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  • Binary Search Tree Implementation

    - by Gabe
    I've searched the forum, and tried to implement the code in the threads I found. But I've been working on this real simple program since about 10am, and can't solve the seg. faults for the life of me. Any ideas on what I'm doing wrong would be greatly appreciated. BST.h (All the implementation problems should be in here.) #ifndef BST_H_ #define BST_H_ #include <stdexcept> #include <iostream> #include "btnode.h" using namespace std; /* A class to represent a templated binary search tree. */ template <typename T> class BST { private: //pointer to the root node in the tree BTNode<T>* root; public: //default constructor to make an empty tree BST(); /* You have to document these 4 functions */ void insert(T value); bool search(const T& value) const; bool search(BTNode<T>* node, const T& value) const; void printInOrder() const; void remove(const T& value); //function to print out a visual representation //of the tree (not just print the tree's values //on a single line) void print() const; private: //recursive helper function for "print()" void print(BTNode<T>* node,int depth) const; }; /* Default constructor to make an empty tree */ template <typename T> BST<T>::BST() { root = NULL; } template <typename T> void BST<T>::insert(T value) { BTNode<T>* newNode = new BTNode<T>(value); cout << newNode->data; if(root == NULL) { root = newNode; return; } BTNode<T>* current = new BTNode<T>(NULL); current = root; current->data = root->data; while(true) { if(current->left == NULL && current->right == NULL) break; if(current->right != NULL && current->left != NULL) { if(newNode->data > current->data) current = current->right; else if(newNode->data < current->data) current = current->left; } else if(current->right != NULL && current->left == NULL) { if(newNode->data < current->data) break; else if(newNode->data > current->data) current = current->right; } else if(current->right == NULL && current->left != NULL) { if(newNode->data > current->data) break; else if(newNode->data < current->data) current = current->left; } } if(current->data > newNode->data) current->left = newNode; else current->right = newNode; return; } //public helper function template <typename T> bool BST<T>::search(const T& value) const { return(search(root,value)); //start at the root } //recursive function template <typename T> bool BST<T>::search(BTNode<T>* node, const T& value) const { if(node == NULL || node->data == value) return(node != NULL); //found or couldn't find value else if(value < node->data) return search(node->left,value); //search left subtree else return search(node->right,value); //search right subtree } template <typename T> void BST<T>::printInOrder() const { //print out the value's in the tree in order // //You may need to use this function as a helper //and create a second recursive function //(see "print()" for an example) } template <typename T> void BST<T>::remove(const T& value) { if(root == NULL) { cout << "Tree is empty. No removal. "<<endl; return; } if(!search(value)) { cout << "Value is not in the tree. No removal." << endl; return; } BTNode<T>* current; BTNode<T>* parent; current = root; parent->left = NULL; parent->right = NULL; cout << root->left << "LEFT " << root->right << "RIGHT " << endl; cout << root->data << " ROOT" << endl; cout << current->data << "CURRENT BEFORE" << endl; while(current != NULL) { cout << "INTkhkjhbljkhblkjhlk " << endl; if(current->data == value) break; else if(value > current->data) { parent = current; current = current->right; } else { parent = current; current = current->left; } } cout << current->data << "CURRENT AFTER" << endl; // 3 cases : //We're looking at a leaf node if(current->left == NULL && current->right == NULL) // It's a leaf { if(parent->left == current) parent->left = NULL; else parent->right = NULL; delete current; cout << "The value " << value << " was removed." << endl; return; } // Node with single child if((current->left == NULL && current->right != NULL) || (current->left != NULL && current->right == NULL)) { if(current->left == NULL && current->right != NULL) { if(parent->left == current) { parent->left = current->right; cout << "The value " << value << " was removed." << endl; delete current; } else { parent->right = current->right; cout << "The value " << value << " was removed." << endl; delete current; } } else // left child present, no right child { if(parent->left == current) { parent->left = current->left; cout << "The value " << value << " was removed." << endl; delete current; } else { parent->right = current->left; cout << "The value " << value << " was removed." << endl; delete current; } } return; } //Node with 2 children - Replace node with smallest value in right subtree. if (current->left != NULL && current->right != NULL) { BTNode<T>* check; check = current->right; if((check->left == NULL) && (check->right == NULL)) { current = check; delete check; current->right = NULL; cout << "The value " << value << " was removed." << endl; } else // right child has children { //if the node's right child has a left child; Move all the way down left to locate smallest element if((current->right)->left != NULL) { BTNode<T>* leftCurrent; BTNode<T>* leftParent; leftParent = current->right; leftCurrent = (current->right)->left; while(leftCurrent->left != NULL) { leftParent = leftCurrent; leftCurrent = leftCurrent->left; } current->data = leftCurrent->data; delete leftCurrent; leftParent->left = NULL; cout << "The value " << value << " was removed." << endl; } else { BTNode<T>* temp; temp = current->right; current->data = temp->data; current->right = temp->right; delete temp; cout << "The value " << value << " was removed." << endl; } } return; } } /* Print out the values in the tree and their relationships visually. Sample output: 22 18 15 10 9 5 3 1 */ template <typename T> void BST<T>::print() const { print(root,0); } template <typename T> void BST<T>::print(BTNode<T>* node,int depth) const { if(node == NULL) { std::cout << std::endl; return; } print(node->right,depth+1); for(int i=0; i < depth; i++) { std::cout << "\t"; } std::cout << node->data << std::endl; print(node->left,depth+1); } #endif main.cpp #include "bst.h" #include <iostream> using namespace std; int main() { BST<int> tree; cout << endl << "LAB #13 - BINARY SEARCH TREE PROGRAM" << endl; cout << "----------------------------------------------------------" << endl; // Insert. cout << endl << "INSERT TESTS" << endl; // No duplicates allowed. tree.insert(0); tree.insert(5); tree.insert(15); tree.insert(25); tree.insert(20); // Search. cout << endl << "SEARCH TESTS" << endl; int x = 0; int y = 1; if(tree.search(x)) cout << "The value " << x << " is on the tree." << endl; else cout << "The value " << x << " is NOT on the tree." << endl; if(tree.search(y)) cout << "The value " << y << " is on the tree." << endl; else cout << "The value " << y << " is NOT on the tree." << endl; // Removal. cout << endl << "REMOVAL TESTS" << endl; tree.remove(0); tree.remove(1); tree.remove(20); // Print. cout << endl << "PRINTED DIAGRAM OF BINARY SEARCH TREE" << endl; cout << "----------------------------------------------------------" << endl; tree.print(); cout << endl << "Program terminated. Goodbye." << endl << endl; } BTNode.h #ifndef BTNODE_H_ #define BTNODE_H_ #include <iostream> /* A class to represent a node in a binary search tree. */ template <typename T> class BTNode { public: //constructor BTNode(T d); //the node's data value T data; //pointer to the node's left child BTNode<T>* left; //pointer to the node's right child BTNode<T>* right; }; /* Simple constructor. Sets the data value of the BTNode to "d" and defaults its left and right child pointers to NULL. */ template <typename T> BTNode<T>::BTNode(T d) : left(NULL), right(NULL) { data = d; } #endif Thanks.

