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• Can't access some websites using Ubuntu 13.10

  - by Adame Doe

  Something's wrong with Ubuntu. Since I've upgraded to 13.10, I can't access some websites for no apparent reason. I've tried everything imaginable to solve this problem : Made sure that MTUs are the same, Disabled IPv6 in both the network manager and used browsers, Deactivated my network keys, DMZed my computer, Used other DNS like Google and OpenDNS, Checked that no firewall was running my computer ... And it's the same result. I even tried to reinstall Ubuntu a couple of times, but no luck. The most annoying thing about it is I can't access wordpress.org! So, there's no way it could be an ISP restriction of some kind. When I use a VPN, I can access pretty much anything. I'm really frustrated because I have to use wordpress.org very often. Any clue? ifconfig adame@adame-ws:~$ ifconfig eth0 Link encap:Ethernet HWaddr 00:26:18:3d:b0:7c inet addr:10.42.0.1 Bcast:10.42.0.255 Mask:255.255.255.0 inet6 addr: fe80::226:18ff:fe3d:b07c/64 Scope:Link UP BROADCAST RUNNING MULTICAST MTU:1500 Metric:1 RX packets:8024 errors:0 dropped:0 overruns:0 frame:0 TX packets:7966 errors:0 dropped:0 overruns:0 carrier:0 collisions:0 txqueuelen:1000 RX bytes:684480 (684.4 KB) TX bytes:616608 (616.6 KB) lo Link encap:Local Loopback inet addr:127.0.0.1 Mask:255.0.0.0 inet6 addr: ::1/128 Scope:Host UP LOOPBACK RUNNING MTU:65536 Metric:1 RX packets:8222 errors:0 dropped:0 overruns:0 frame:0 TX packets:8222 errors:0 dropped:0 overruns:0 carrier:0 collisions:0 txqueuelen:0 RX bytes:568269 (568.2 KB) TX bytes:568269 (568.2 KB) wlan0 Link encap:Ethernet HWaddr 00:19:70:40:85:eb inet addr:192.168.2.3 Bcast:192.168.2.255 Mask:255.255.255.0 inet6 addr: fe80::219:70ff:fe40:85eb/64 Scope:Link UP BROADCAST RUNNING MULTICAST MTU:1464 Metric:1 RX packets:123705 errors:0 dropped:0 overruns:0 frame:0 TX packets:98141 errors:0 dropped:0 overruns:0 carrier:0 collisions:0 txqueuelen:1000 RX bytes:94963545 (94.9 MB) TX bytes:10387470 (10.3 MB) /etc/hosts 127.0.0.1 localhost 127.0.1.1 adame-ws ::1 ip6-localhost ip6-loopback fe00::0 ip6-localnet ff00::0 ip6-mcastprefix ff02::1 ip6-allnodes ff02::2 ip6-allrouters tracepath wordpress.org 1: adame-ws.local 0.092ms pmtu 1500 1: 192.168.2.1 1.300ms asymm 2 1: 192.168.2.1 1.060ms asymm 2 2: no reply 3: no reply 4: no reply 5: no reply 6: no reply 7: no reply 8: no reply ... keep on going like that ping wordpress.org adame@adame-ws:~$ ping wordpress.org PING wordpress.org (66.155.40.250) 56(84) bytes of data. --- wordpress.org ping statistics --- 10 packets transmitted, 0 received, 100% packet loss, time 9071ms

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• need to align part of list item to right of li - using CSS3 Jquery column-layout

