C++ 如何提取用户'；使用c/c++；

我的linux应用程序支持LDAP，并使用openldap库连接到openldap服务器。对于特定功能，我需要显示每个用户的所有组的名称。当我瞄准微软的广告时，我通过寻找Memberof很容易得到结果。然而，我似乎无法在OpenLDAP中获得任何东西。希望这里有一些很酷的黑客能告诉我对示例代码的正确修改

提前谢谢

/*
 * authusername could be the name of the user that we are investigating, set it to NULL if we want to list all the users in the LDAP
 * ldaploginattr could be the search criteria that uniquely identifies a user like samAccountname or UID or UniquePrincipalName
 * ldapbasedn should be the DN scope like DC=example,DC=local
 * ldapgroupid should be the identifier for user's group enlistments like MemberOf in Microsoft AD
 * */

int ldap_to_cache (LDAP *ld, const char *authusername, const char *ldaploginattr , const char *ldapbasedn , const char *ldapgroupid)
{
    int ldap_search_result = -1,ldap_bind_result = 0;
    int num_entries_returned = 0;
    char bind_filter[512] = "";
    int cnt = 1, ret = -11; // return -1 on failure
    LDAPMessage *pMsg = NULL, *entry = NULL;

    TRACE ;

    snprintf(bind_filter, sizeof(bind_filter) -1, "(&(objectclass=person)(%s=%s))", ldaploginattr, (authusername == NULL) ? "*" : authusername);

    /* search for all the users that have the desired ldaploginattr like samAccountname or UID or UPN*/
    ldap_search_result = ldap_search_ext_s(ld, ldapbasedn, LDAP_SCOPE_SUBTREE, bind_filter, NULL, 0, NULL, NULL, NULL, 0, &pMsg);

    TRACE ;

    if (ldap_search_result != LDAP_SUCCESS ) {

    printf ("error: %d:%s\n", ldap_search_result, ldap_err2string(ldap_search_result) );

        if(pMsg == NULL)    
                return ret;

    }


    TRACE ;

    num_entries_returned = ldap_count_entries(ld, pMsg);  // if we were called with authusername = NULL, then we create a list of all users in the LDAP Directory
    ret = num_entries_returned;

    TRACE ;

    for (entry = ldap_first_entry(ld, pMsg); entry != NULL; entry = ldap_next_entry(ld, entry))
    {
        int i = 0, j = 0;
        char LDAPuser[512] = "", LDAP_user_DN[512] = "", LDAP_user_group_memberships[512] = ""; 
        char *gdn = NULL, **dn_res, **attrib_vals; 

        memset(LDAP_user_group_memberships, '\0', 512);
        char *ptr = LDAP_user_group_memberships;

        /* retrieve the names of all the users in LDAP that match the desired attribute*/
        attrib_vals = ldap_get_values(ld, entry, ldaploginattr);


        for(i = 0; attrib_vals[i] != NULL; i++)
            snprintf (LDAPuser , 512 -1 , "%s" , attrib_vals[i]);

        gdn = ldap_get_dn(ld, entry);
        snprintf(LDAP_user_DN, 512 -1, "%s",gdn);  

        printf ("LDAP user #%d : %s\n" , cnt, LDAPuser);
        printf ("userDN: %s\n" , LDAP_user_DN);
        printf ("group memberships: (%s):\n" , ldapgroupid);

        dn_res = ldap_get_values(ld, entry, ldapgroupid);

        if(dn_res != NULL) {

            for(j = 0; dn_res[j] != NULL; j++) {
                printf ("%s: %s\n" , ldapgroupid, dn_res[j]);
            }
        }

        cnt++;

        if(i > 0)
            ldap_value_free(attrib_vals);                   
        if(j > 0)       
            ldap_value_free(dn_res);        
        if (gdn != NULL )
            ldap_memfree (gdn) ;
    }

    TRACE ;

    if (pMsg != NULL)
        ldap_msgfree(pMsg);

    TRACE ;

    return cnt;
}

---

RFC2307bis成员资格存储如下：

dn: cn=abuild,ou=groups,o=company
objectClass: groupOfNames
objectClass: posixGroup
cn: abuild
description: Build System Users
gidNumber: 399
member: uid=abuild,ou=groups,o=company
member: uid=joe,ou=users,o=company

因此，代码是：

static const char *const attrs[] = {"cn", NULL};
char filter[] = "(&(objectClass=posixGroup)(member=uid\\3Djoe,ou\\3Dusers,o\\3Dcompany))";
LDAPMessage *result;
int ret;
ret = ldap_search_ext_s(conn, searchroot, LDAP_SCOPE_SUBTREE,
      filter, (char **)attrs, false, NULL, NULL, NULL, LDAP_MAXINT,
      &result);

---

不太确定我是否理解正确。但这是否意味着如果我设置：charfilter[]=“（&（objectClass=posixGroup）（member=*）”；结果是否会包含LDAP中所有组的cn？