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  • Accessing Active Directory Role Membership through LDAP using SQL Server 2005

    - by David Neale
    I would like to get a list of Active Directory users along with the security groups they are members of using SQL Server 2005 linked servers. I have the query working to retrieve records but I'm not sure how to access the memberOf attribute (it is a multi-value LDAP attribute). I have this temporary to store the information: DROP TABLE #ADUSERGROUPS CREATE TABLE #ADUSERGROUPS ( sAMAccountName varchar(30), UserGroup varchar(50) ) Each group/user association should be one row. This is my SELECT statement: SELECT sAMAccountName,memberOf FROM OpenQuery(ADSI, '<LDAP://hqdc04/DC=nt,DC=avs>; (&(objectClass=User)(sAMAccountName=9695)(sn=*)(mail=*)(userAccountControl=512)); sAMAccountName,memberOf;subtree') I get this error msg: OLE DB error trace [OLE/DB Provider 'ADSDSOObject' IRowset::GetData returned 0x40eda: Data status returned from the provider: [COLUMN_NAME=memberOf STATUS=DBSTATUS_E_CANTCONVERTVALUE], [COLUMN_NAME=sAMAccountName STATUS=DBSTATUS_S_OK]]. Msg 7346, Level 16, State 2, Line 2 Could not get the data of the row from the OLE DB provider 'ADSDSOObject'. Could not convert the data value due to reasons other than sign mismatch or overflow.

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  • Cisco ASA: Allowing and Denying VPN Access based on membership to an AD group

    - by milkandtang
    I have a Cisco ASA 5505 connecting to an Active Directory server for VPN authentication. Usually we'd restrict this to a particular OU, but in this case users which need access are spread across multiple OUs. So, I'd like to use a group to specify which users have remote access. I've created the group and added the users, but I'm having trouble figuring out how to deny users which aren't in that group. Right now, if someone connects they get assigned the correct group policy "companynamera" if they are in that group, so the LDAP mapping is working. However, users who are not in that group still authenticate fine, and their group policy becomes the LDAP path of their first group, i.e. CN=Domain Users,CN=Users,DC=example,DC=com, and then are still allowed access. How do I add a filter so that I can map everything that isn't "companynamera" to no access? Config I'm using (with some stuff such as ACLs and mappings removed, since they are just noise here): gateway# show run : Saved : ASA Version 8.2(1) ! hostname gateway domain-name corp.company-name.com enable password gDZcqZ.aUC9ML0jK encrypted passwd gDZcqZ.aUC9ML0jK encrypted names name 192.168.0.2 dc5 description FTP Server name 192.168.0.5 dc2 description Everything server name 192.168.0.6 dc4 description File Server name 192.168.0.7 ts1 description Light Use Terminal Server name 192.168.0.8 ts2 description Heavy Use Terminal Server name 4.4.4.82 primary-frontier name 5.5.5.26 primary-eschelon name 172.21.18.5 dmz1 description Kerio Mail Server and FTP Server name 4.4.4.84 ts-frontier name 4.4.4.85 vpn-frontier name 5.5.5.28 ts-eschelon name 5.5.5.29 vpn-eschelon name 5.5.5.27 email-eschelon name 4.4.4.83 guest-frontier name 4.4.4.86 email-frontier dns-guard ! interface Vlan1 nameif inside security-level 100 ip address 192.168.0.254 255.255.255.0 ! interface Vlan2 description Frontier FiOS nameif outside security-level 0 ip address primary-frontier 255.255.255.0 ! interface Vlan3 description Eschelon T1 nameif backup security-level 0 ip address primary-eschelon 255.255.255.248 ! interface Vlan4 nameif dmz security-level 50 ip address 172.21.18.254 255.255.255.0 ! interface Vlan5 nameif guest security-level 25 ip address 172.21.19.254 255.255.255.0 ! interface Ethernet0/0 switchport access vlan 2 ! interface Ethernet0/1 switchport access vlan 3 ! interface Ethernet0/2 switchport access vlan 4 ! interface Ethernet0/3 switchport access vlan 5 ! interface Ethernet0/4 ! interface Ethernet0/5 ! interface Ethernet0/6 ! interface Ethernet0/7 ! ftp mode passive clock timezone PST -8 clock summer-time PDT recurring dns domain-lookup inside dns server-group DefaultDNS name-server dc2 domain-name corp.company-name.com same-security-traffic permit intra-interface access-list companyname_splitTunnelAcl standard permit 192.168.0.0 255.255.255.0 access-list companyname_splitTunnelAcl standard permit 172.21.18.0 255.255.255.0 access-list inside_nat0_outbound extended permit ip any 172.21.20.0 255.255.255.0 access-list inside_nat0_outbound extended permit ip any 172.21.18.0 255.255.255.0 access-list bypassingnat_dmz extended permit ip 172.21.18.0 255.255.255.0 192.168.0.0 255.255.255.0 pager lines 24 logging enable logging buffer-size 12288 logging buffered warnings logging asdm notifications mtu inside 1500 mtu outside 1500 mtu backup 1500 mtu dmz 1500 mtu guest 1500 ip local pool VPNpool 172.21.20.50-172.21.20.59 mask 255.255.255.0 no failover icmp unreachable rate-limit 1 burst-size 1 no asdm history enable arp timeout 14400 global (outside) 1 interface global (outside) 2 email-frontier global (outside) 3 guest-frontier global (backup) 1 interface global (dmz) 1 interface nat (inside) 0 access-list inside_nat0_outbound nat (inside) 2 dc5 255.255.255.255 nat (inside) 1 192.168.0.0 255.255.255.0 nat (dmz) 0 access-list bypassingnat_dmz nat (dmz) 2 dmz1 255.255.255.255 nat (dmz) 1 172.21.18.0 255.255.255.0 access-group outside_access_in in interface outside access-group dmz_access_in in interface dmz route outside 0.0.0.0 0.0.0.0 4.4.4.1 1 track 1 route backup 0.0.0.0 0.0.0.0 5.5.5.25 254 timeout xlate 3:00:00 timeout conn 1:00:00 half-closed 0:10:00 udp 0:02:00 icmp 0:00:02 timeout sunrpc 0:10:00 h323 0:05:00 h225 1:00:00 mgcp 0:05:00 mgcp-pat 0:05:00 timeout sip 0:30:00 sip_media 0:02:00 sip-invite 0:03:00 sip-disconnect 0:02:00 timeout sip-provisional-media 0:02:00 uauth 0:05:00 absolute timeout tcp-proxy-reassembly 0:01:00 ldap attribute-map RemoteAccessMap map-name memberOf IETF-Radius-Class map-value memberOf CN=RemoteAccess,CN=Users,DC=corp,DC=company-name,DC=com companynamera dynamic-access-policy-record DfltAccessPolicy aaa-server ActiveDirectory protocol ldap aaa-server ActiveDirectory (inside) host dc2 ldap-base-dn dc=corp,dc=company-name,dc=com ldap-scope subtree ldap-login-password * ldap-login-dn cn=administrator,ou=Admins,dc=corp,dc=company-name,dc=com server-type microsoft aaa-server ADRemoteAccess protocol ldap aaa-server ADRemoteAccess (inside) host dc2 ldap-base-dn dc=corp,dc=company-name,dc=com ldap-scope subtree ldap-login-password * ldap-login-dn cn=administrator,ou=Admins,dc=corp,dc=company-name,dc=com server-type microsoft ldap-attribute-map RemoteAccessMap aaa authentication enable console LOCAL aaa authentication ssh console LOCAL http server enable http 192.168.0.0 255.255.255.0 inside no snmp-server location no snmp-server contact snmp-server enable traps snmp authentication linkup linkdown coldstart sla monitor 123 type echo protocol ipIcmpEcho 4.4.4.1 interface outside num-packets 3 frequency 10 sla monitor schedule 123 life forever start-time now crypto ipsec transform-set ESP-3DES-SHA esp-3des esp-sha-hmac crypto ipsec security-association lifetime seconds 28800 crypto ipsec security-association lifetime kilobytes 4608000 crypto dynamic-map outside_dyn_map 20 set pfs crypto dynamic-map outside_dyn_map 20 set transform-set ESP-3DES-SHA crypto map outside_map 65535 ipsec-isakmp dynamic outside_dyn_map crypto map outside_map interface outside crypto isakmp enable outside crypto isakmp policy 10 authentication pre-share encryption 3des hash sha group 2 lifetime 86400 ! track 1 rtr 123 reachability telnet timeout 5 ssh 192.168.0.0 255.255.255.0 inside ssh timeout 5 ssh version 2 console timeout 0 management-access inside dhcpd auto_config outside ! threat-detection basic-threat threat-detection statistics access-list no threat-detection statistics tcp-intercept webvpn group-policy companynamera internal group-policy companynamera attributes wins-server value 192.168.0.5 dns-server value 192.168.0.5 vpn-tunnel-protocol IPSec password-storage enable split-tunnel-policy tunnelspecified split-tunnel-network-list value companyname_splitTunnelAcl default-domain value corp.company-name.com split-dns value corp.company-name.com group-policy companyname internal group-policy companyname attributes wins-server value 192.168.0.5 dns-server value 192.168.0.5 vpn-tunnel-protocol IPSec password-storage enable split-tunnel-policy tunnelspecified split-tunnel-network-list value companyname_splitTunnelAcl default-domain value corp.company-name.com split-dns value corp.company-name.com username admin password IhpSqtN210ZsNaH. encrypted privilege 15 tunnel-group companyname type remote-access tunnel-group companyname general-attributes address-pool VPNpool authentication-server-group ActiveDirectory LOCAL default-group-policy companyname tunnel-group companyname ipsec-attributes pre-shared-key * tunnel-group companynamera type remote-access tunnel-group companynamera general-attributes address-pool VPNpool authentication-server-group ADRemoteAccess LOCAL default-group-policy companynamera tunnel-group companynamera ipsec-attributes pre-shared-key * ! class-map type inspect ftp match-all ftp-inspection-map class-map inspection_default match default-inspection-traffic ! ! policy-map type inspect ftp ftp-inspection-map parameters class ftp-inspection-map policy-map type inspect dns migrated_dns_map_1 parameters message-length maximum 512 policy-map global_policy class inspection_default inspect dns migrated_dns_map_1 inspect ftp inspect h323 h225 inspect h323 ras inspect http inspect ils inspect netbios inspect rsh inspect rtsp inspect skinny inspect sqlnet inspect sunrpc inspect tftp inspect sip inspect xdmcp inspect icmp inspect icmp error inspect esmtp inspect pptp ! service-policy global_policy global prompt hostname context Cryptochecksum:487525494a81c8176046fec475d17efe : end gateway# Thanks so much!