  - by Brad

  Using this jquery script to acheive CSS3 3-columns, to display a list of members alphabetically. I need it to display this way, which is does: A D B E C F Here is what I am using http://www.csscripting.com/css-multi-column/example6.php? (using this js file http://www.csscripting.com/js/v1.0beta/css3-multi-column.js) To the right of each member, it has their phone extension, which I want to float to the right, so it easy to read. I tried putting the phone extension within a div and span and when I do that, it tends to screw up at the last item in each column, by placing the person's name correctly, but their extension is the very first item in the next column. Screenshot: http://cl.ly/fq4 of what it is doing HTML Code: <div class="Article3Col"> <ul> <li>Doe, John <div style="float:right;"> 8317 </div> </li> <li>Doe, Sally <div style="float:right;"> 8729 </div> </li> <li>Doe, John <div style="float:right;"> 8317 </div> </li> <li>Doe, Sally <div style="float:right;"> 8729 </div> </li> <li>Doe, John <div style="float:right;"> 8317 </div> </li> <li>Doe, Sally <div style="float:right;"> 8729 </div> </li> <li>Doe, John <div style="float:right;"> 8317 </div> </li> <li>Doe, Sally <div style="float:right;"> 8729 </div> </li> <li>Doe, John <div style="float:right;"> 8317 </div> </li> <li>Doe, Sally <div style="float:right;"> 8729 </div> </li> <li>Doe, John <div style="float:right;"> 8317 </div> </li> <li>Doe, Sally <div style="float:right;"> 8729 </div> </li> <li>Doe, John <div style="float:right;"> 8317 </div> </li> <li>Doe, Sally <div style="float:right;"> 8729 </div> </li> <li>Doe, John <div style="float:right;"> 8317 </div> </li> <li>Doe, Sally <div style="float:right;"> 8729 </div> </li> <li>Doe, John <div style="float:right;"> 8317 </div> </li> <li>Doe, Sally <div style="float:right;"> 8729 </div> </li> </ul> </div> CSS: .Article3Col { column-count:3; } Any help is appreciated.

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• Search 2 Columns with 1 Input Field

  - by Norbert

  I have a db with two columns: first name and last name. The first name can have multiple words. Last name can contain hyphenated words. Is there a way to search both columns with only one input box? Database ID `First Name` `Last Name` 1 John Peter Doe 2 John Fubar 3 Michael Doe Search john peter returns id 1 john returns id 1,2 doe returns id 1,3 john doe returns id 1 peter john returns id 1 peter doe returns id 1 doe john returns id 1 I previously tried the following. Searching for John Doe: SELECT * FROM names WHERE ( `first` LIKE '%john%' OR `first` LIKE '%doe%' OR `last` LIKE '%john%' OR `last` LIKE '%doe%' ) which returns both 1 and 3

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• MS Outlook Voting mismatch

  - by Robert Ilbrink

  I did send out an Outlook vote to hundreds of employees, using their email address (which is mostly [email protected], but there are many exceptions). The incoming votes are not matched against the email address, but against the display name. Unfortunately, the display name has no real standard either. So instead of seeing this: [email protected] Voted: Yes [email protected] Voted: No I see this: [email protected] [email protected] Doe, Johnathan Philip Voted: Yes Doe - Peeters, Marian Voted: No In the actual list I see the addresses that I sent the vote to PLUS extra lines with the votes that came back. Is there a quick way to match my "send" list with the "received" list? One thing I thought of was to dump the global address book in a file and in Excel use =vlookup. But that seems a lot of work (and I am not even sure that I have the authorization to dump the address book).

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• Uploadify works for Visual Studio but not for IIS 7(same machines), using Forms authentication. Doe

  - by Marc

  I'm using the Uploadify jQuery control for client-side uploads. I think my IIS 7 configuration has issues with it. The uploadify POST immediately returns a HTTP 1.1 302 Found, back to my login page. I've tried to allow anonymous access to the uploading section(subfolder) plus the page(script) that processes the image in the web.config, using the location node(configuration ... location). Seems like the Uploadify post is immediately blocked. Again, this worked fine just using Visual Studio 2008, but when I run the site on the same machine I get the redirect. Your thoughts/ideas are very welcomed!

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• How doe we name test methods where we are checking for more than one condition?

  - by Sandbox

  I follow the technique specified in Roy Osherove's The Art Of Unit Testing book while naming test methods - MethodName_Scenario_Expectation. It suits perfectly well for my 'unit' tests. But,for tests that I write in 'controller' or 'coordinator' class, there isn't necessarily a method which I want to test. For these tests, I generate multiple conditions which make up one scenario and then I verify the expectation. For example, I may set some properties on different instances, generate an event and then verify that my expectations from controller/coordinator is being met. Now, my controller handles events using a private event handler. Here my scenario is that, I set some properties, say 3 condition1,condition2 and condition3 Also, my scenario includes an event is raised I don't have a method name as my event handler is private. How do I name such a test method?

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• Why doe my UITableView only show two rows of each section?

