc=domain
principal.ldap.roles.by.user.search.filter=(&(objectClass=person)(sAMAccountName={0}))
principal.ldap.security.credentials=AD PWD
principal.ldap.security.principal=CN=AD account,cn=Users,dc=domain
principal.ldap.server=ldap://192.168.99.11:389
principal.ldap.user.attribute=sAMAccountName
principal.ldap.user.search.base=dc=domain
principal.ldap.user.search.filter=(&(objectClass=user)(objectCategory=person))
principal.ldap.username.attribute=cn
principal.ldap.username.search.base=dc=domain
principal.ldap.username.search.filter=(&(objectclass=person)(sAMAccountName={0}))
principal.ldap.users.by.role.attribute=member
principal.ldap.users.by.role.search.base=dc=domain
principal.ldap.users.by.role.search.filter=(&(objectClass=group)(cn={0}))
principal.ldap.users.from.roles=false
Pls help, thanks.