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  • Scala: working around the "illegal cyclic reference"

    - by Paul Milovanov
    Hi all, I'm trying to implement a HashMap-based tree that'd support O(1) subtree lookup for a given root key. To that goal, I'm trying to do the following: scala> type Q = HashMap[Char, Q] <console>:6: error: illegal cyclic reference involving type Q type Q = HashMap[Char, Q] ^ So the question is, is there a way for me to do something of the sort without resorting to the ugly HashMap[Char, Any] with subsequent casting of values to HashMap[Char, Any]? Now, I also see that I can use something like the following to avoid the cyclic-reference error, and it might even be cleaner -- but it'd be nice to find out how to correctly do it the first way, just for the educational value. import collections.mutable.HashMap class LTree { val children = new HashMap[Char, LTree] } Thanks a bunch.

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  • Acquiring AD OU list.

    - by Stephen Murby
    Hi, I am looking to be able to pull a list of current OU's from Active Directory I have been looking at some example code online for sometime, but O don't seem to be able to get this to work. string defaultNamingContext; DirectoryEntry rootDSE = new DirectoryEntry("LDAP://RootDSE"); defaultNamingContext = rootDSE.Properties["defaultNamingContext"].Value.ToString(); DirectorySearcher ouSearch = new DirectorySearcher(rootDSE, "(objectClass=organizationalUnit)", null, SearchScope.Subtree); MessageBox.Show(rootDSE.ToString()); try { SearchResultCollection collectedResult = ouSearch.FindAll(); foreach (SearchResult temp in collectedResult) { comboBox1.Items.Add(temp.Properties["name"][0]); DirectoryEntry ou = temp.GetDirectoryEntry(); } The error I get is There provider does not support searching and cannot search LDAP://RootDSE Any Ideas? for each of those returned search results I want to add them to a combo box. (shouldn't be too hard)

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  • Active Directory List OU's

    - by Stephen Murby
    I have this code currently, string defaultNamingContext; DirectoryEntry rootDSE = new DirectoryEntry("LDAP://RootDSE"); defaultNamingContext = rootDSE.Properties["defaultNamingContext"].Value.ToString(); rootDSE = new DirectoryEntry("LDAP://" + defaultNamingContext); //DirectoryEntry domain = new DirectoryEntry((string)"LDAP://" + defaultNamingContext); DirectorySearcher ouSearch = new DirectorySearcher(rootDSE,"(objectCategory=Organizational-Unit)", null, SearchScope.Subtree); MessageBox.Show(rootDSE.Path.ToString()); try { SearchResultCollection collectedResult = ouSearch.FindAll(); foreach (SearchResult temp in collectedResult) { comboBox1.Items.Add(temp.Properties["name"][0]); DirectoryEntry ou = temp.GetDirectoryEntry(); } } When i use the debugger i can see that rootDSE.Path is infact pointing to the right place, in this case "DC=g-t-p,DC=Local" but the directory searcher doesn't find any results. Can anyone help?