  - by Mike Owens

  I have a UITableView and when I build it only two rows will be displayed. Each section has more than two cells to be displayed, I am confused since they are all done the same?`#import #import "Store.h" import "VideoViewController.h" @implementation Store @synthesize listData; // Implement viewDidLoad to do additional setup after loading the view, typically from a nib. - (void)viewDidLoad { [self createTableData]; [super viewDidLoad]; } (void)didReceiveMemoryWarning { // Releases the view if it doesn't have a superview. [super didReceiveMemoryWarning]; // Release any cached data, images, etc that aren't in use. } (void)viewDidUnload { //self.listData = nil; //[super viewDidUnload]; // Release any retained subviews of the main view. // e.g. self.myOutlet = nil; } pragma mark - pragma mark Table View Data Source Methods // Customize the number of sections in the table view. - (NSInteger)numberOfSectionsInTableView:(UITableView *)tableView { return [videoSections count]; } //Get number of rows -(NSInteger)tableView:(UITableView *)tableView numberOfRowsInSection:(NSInteger)section { return [self.listData count]; } -(UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath { static NSString *StoreTableIdentifier = @"StoreTableIdentifier"; UITableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:StoreTableIdentifier]; if (cell == nil) { cell = [[[UITableViewCell alloc] initWithStyle:UITableViewCellStyleDefault reuseIdentifier:StoreTableIdentifier] autorelease]; } cell.textLabel.text = [[[listData objectAtIndex:indexPath.section] objectAtIndex:indexPath.row] objectForKey:@"name"]; //Change font and color of tableView cell.accessoryType = UITableViewCellAccessoryDisclosureIndicator; cell.textLabel.font=[UIFont fontWithName:@"Georgia" size:16.0]; cell.textLabel.textColor = [UIColor brownColor]; return cell; } -(NSString *)tableView: (UITableView *)tableView titleForHeaderInSection: (NSInteger) section { return [videoSections objectAtIndex:section]; } -(void)tableView: (UITableView *)tableView didSelectRowAtIndexPath: (NSIndexPath *)indexPath { VideoViewController *videoViewController = [[VideoViewController alloc] initWithNibName: @"VideoViewController" bundle:nil]; videoViewController.detailURL = [[NSURL alloc] initWithString: [[[listData objectAtIndex:indexPath.section] objectAtIndex:indexPath.row] objectForKey:@"url"]]; videoViewController.title = [[[listData objectAtIndex:indexPath.section] objectAtIndex:indexPath.row] objectForKey:@"name"]; [self.navigationController pushViewController:videoViewController animated:YES]; [videoViewController release]; } pragma mark Table View Methods //Data in table cell -(void) createTableData { NSMutableArray *beginningVideos; NSMutableArray *intermediateVideos; videoSections = [[NSMutableArray alloc] initWithObjects: @"Beginning Videos", @"Intermediate Videos", nil]; beginningVideos = [[NSMutableArray alloc] init]; intermediateVideos = [[NSMutableArray alloc] init]; [beginningVideos addObject:[[NSMutableDictionary alloc] initWithObjectsAndKeys:@"Shirts", @"name", @"http://www.andalee.com/iPhoneVideos/testMovie.m4v", @"url", nil]]; [beginningVideos addObject:[[NSMutableDictionary alloc] initWithObjectsAndKeys:@"Posters", @"name", @"http://devimages.apple.com/iphone/samples/bipbopall.html", @"url", nil]]; [beginningVideos addObject:[[NSMutableDictionary alloc] initWithObjectsAndKeys:@"Stickers",@"name", @"http://www.andalee.com/iPhoneVideos/mov.MOV",@"url",nil]]; [beginningVideos addObject:[[NSMutableDictionary alloc] initWithObjectsAndKeys:@"Egyptian",@"name", @"http://www.andalee.com/iPhoneVideos/2ndMovie.MOV",@"url",nil]]; [intermediateVideos addObject:[[NSMutableDictionary alloc] initWithObjectsAndKeys:@"Drum Solo", @"name", @"http://www.andalee.com", @"url", nil]]; [intermediateVideos addObject:[[NSMutableDictionary alloc] initWithObjectsAndKeys:@"Veil", @"name", @"http://www.andalee.com", @"url", nil]]; [intermediateVideos addObject:[[NSMutableDictionary alloc] initWithObjectsAndKeys:@"Three Quarter Shimmy",@"name", @"http://www.andalee.com", @"url",nil]]; listData = [[NSMutableArray alloc] initWithObjects:beginningVideos, intermediateVideos, nil]; [beginningVideos release]; [intermediateVideos release]; } (void)dealloc { [listData release]; [videoSections release]; [super dealloc]; } @end `