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  • Finding the heaviest length-constrained path in a weighted Binary Tree

    - by Hristo
    UPDATE I worked out an algorithm that I think runs in O(n*k) running time. Below is the pseudo-code: routine heaviestKPath( T, k ) // create 2D matrix with n rows and k columns with each element = -8 // we make it size k+1 because the 0th column must be all 0s for a later // function to work properly and simplicity in our algorithm matrix = new array[ T.getVertexCount() ][ k + 1 ] (-8); // set all elements in the first column of this matrix = 0 matrix[ n ][ 0 ] = 0; // fill our matrix by traversing the tree traverseToFillMatrix( T.root, k ); // consider a path that would arc over a node globalMaxWeight = -8; findArcs( T.root, k ); return globalMaxWeight end routine // node = the current node; k = the path length; node.lc = node’s left child; // node.rc = node’s right child; node.idx = node’s index (row) in the matrix; // node.lc.wt/node.rc.wt = weight of the edge to left/right child; routine traverseToFillMatrix( node, k ) if (node == null) return; traverseToFillMatrix(node.lc, k ); // recurse left traverseToFillMatrix(node.rc, k ); // recurse right // in the case that a left/right child doesn’t exist, or both, // let’s assume the code is smart enough to handle these cases matrix[ node.idx ][ 1 ] = max( node.lc.wt, node.rc.wt ); for i = 2 to k { // max returns the heavier of the 2 paths matrix[node.idx][i] = max( matrix[node.lc.idx][i-1] + node.lc.wt, matrix[node.rc.idx][i-1] + node.rc.wt); } end routine // node = the current node, k = the path length routine findArcs( node, k ) if (node == null) return; nodeMax = matrix[node.idx][k]; longPath = path[node.idx][k]; i = 1; j = k-1; while ( i+j == k AND i < k ) { left = node.lc.wt + matrix[node.lc.idx][i-1]; right = node.rc.wt + matrix[node.rc.idx][j-1]; if ( left + right > nodeMax ) { nodeMax = left + right; } i++; j--; } // if this node’s max weight is larger than the global max weight, update if ( globalMaxWeight < nodeMax ) { globalMaxWeight = nodeMax; } findArcs( node.lc, k ); // recurse left findArcs( node.rc, k ); // recurse right end routine Let me know what you think. Feedback is welcome. I think have come up with two naive algorithms that find the heaviest length-constrained path in a weighted Binary Tree. Firstly, the description of the algorithm is as follows: given an n-vertex Binary Tree with weighted edges and some value k, find the heaviest path of length k. For both algorithms, I'll need a reference to all vertices so I'll just do a simple traversal of the Tree to have a reference to all vertices, with each vertex having a reference to its left, right, and parent nodes in the tree. Algorithm 1 For this algorithm, I'm basically planning on running DFS from each node in the Tree, with consideration to the fixed path length. In addition, since the path I'm looking for has the potential of going from left subtree to root to right subtree, I will have to consider 3 choices at each node. But this will result in a O(n*3^k) algorithm and I don't like that. Algorithm 2 I'm essentially thinking about using a modified version of Dijkstra's Algorithm in order to consider a fixed path length. Since I'm looking for heaviest and Dijkstra's Algorithm finds the lightest, I'm planning on negating all edge weights before starting the traversal. Actually... this doesn't make sense since I'd have to run Dijkstra's on each node and that doesn't seem very efficient much better than the above algorithm. So I guess my main questions are several. Firstly, do the algorithms I've described above solve the problem at hand? I'm not totally certain the Dijkstra's version will work as Dijkstra's is meant for positive edge values. Now, I am sure there exist more clever/efficient algorithms for this... what is a better algorithm? I've read about "Using spine decompositions to efficiently solve the length-constrained heaviest path problem for trees" but that is really complicated and I don't understand it at all. Are there other algorithms that tackle this problem, maybe not as efficiently as spine decomposition but easier to understand? Thanks.

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  • Finding backedges in a graph, with special conditions.

    - by Morteza M.
    There is a vertex v, such that from the subtree rooted at v, there are at least two backedges to proper ancestors of v. The problem is finding whether such backedges exist or not ( finding v is not important at all). I run DFS algorithm, I can find backedges and save them in an array. I know which backedges in this array belong to a common tree. but I have problem on matching this condition in O(E) time. can anyone help me with that?

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  • elisp newbie question: Can't find 'filename' function definition in org.el?

    - by Dave Paroulek
    I really love org-mode in emacs and want to customize a few things. While reading thru org.el, I'm finding several references to filename but can't find filename using describe-function? I'm sure there's a simple answer, but I'm just learning elisp and it's not obvious. Any insight into where filename is defined? And/or if it's not a function, what is it? For example, filename on line 25502: (filename (if to-buffer (expand-file-name (concat (file-name-sans-extension (or (and subtree-p (org-entry-get (region-beginning) "EXPORT_FILE_NAME" t)) (file-name-nondirectory buffer-file-name))) "." html-extension) (file-name-as-directory (or pub-dir (org-export-directory :html opt-plist))))))

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  • postgresql syntax while exists loop

    - by veilig
    I'm working at function from Joe Celkos book - Trees and Hierarchies in SQL for Smarties I'm trying to delete a subtree from an adjacency list but part my function is not working yet. WHILE EXISTS –– mark leaf nodes (SELECT * FROM OrgChart WHERE boss_emp_nbr = -99999 AND emp_nbr > -99999) LOOP –– get list of next level subordinates DELETE FROM WorkingTable; INSERT INTO WorkingTable SELECT emp_nbr FROM OrgChart WHERE boss_emp_nbr = -99999; –– mark next level of subordinates UPDATE OrgChart SET emp_nbr = -99999 WHERE boss_emp_nbr IN (SELECT emp_nbr FROM WorkingTable); END LOOP; my question: is the WHILE EXISTS correct for use w/ postgresql? I appear to be stumbling and getting caught in an infinite loop in this part. Perhaps there is a more correct syntax I am unaware of.

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  • Is there a tool for detecting Visual Studio projects with duplicate GUIDs?

    - by sharptooth
    When creating new Visual Studio C++ projects there're two ways: either run the wizard and then painfully change all the necessary settings in the project or just copy and existing project, rename everything there and add files into it. The second variant is great except that the .vcproj file stores a project GUID. This GUID is used to track project dependencies and the startup project when two or more projects are in one solution. If any two projects in one solution have identical GUIDs problems can arise - dependencies are lost and the startup prject is reset on next solution reload. Clearly there's a need for a tool that would scan the filesystem subtree and detect projects with identical GUIDs here. Before I start writing one ... is there a ready tool for that?

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  • How to find largest common sub-tree in the given two binary search trees?

    - by Bhushan
    Two BSTs (Binary Search Trees) are given. How to find largest common sub-tree in the given two binary trees? EDIT 1: Here is what I have thought: Let, r1 = current node of 1st tree r2 = current node of 2nd tree There are some of the cases I think we need to consider: Case 1 : r1.data < r2.data 2 subproblems to solve: first, check r1 and r2.left second, check r1.right and r2 Case 2 : r1.data > r2.data 2 subproblems to solve: - first, check r1.left and r2 - second, check r1 and r2.right Case 3 : r1.data == r2.data Again, 2 cases to consider here: (a) current node is part of largest common BST compute common subtree size rooted at r1 and r2 (b)current node is NOT part of largest common BST 2 subproblems to solve: first, solve r1.left and r2.left second, solve r1.right and r2.right I can think of the cases we need to check, but I am not able to code it, as of now. And it is NOT a homework problem. Does it look like?

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  • How to find longest common substring using trees?

    - by user384706
    The longest common substring problem according to wiki can be solved using a suffix tree. From wiki: The longest common substrings of a set of strings can be found by building a generalised suffix tree for the strings, and then finding the deepest internal nodes which have leaf nodes from all the strings in the subtree below it I don't get this. Example: if I have: ABCDE and XABCZ then the suffix tree is (some branches from XABCZ omitted due to space): The longest common substring is ABC but it is not I can not see how the description of wiki helps here. ABC is not the deepest internal nodes with leaf nodes. Any help to understand how this works?

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  • How to save an order (permutation) in an sql db

    - by Bendlas
    I have a tree structure in an sql table like so: CREATE TABLE containers ( container_id serial NOT NULL PRIMARY KEY, parent integer REFERENCES containers (container_id)) Now i want to define an ordering between nodes with the same parent. I Have thought of adding a node_index column, to ORDER BY, but that seem suboptimal, since that involves modifying the index of a lot of nodes when modifying the stucture. That could include adding, removing, reordering or moving nodes from some subtree to another. Is there a sql datatype for an ordered sequence, or an efficient way to emulate one? Doesn't need to be fully standard sql, I just need a solution for mssql and hopefully postgresql EDIT To make it clear, the ordering is arbitrary. Actually, the user will be able to drag'n'drop tree nodes in the GUI

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  • Get machine name from Active Directory

    - by Stephen Murby
    I have performed an "LDAP://" query to get a list of computers within a specified OU, my issue is not being able to collect just the computer "name" or even "cn". DirectoryEntry toShutdown = new DirectoryEntry("LDAP://" + comboBox1.Text.ToString()); DirectorySearcher machineSearch = new DirectorySearcher(toShutdown); //machineSearch.Filter = "(objectCatergory=computer)"; machineSearch.Filter = "(objectClass=computer)"; machineSearch.SearchScope = SearchScope.Subtree; machineSearch.PropertiesToLoad.Add("name"); SearchResultCollection allMachinesCollected = machineSearch.FindAll(); Methods myMethods = new Methods(); string pcName; foreach (SearchResult oneMachine in allMachinesCollected) { //pcName = oneMachine.Properties.PropertyNames.ToString(); pcName = oneMachine.Properties["name"].ToString(); MessageBox.Show(pcName); } Help much appreciated.

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  • Recursive move utility on Unix?

    - by Thomas Vander Stichele
    Sometimes I have two trees that used to have the same content, but have grown out of sync (because I moved disks around or whatever). A good example is a tree where I mirror upstream packages from Fedora. I want to merge those two trees again by moving all of the files from tree1 into tree2. Usually I do this with: rsync -arv tree1/* tree2 Then delete tree1. However, this takes an awful lot of time and disk space, and it would be much easier to be able to do: mv -r tree1/* tree2 In other words, a recursive move. It would be faster because first of all it would not even copy, just move the inodes, and second I wouldn't need a delete at the end. Does this exist ? As a test case, consider the following sequence of commands: $ mkdir -p a/b $ touch a/b/c1 $ rsync -arv a/ a2 sending incremental file list created directory ./ b/ b/c1 b/c2 sent 173 bytes received 57 bytes 460.00 bytes/sec total size is 0 speedup is 0.00 $ touch a/b/c2 What command would now have the effect of moving a/b/c2 to a2/b/c2 and then deleting the a subtree (since everything in it is already in the destination tree) ?

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  • How can I make Mac OS X Address Book display a person’s home address from an LDAP server?

    - by Arcturus
    Hi, (I've posted this question on Stack Overflow first, but someone told me it belonged here.) I have a custom LDAP server, which I can customize to generate whichever object class and attributes I need. I'm trying to display people from that server in the Mac OS X address book. Names and organizations display correctly, as well as work-related phone and address. However, I've never been able to have a home address displayed in the address book. This is an example of output from running a ldapsearch: # extended LDIF # # LDAPv3 # base <dc=example,dc=com> with scope subtree # filter: (givenName=Joh*) # requesting: ALL # # 10041, example.com dn: uid=10041,dc=example,dc=com objectclass: top objectclass: person objectclass: organizationalPerson objectclass: inetOrgPerson objectclass: mozillaOrgPerson uid: 10041 cn: John Doe givenName: John sn: Doe o: Acme telephoneNumber: 500 00 00 mobile: 500 00 00 mail: [email protected] street: Baker St postalCode: 10098 l: New York c: US homePostalAddress: White St mozillaHomePostalCode: 10098 mozillaHomeLocalityName: New York mozillaHomeCountryName: US # search result search: 2 result: 0 Success # numResponses: 2 # numEntries: 1 Every piece of information shows up in the address book up to here: homePostalAddress: White St mozillaHomePostalCode: 10098 mozillaHomeLocalityName: New York mozillaHomeCountryName: US Which object class or attribute name should I use to have the home address show up in the Mac OS X address book?

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  • Authenticate users with Zimbra LDAP Server from other CentOS clients

    - by efesaid
    I'am wondering that how can integrate my database,web,backup etc.. centos servers with Zimbra LDAP Server. Does it require more advanced configuration than standart ldap authentication ? My zimbra server version is [zimbra@zimbra ~]$ zmcontrol -v Release 8.0.5_GA_5839.RHEL6_64_20130910123908 RHEL6_64 FOSS edition. My LDAP Server status is [zimbra@ldap ~]$ zmcontrol status Host ldap.domain.com ldap Running snmp Running stats Running zmconfigd Running I already installed nss-pam-ldapd packages to my servers. [root@www]# rpm -qa | grep ldap nss-pam-ldapd-0.7.5-18.2.el6_4.x86_64 apr-util-ldap-1.3.9-3.el6_0.1.x86_64 pam_ldap-185-11.el6.x86_64 openldap-2.4.23-32.el6_4.1.x86_64 My /etc/nslcd.conf is [root@www]# tail -n 7 /etc/nslcd.conf uid nslcd gid ldap # This comment prevents repeated auto-migration of settings. uri ldap://ldap.domain.com base dc=domain,dc=com binddn uid=zimbra,cn=admins,cn=zimbra bindpw **pass** ssl no tls_cacertdir /etc/openldap/cacerts When i run [root@www ~]# id username id: username: No such user But i am sure that username user exist on ldap server. EDIT : When i run ldapsearch command i got all result with credentials and dn. [root@www ~]# ldapsearch -H ldap://ldap.domain.com:389 -w **pass** -D uid=zimbra,cn=admins,cn=zimbra -x 'objectclass=*' # extended LDIF # # LDAPv3 # base <dc=domain,dc=com> (default) with scope subtree # filter: objectclass=* # requesting: ALL # # domain.com dn: dc=domain,dc=com zimbraDomainType: local zimbraDomainStatus: active . . .