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• How to Programmatically Split Data Using VBA Using Specific Logic

  - by Charlene

  This is an addition to my previous post here. The code that was previously supplied to me worked like a charm, but I am having issues modifying it adding some additional logic. I am creating a macro in VBA to do the following. I have raw order data that I need to transform based on some logic. Raw Data: order-id product-num date buyer-name prod-name qty-purc sales-tax freight order-st 0000000000-00 10000000000000 5/29/2014 John Doe Product 0 1 1.00 1.50 GA 0000000000-00 10000000000001 5/29/2014 John Doe Product 1 2 1.00 1.50 GA 0000000000-00 10000000000002 5/29/2014 John Doe Product 2 1 1.00 2.00 GA 0000000000-01 10000000000002 5/30/2014 Jane Doe Product 2 1 0.00 0.00 PA 0000000000-01 10000000000003 5/30/2014 Jane Doe Product 3 1 0.00 0.00 PA Desired Outcome: HDR 0000000000-00 John Doe 5/29/2014 CHG Tax 3.00 CHG Freight 5.00 ITM 10000000000000 Product 0 1 ITM 10000000000001 Product 1 2 ITM 10000000000002 Product 2 1 HDR 0000000000-01 Jane Doe 5/30/2014 ITM 10000000000002 Product 2 1 ITM 10000000000003 Product 3 1 The "CHG" rows are created based on the following logic; if the order-st is CA or GA, add the total of sales-tax and freight for each of the rows with the same order-id. If the order-st is NOT CA or GA, no CHG rows should be created. Any help would be appreciated - let me know if I left any details out!

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• How to Programmatically Split and Manipulate Rows of Data From Excel

  - by Charlene

  I am hoping one of you will be able to help get me started on this issue. I need to create some sort of macro or VBA code to split and manipulate rows of data in Excel. For this example, we have 5 rows of data. The first 3 rows are item information for Order # 0000000000-00 and the last 2 rows are item information for order # 0000000000-01. I need one row ("HDR") for each order number, and one row ("ITM") for each product per order. I have included an example below showing the data I will receive and the desired outcome. Raw Data: order-id product-num date buyer-name product-name quantity-purchased 0000000000-00 10000000000000 5/29/2014 John Doe Product 0 1 0000000000-00 10000000000001 5/29/2014 John Doe Product 1 2 0000000000-00 10000000000002 5/29/2014 John Doe Product 2 1 0000000000-01 10000000000002 5/30/2014 Jane Doe Product 2 1 0000000000-01 10000000000003 5/30/2014 Jane Doe Product 3 1 Desired Outcome: HDR 0000000000-00 John Doe 5/29/2014 ITM 10000000000000 Product 0 1 ITM 10000000000001 Product 1 2 ITM 10000000000002 Product 2 1 HDR 0000000000-01 Jane Doe 5/30/2014 ITM 10000000000002 Product 2 1 ITM 10000000000003 Product 3 1 Any and all help would be much appreciated!!! Thank you.

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• How to have a consistent mail receiver name displayed in Outlook 2007 in the inbox folder

  - by usr

  Currently the same sender address appears differently in my inbox for different mails. The variations are: [email protected] Doe, John That is because I received some mails before I changed the name to "John Doe" and some after. It is now very annoying to find mails from John because I have to look at two places in my inbox after I sort them by sender. How can I make Outlook always sho "Doe, John"?

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• How- XLST Transformation

  - by Yuan Ray

  Just wanted to ask on how to get the author names in the given xml sample below and put an attribut of eq="yes". EQ means Equal Contributors. This is the XML. <ArticleFootnote Type="Misc"> <Para>John Doe and Jane Doe are equal contributors.</Para> </ArticleFootnote> This should be the output in other form of XML. <AuthorGroups> <Authors eq="yes">John Doe</Authors> <Authors eq="yes">Jane Doe</Authors> </AuthorGroups> Assuming that JOhn Doe and Jane Doe are already defined in the list of authors but after the transformation, author tag should have the attribute eq="yes". Please help as I don't know much writing in xlst. Thanks in advance.