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  • org-sort multi: date/time (?d ?t) | priority (?p) | title (?a)

    - by lawlist
    Is anyone aware of an org-sort function / modification that can refile / organize a group of TODO so that it sorts them by three (3) criteria: first sort by due date, second sort by priority, and third sort by by title of the task? EDIT: I believe that org-sort by deadline (?d) has a bug that cannot properly handle undated tasks. I am working on a workaround (i.e., moving the undated todo to a different heading before the deadline (?d) sort occurs), but perhaps the best thing to do would be to try and fix the original sorting function. Development of the workaround can be found in this thread (i.e., moving the undated tasks to a different heading in one fell swoop): How to automate org-refile for multiple todo EDIT: Apparently, the following code (ancient history) that I found on the internet was eventually modified and included as a part of org-sort-entries. Unfortunately, undated todo are not properly sorted when sorting by deadline -- i.e., they are mixed in with the dated todo. ;; multiple sort (defun org-sort-multi (&rest sort-types) "Multiple sorts on a certain level of an outline tree, or plain list items. SORT-TYPES is a list where each entry is either a character or a cons pair (BOOL . CHAR), where BOOL is whether or not to sort case-sensitively, and CHAR is one of the characters defined in `org-sort-entries-or-items'. Entries are applied in back to front order. Example: To sort first by TODO status, then by priority, then by date, then alphabetically (case-sensitive) use the following call: (org-sort-multi '(?d ?p ?t (t . ?a)))" (interactive) (dolist (x (nreverse sort-types)) (when (char-valid-p x) (setq x (cons nil x))) (condition-case nil (org-sort-entries (car x) (cdr x)) (error nil)))) ;; sort current level (defun lawlist-sort (&rest sort-types) "Sort the current org level. SORT-TYPES is a list where each entry is either a character or a cons pair (BOOL . CHAR), where BOOL is whether or not to sort case-sensitively, and CHAR is one of the characters defined in `org-sort-entries-or-items'. Entries are applied in back to front order. Defaults to \"?o ?p\" which is sorted by TODO status, then by priority" (interactive) (when (equal mode-name "Org") (let ((sort-types (or sort-types (if (or (org-entry-get nil "TODO") (org-entry-get nil "PRIORITY")) '(?d ?t ?p) ;; date, time, priority '((nil . ?a)))))) (save-excursion (outline-up-heading 1) (let ((start (point)) end) (while (and (not (bobp)) (not (eobp)) (<= (point) start)) (condition-case nil (outline-forward-same-level 1) (error (outline-up-heading 1)))) (unless (> (point) start) (goto-char (point-max))) (setq end (point)) (goto-char start) (apply 'org-sort-multi sort-types) (goto-char end) (when (eobp) (forward-line -1)) (when (looking-at "^\\s-*$") ;; (delete-line) ) (goto-char start) ;; (dotimes (x ) (org-cycle)) ))))) EDIT: Here is a more modern version of multi-sort, which is likely based upon further development of the above-code: (defun org-sort-all () (interactive) (save-excursion (goto-char (point-min)) (while (re-search-forward "^\* " nil t) (goto-char (match-beginning 0)) (condition-case err (progn (org-sort-entries t ?a) (org-sort-entries t ?p) (org-sort-entries t ?o) (forward-line)) (error nil))) (goto-char (point-min)) (while (re-search-forward "\* PROJECT " nil t) (goto-char (line-beginning-position)) (ignore-errors (org-sort-entries t ?a) (org-sort-entries t ?p) (org-sort-entries t ?o)) (forward-line)))) EDIT: The best option will be to fix sorting of deadlines (?d) so that undated todo are moved to the bottom of the outline, instead of mixed in with the dated todo. Here is an excerpt from the current org.el included within Emacs Trunk (as of July 1, 2013): (defun org-sort (with-case) "Call `org-sort-entries', `org-table-sort-lines' or `org-sort-list'. Optional argument WITH-CASE means sort case-sensitively." (interactive "P") (cond ((org-at-table-p) (org-call-with-arg 'org-table-sort-lines with-case)) ((org-at-item-p) (org-call-with-arg 'org-sort-list with-case)) (t (org-call-with-arg 'org-sort-entries with-case)))) (defun org-sort-remove-invisible (s) (remove-text-properties 0 (length s) org-rm-props s) (while (string-match org-bracket-link-regexp s) (setq s (replace-match (if (match-end 2) (match-string 3 s) (match-string 1 s)) t t s))) s) (defvar org-priority-regexp) ; defined later in the file (defvar org-after-sorting-entries-or-items-hook nil "Hook that is run after a bunch of entries or items have been sorted. When children are sorted, the cursor is in the parent line when this hook gets called. When a region or a plain list is sorted, the cursor will be in the first entry of the sorted region/list.") (defun org-sort-entries (&optional with-case sorting-type getkey-func compare-func property) "Sort entries on a certain level of an outline tree. If there is an active region, the entries in the region are sorted. Else, if the cursor is before the first entry, sort the top-level items. Else, the children of the entry at point are sorted. Sorting can be alphabetically, numerically, by date/time as given by a time stamp, by a property or by priority. The command prompts for the sorting type unless it has been given to the function through the SORTING-TYPE argument, which needs to be a character, \(?n ?N ?a ?A ?t ?T ?s ?S ?d ?D ?p ?P ?o ?O ?r ?R ?f ?F). Here is the precise meaning of each character: n Numerically, by converting the beginning of the entry/item to a number. a Alphabetically, ignoring the TODO keyword and the priority, if any. o By order of TODO keywords. t By date/time, either the first active time stamp in the entry, or, if none exist, by the first inactive one. s By the scheduled date/time. d By deadline date/time. c By creation time, which is assumed to be the first inactive time stamp at the beginning of a line. p By priority according to the cookie. r By the value of a property. Capital letters will reverse the sort order. If the SORTING-TYPE is ?f or ?F, then GETKEY-FUNC specifies a function to be called with point at the beginning of the record. It must return either a string or a number that should serve as the sorting key for that record. Comparing entries ignores case by default. However, with an optional argument WITH-CASE, the sorting considers case as well." (interactive "P") (let ((case-func (if with-case 'identity 'downcase)) (cmstr ;; The clock marker is lost when using `sort-subr', let's ;; store the clocking string. (when (equal (marker-buffer org-clock-marker) (current-buffer)) (save-excursion (goto-char org-clock-marker) (looking-back "^.*") (match-string-no-properties 0)))) start beg end stars re re2 txt what tmp) ;; Find beginning and end of region to sort (cond ((org-region-active-p) ;; we will sort the region (setq end (region-end) what "region") (goto-char (region-beginning)) (if (not (org-at-heading-p)) (outline-next-heading)) (setq start (point))) ((or (org-at-heading-p) (condition-case nil (progn (org-back-to-heading) t) (error nil))) ;; we will sort the children of the current headline (org-back-to-heading) (setq start (point) end (progn (org-end-of-subtree t t) (or (bolp) (insert "\n")) (org-back-over-empty-lines) (point)) what "children") (goto-char start) (show-subtree) (outline-next-heading)) (t ;; we will sort the top-level entries in this file (goto-char (point-min)) (or (org-at-heading-p) (outline-next-heading)) (setq start (point)) (goto-char (point-max)) (beginning-of-line 1) (when (looking-at ".*?\\S-") ;; File ends in a non-white line (end-of-line 1) (insert "\n")) (setq end (point-max)) (setq what "top-level") (goto-char start) (show-all))) (setq beg (point)) (if (>= beg end) (error "Nothing to sort")) (looking-at "\\(\\*+\\)") (setq stars (match-string 1) re (concat "^" (regexp-quote stars) " +") re2 (concat "^" (regexp-quote (substring stars 0 -1)) "[ \t\n]") txt (buffer-substring beg end)) (if (not (equal (substring txt -1) "\n")) (setq txt (concat txt "\n"))) (if (and (not (equal stars "*")) (string-match re2 txt)) (error "Region to sort contains a level above the first entry")) (unless sorting-type (message "Sort %s: [a]lpha [n]umeric [p]riority p[r]operty todo[o]rder [f]unc [t]ime [s]cheduled [d]eadline [c]reated A/N/P/R/O/F/T/S/D/C means reversed:" what) (setq sorting-type (read-char-exclusive)) (and (= (downcase sorting-type) ?f) (setq getkey-func (org-icompleting-read "Sort using function: " obarray 'fboundp t nil nil)) (setq getkey-func (intern getkey-func))) (and (= (downcase sorting-type) ?r) (setq property (org-icompleting-read "Property: " (mapcar 'list (org-buffer-property-keys t)) nil t)))) (message "Sorting entries...") (save-restriction (narrow-to-region start end) (let ((dcst (downcase sorting-type)) (case-fold-search nil) (now (current-time))) (sort-subr (/= dcst sorting-type) ;; This function moves to the beginning character of the "record" to ;; be sorted. (lambda nil (if (re-search-forward re nil t) (goto-char (match-beginning 0)) (goto-char (point-max)))) ;; This function moves to the last character of the "record" being ;; sorted. (lambda nil (save-match-data (condition-case nil (outline-forward-same-level 1) (error (goto-char (point-max)))))) ;; This function returns the value that gets sorted against. (lambda nil (cond ((= dcst ?n) (if (looking-at org-complex-heading-regexp) (string-to-number (match-string 4)) nil)) ((= dcst ?a) (if (looking-at org-complex-heading-regexp) (funcall case-func (match-string 4)) nil)) ((= dcst ?t) (let ((end (save-excursion (outline-next-heading) (point)))) (if (or (re-search-forward org-ts-regexp end t) (re-search-forward org-ts-regexp-both end t)) (org-time-string-to-seconds (match-string 0)) (org-float-time now)))) ((= dcst ?c) (let ((end (save-excursion (outline-next-heading) (point)))) (if (re-search-forward (concat "^[ \t]*\\[" org-ts-regexp1 "\\]") end t) (org-time-string-to-seconds (match-string 0)) (org-float-time now)))) ((= dcst ?s) (let ((end (save-excursion (outline-next-heading) (point)))) (if (re-search-forward org-scheduled-time-regexp end t) (org-time-string-to-seconds (match-string 1)) (org-float-time now)))) ((= dcst ?d) (let ((end (save-excursion (outline-next-heading) (point)))) (if (re-search-forward org-deadline-time-regexp end t) (org-time-string-to-seconds (match-string 1)) (org-float-time now)))) ((= dcst ?p) (if (re-search-forward org-priority-regexp (point-at-eol) t) (string-to-char (match-string 2)) org-default-priority)) ((= dcst ?r) (or (org-entry-get nil property) "")) ((= dcst ?o) (if (looking-at org-complex-heading-regexp) (- 9999 (length (member (match-string 2) org-todo-keywords-1))))) ((= dcst ?f) (if getkey-func (progn (setq tmp (funcall getkey-func)) (if (stringp tmp) (setq tmp (funcall case-func tmp))) tmp) (error "Invalid key function `%s'" getkey-func))) (t (error "Invalid sorting type `%c'" sorting-type)))) nil (cond ((= dcst ?a) 'string<) ((= dcst ?f) compare-func) ((member dcst '(?p ?t ?s ?d ?c)) '<))))) (run-hooks 'org-after-sorting-entries-or-items-hook) ;; Reset the clock marker if needed (when cmstr (save-excursion (goto-char start) (search-forward cmstr nil t) (move-marker org-clock-marker (point)))) (message "Sorting entries...done"))) (defun org-do-sort (table what &optional with-case sorting-type) "Sort TABLE of WHAT according to SORTING-TYPE. The user will be prompted for the SORTING-TYPE if the call to this function does not specify it. WHAT is only for the prompt, to indicate what is being sorted. The sorting key will be extracted from the car of the elements of the table. If WITH-CASE is non-nil, the sorting will be case-sensitive." (unless sorting-type (message "Sort %s: [a]lphabetic, [n]umeric, [t]ime. A/N/T means reversed:" what) (setq sorting-type (read-char-exclusive))) (let ((dcst (downcase sorting-type)) extractfun comparefun) ;; Define the appropriate functions (cond ((= dcst ?n) (setq extractfun 'string-to-number comparefun (if (= dcst sorting-type) '< '>))) ((= dcst ?a) (setq extractfun (if with-case (lambda(x) (org-sort-remove-invisible x)) (lambda(x) (downcase (org-sort-remove-invisible x)))) comparefun (if (= dcst sorting-type) 'string< (lambda (a b) (and (not (string< a b)) (not (string= a b))))))) ((= dcst ?t) (setq extractfun (lambda (x) (if (or (string-match org-ts-regexp x) (string-match org-ts-regexp-both x)) (org-float-time (org-time-string-to-time (match-string 0 x))) 0)) comparefun (if (= dcst sorting-type) '< '>))) (t (error "Invalid sorting type `%c'" sorting-type))) (sort (mapcar (lambda (x) (cons (funcall extractfun (car x)) (cdr x))) table) (lambda (a b) (funcall comparefun (car a) (car b))))))