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• To ref or not to ref

  - by nmarun

  So the question is what is the point of passing a reference type along with the ref keyword? I have an Employee class as below: 1: public class Employee 2: { 3: public string FirstName { get; set; } 4: public string LastName { get; set; } 5:  6: public override string ToString() 7: { 8: return string.Format("{0}-{1}", FirstName, LastName); 9: } 10: } In my calling class, I say: 1: class Program 2: { 3: static void Main() 4: { 5: Employee employee = new Employee 6: { 7: FirstName = "John", 8: LastName = "Doe" 9: }; 10: Console.WriteLine(employee); 11: CallSomeMethod(employee); 12: Console.WriteLine(employee); 13: } 14:  15: private static void CallSomeMethod(Employee employee) 16: { 17: employee.FirstName = "Smith"; 18: employee.LastName = "Doe"; 19: } 20: }   After having a look at the code, you’ll probably say, Well, an instance of a class gets passed as a reference, so any changes to the instance inside the CallSomeMethod, actually modifies the original object. Hence the output will be ‘John-Doe’ on the first call and ‘Smith-Doe’ on the second. And you’re right: So the question is what’s the use of passing this Employee parameter as a ref? 1: class Program 2: { 3: static void Main() 4: { 5: Employee employee = new Employee 6: { 7: FirstName = "John", 8: LastName = "Doe" 9: }; 10: Console.WriteLine(employee); 11: CallSomeMethod(ref employee); 12: Console.WriteLine(employee); 13: } 14:  15: private static void CallSomeMethod(ref Employee employee) 16: { 17: employee.FirstName = "Smith"; 18: employee.LastName = "Doe"; 19: } 20: } The output is still the same: Ok, so is there really a need to pass a reference type using the ref keyword? I’ll remove the ‘ref’ keyword and make one more change to the CallSomeMethod method. 1: class Program 2: { 3: static void Main() 4: { 5: Employee employee = new Employee 6: { 7: FirstName = "John", 8: LastName = "Doe" 9: }; 10: Console.WriteLine(employee); 11: CallSomeMethod(employee); 12: Console.WriteLine(employee); 13: } 14:  15: private static void CallSomeMethod(Employee employee) 16: { 17: employee = new Employee 18: { 19: FirstName = "Smith", 20: LastName = "John" 21: }; 22: } 23: } In line 17 you’ll see I’ve ‘new’d up the incoming Employee parameter and then set its properties to new values. The output tells me that the original instance of the Employee class does not change. Huh? But an instance of a class gets passed by reference, so why did the values not change on the original instance or how do I keep the two instances in-sync all the times? Aah, now here’s the answer. In order to keep the objects in sync, you pass them using the ‘ref’ keyword. 1: class Program 2: { 3: static void Main() 4: { 5: Employee employee = new Employee 6: { 7: FirstName = "John", 8: LastName = "Doe" 9: }; 10: Console.WriteLine(employee); 11: CallSomeMethod(ref employee); 12: Console.WriteLine(employee); 13: } 14:  15: private static void CallSomeMethod(ref Employee employee) 16: { 17: employee = new Employee 18: { 19: FirstName = "Smith", 20: LastName = "John" 21: }; 22: } 23: } Viola! Now, to prove it beyond doubt, I said, let me try with another reference type: string. 1: class Program 2: { 3: static void Main() 4: { 5: string name = "abc"; 6: Console.WriteLine(name); 7: CallSomeMethod(ref name); 8: Console.WriteLine(name); 9: } 10:  11: private static void CallSomeMethod(ref string name) 12: { 13: name = "def"; 14: } 15: } The output was as expected, first ‘abc’ and then ‘def’ - proves the 'ref' keyword works here as well. Now, what if I remove the ‘ref’ keyword? The output should still be the same as the above right, since string is a reference type? 1: class Program 2: { 3: static void Main() 4: { 5: string name = "abc"; 6: Console.WriteLine(name); 7: CallSomeMethod(name); 8: Console.WriteLine(name); 9: } 10:  11: private static void CallSomeMethod(string name) 12: { 13: name = "def"; 14: } 15: } Wrong, the output shows ‘abc’ printed twice. Wait a minute… now how could this be? This is because string is an immutable type. This means that any time you modify an instance of string, new memory address is allocated to the instance. The effect is similar to ‘new’ing up the Employee instance inside the CallSomeMethod in the absence of the ‘ref’ keyword. Verdict: ref key came to the rescue and saved the planet… again!

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• De-share third level .name domain

  - by Enzo

  Sometime ago I registered a third level .name domain of the type john.doe.name (this is just an example-my domain is not actually john.doe.name). I just realised that I could have registered the entire second level domain doe.name, which would grant me control of the third level one anyway. Since I already registered the third level, doe.name is now "shared". I have 2 questions: 1) How do I check if I am the only one using the shared second level domain? (normal whois lookup doesn't give any result) 2) Can I "de-share" the domain and buy the entire second level domain? Cheers!