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  • Checksum for Protecting Read-Only Documents

    - by Kim
    My father owns a small business and has to hand over several year's worth of financial documents to his insurance's auditor. He's asked me to go through and make sure everything is "read-only" so the data (the files) absolutely, positively cannot be modified or manipulated (he's a bit paranoid). We're talking about 20,000 documents (emails, spreadsheets, etc.). My first inclination was to place everything inside of one root folder ("mydadsdocs/") and then write a script that recursively traversed its directory subtree and set the file permissions to read-only. But then I got to thinking: that's a lot of work for me to do to satisfy an old man who is just being paranoid, and afterall, if someone really wanted to modify a read-only file, it would be pretty easy to change file permissions anyways, soo.... Is there like a checksum I could run on the root folder, something that was very quick and easy, and that would basically "stamp" the data in that folder so if someone did change it, my father would have someone of knowing/proving it? If so, how? If not, any other recommendations that are quick, cheap (free) and effective?

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  • Openldap with ppolicy

    - by nitins
    We have working installation of OpenLDAP version 2.4 which is using shadowAccount attributes. I want to enable ppolicy overlays. I have gone through the steps provided at OpenLDAP and ppolicy howto. I have made the changes to slapd.conf and imported the password policy. On restart OpenLDAP is working fine and I can see the password policy when I do a ldapsearch. The user object looks like given below. # extended LDIF # # LDAPv3 # base <dc=xxxxx,dc=in> with scope subtree # filter: uid=testuser # requesting: ALL # # testuser, People, xxxxxx.in dn: uid=testuser,ou=People,dc=xxxxx,dc=in uid: testuser cn: testuser objectClass: account objectClass: posixAccount objectClass: top objectClass: shadowAccount shadowMax: 90 shadowWarning: 7 loginShell: /bin/bash uidNumber: 569 gidNumber: 1005 homeDirectory: /data/testuser userPassword:: xxxxxxxxxxxxx shadowLastChange: 15079 The password policy is given below. # default, policies, xxxxxx.in dn: cn=default,ou=policies,dc=xxxxxx,dc=in objectClass: top objectClass: device objectClass: pwdPolicy cn: default pwdAttribute: userPassword pwdMaxAge: 7776002 pwdExpireWarning: 432000 pwdInHistory: 0 pwdCheckQuality: 1 pwdMinLength: 8 pwdMaxFailure: 5 pwdLockout: TRUE pwdLockoutDuration: 900 pwdGraceAuthNLimit: 0 pwdFailureCountInterval: 0 pwdMustChange: TRUE pwdAllowUserChange: TRUE pwdSafeModify: FALSE I do not what should be done after this. How can the shadowAccount attributes be replaced with the password policy.

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  • directory services group query changing randomly

    - by yamspog
    I am receiving an unusual behaviour in my asp.net application. I have code that uses Directory Services to find the AD groups for a given, authenticated user. The code goes something like ... string username = "user"; string domain = "LDAP://DC=domain,DC=com"; DirectorySearcher search = new DirectorySearcher(domain); search.Filter = "(SAMAccountName=" + username + ")"; And then I query and get the list of groups for the given user. The problem is that the code was receiving the list of groups as a list of strings. With our latest release of the software, we are starting to receive the list of groups as a byte[]. The system will return string, suddenly return byte[] and then with a reboot it returns string again. Anyone have any ideas? code sample: DirectoryEntry dirEntry = new DirectoryEntry("LDAP://" + ldapSearchBase); DirectorySearcher userSearcher = new DirectorySearcher(dirEntry) { SearchScope = SearchScope.Subtree, CacheResults = false, Filter = ("(" + txtLdapSearchNameFilter.Text + "=" + userName + ")") }; userResult = userSearcher.FindOne(); ResultPropertyValueCollection valCol = userResult.Properties["memberOf"]; foreach (object val in valCol) { if (val is string) { distName = val.ToString(); } else { distName = enc.GetString((Byte[])val); } }

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  • Scripted forwarding for Outlook 2003

    - by John Gardeniers
    We have a staff member in sales who has gone onto a 4 day week (getting ready for retirement), so each Thursday afternoon her email needs to be forwarded to another user and each Friday afternoon it needs to be set back. I'm using the VBS script below to do this, run via the Task Scheduler. Although the script appears to do it's job, based on what I see when I view the user's Exchange settings, Exchange doesn't always recognise that the setting has changed. e.g. Last Thursday the forwarding was a enabled and worked correctly. On Friday the script did it's thing to clear the forwarding but Exchange continued to forward messages all weekend. I found that I can force Exchange to honour the changed setting be merely opening and closing the user's properties in ADUC. Of course I don't want to have to do that. Is there a non-manual way I can have Exchange read and honour the setting? The script (VBS): ' Call this script with the following parameters: ' ' SrcUser - The logon ID of the suer who's account is to be modified ' DstUser - The logon account of the person to who mail is to be forwarded ' Use "reset" to clear the email forwarding SrcUser = WScript.Arguments.Item(0) DstUser = WScript.Arguments.Item(1) SourceUser = SearchDistinguishedName(SrcUser) 'The user login name Set objUser = GetObject("LDAP://" & SourceUser) If DstUser = "reset" then objUser.PutEx 1, "altRecipient", "" Else ForwardTo = SearchDistinguishedName(DstUser)' The contact common name objUser.Put "AltRecipient", ForwardTo End If objUser.SetInfo Public Function SearchDistinguishedName(ByVal vSAN) Dim oRootDSE, oConnection, oCommand, oRecordSet Set oRootDSE = GetObject("LDAP://rootDSE") Set oConnection = CreateObject("ADODB.Connection") oConnection.Open "Provider=ADsDSOObject;" Set oCommand = CreateObject("ADODB.Command") oCommand.ActiveConnection = oConnection oCommand.CommandText = "<LDAP://" & oRootDSE.get("defaultNamingContext") & ">;(&(objectCategory=User)(samAccountName=" & vSAN & "));distinguishedName;subtree" Set oRecordSet = oCommand.Execute On Error Resume Next SearchDistinguishedName = oRecordSet.Fields("DistinguishedName") On Error GoTo 0 oConnection.Close Set oRecordSet = Nothing Set oCommand = Nothing Set oConnection = Nothing Set oRootDSE = Nothing End Function

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  • Selectively allow unsafe html tags in Plone

    - by dhill
    I'm searching for a way to put widgets from several services (PicasaWeb, Yahoo Pipes, Delicious bookmarks, etc.) on the community site I host on Plone (currently 3.2.1). I'm looking for a way to allow a group of users to use dangerous html tags. There are some ways I see, but I don't know how to implement those. One would be changing safe_html for the pages editors own (1). Another would be to allow those tags on some subtree (2). And yet another finding an equivalent of "static text portlet" that would display in the middle panel (3). We could then use some of the composite products (I stumbled upon Collage and CMFContentPanels), to include the unsafe content on other sites. My site has been ridden by advert bots, so I don't want to remove the filtering all together. I don't have an easy (no false positives) way of checking which users are bots, so deploying captcha now wouldn't help either. The question is: How to implement any of those solutions? (I already asked that on plone mailing list without an answer, so I thought I would give it another try here.)

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  • ldapsearch against Active Directory fails

    - by Guacamole
    I am using ldapsearch from OpenLDAP tools to search our corporate Active Directory for my email and phone number. This query is a test to ensure that I can authenticate against the domain so I can set up a linux wiki with NTLM authentication. My theory is that if I can successfully query the AD for information, then I am a step closer to getting my wiki to authenticate against AD (I have instructions to set up moin wiki under ActiveDirectory). The problem is that I can't seem to get the ldapsearch query right. I have seen many tutorials on the net that indicate that -D should be something like -D "Americas\John_Marsharll"; however, I keep getting ldap_bind: Invalid credentials (49) error messages when I use Americas\John_Marshall. The only time I get sensical results is when I query with the parameters below. However, even then, I can't figure out how to get email and phone number. [John_Marsharll@WN7-BG3YSM1 ~]$ ldapsearch -x -h 10.1.1.1 \ -b "cn=Users,dc=Americas" mail telephonenumber -D "cn=John_Marshall,dc=Americas" # extended LDIF # # LDAPv3 # base <cn=Users,dc=Americas> with scope subtree # filter: (objectclass=*) # requesting: mail telephonenumber -D cn=John_Marshall,dc=Americas # # search result search: 2 result: 32 No such object # numResponses: 1 [John_Marshall@WN7-BG3YSM1 ~]$ Can someone give me pointers on what I'm doing wrong with the ldapsearch query above? Our AD ldap server is 10.1.1.1 and the AD domain is "Americas".

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