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• "ldap_add: Naming violation (64)" error when configuring OpenLDAP

  - by user3215

  I am following the Ubuntu server guide to configure OpenLDAP on an Ubuntu 10.04 server, but can not get it to work. When I try to use sudo ldapadd -x -D cn=admin,dc=don,dc=com -W -f frontend.ldif I'm getting the following error: Enter LDAP Password: <entered 'secret' as password> adding new entry "dc=don,dc=com" ldap_add: Naming violation (64) additional info: value of single-valued naming attribute 'dc' conflicts with value present in entry Again when I try to do the same, I'm getting the following error: root@avy-desktop:/home/avy# sudo ldapadd -x -D cn=admin,dc=don,dc=com -W -f frontend.ldif Enter LDAP Password: ldap_bind: Invalid credentials (49) Here is the backend.ldif file: # Load dynamic backend modules dn: cn=module,cn=config objectClass: olcModuleList cn: module olcModulepath: /usr/lib/ldap olcModuleload: back_hdb # Database settings dn: olcDatabase=hdb,cn=config objectClass: olcDatabaseConfig objectClass: olcHdbConfig olcDatabase: {1}hdb olcSuffix: dc=don,dc=com olcDbDirectory: /var/lib/ldap olcRootDN: cn=admin,dc=don,dc=com olcRootPW: secret olcDbConfig: set_cachesize 0 2097152 0 olcDbConfig: set_lk_max_objects 1500 olcDbConfig: set_lk_max_locks 1500 olcDbConfig: set_lk_max_lockers 1500 olcDbIndex: objectClass eq olcLastMod: TRUE olcDbCheckpoint: 512 30 olcAccess: to attrs=userPassword by dn="cn=admin,dc=don,dc=com" write by anonymous auth by self write by * none olcAccess: to attrs=shadowLastChange by self write by * read olcAccess: to dn.base="" by * read olcAccess: to * by dn="cn=admin,dc=don,dc=com" write by * read frontend.ldif file: # Create top-level object in domain dn: dc=don,dc=com objectClass: top objectClass: dcObject objectclass: organization o: Example Organization dc: Example description: LDAP Example # Admin user. dn: cn=admin,dc=don,dc=com objectClass: simpleSecurityObject objectClass: organizationalRole cn: admin description: LDAP administrator userPassword: secret dn: ou=people,dc=don,dc=com objectClass: organizationalUnit ou: people dn: ou=groups,dc=don,dc=com objectClass: organizationalUnit ou: groups dn: uid=john,ou=people,dc=don,dc=com objectClass: inetOrgPerson objectClass: posixAccount objectClass: shadowAccount uid: john sn: Doe givenName: John cn: John Doe displayName: John Doe uidNumber: 1000 gidNumber: 10000 userPassword: password gecos: John Doe loginShell: /bin/bash homeDirectory: /home/john shadowExpire: -1 shadowFlag: 0 shadowWarning: 7 shadowMin: 8 shadowMax: 999999 shadowLastChange: 10877 mail: [email protected] postalCode: 31000 l: Toulouse o: Example mobile: +33 (0)6 xx xx xx xx homePhone: +33 (0)5 xx xx xx xx title: System Administrator postalAddress: initials: JD dn: cn=example,ou=groups,dc=don,dc=com objectClass: posixGroup cn: example gidNumber: 10000 Can anyone help me?

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• How to Configure OpenLDAP on Ubuntu 10.04 Server

  - by user3215

  I am following the Ubuntu server guide to configure OpenLDAP on an Ubuntu 10.04 server, but can not get it to work. When I try to use sudo ldapadd -x -D cn=admin,dc=don,dc=com -W -f frontend.ldif I'm getting the following error: Enter LDAP Password: <entered 'secret' as password> adding new entry "dc=don,dc=com" ldap_add: Naming violation (64) additional info: value of single-valued naming attribute 'dc' conflicts with value present in entry Again when I try to do the same, I'm getting the following error: root@avy-desktop:/home/avy# sudo ldapadd -x -D cn=admin,dc=don,dc=com -W -f frontend.ldif Enter LDAP Password: ldap_bind: Invalid credentials (49) Here is the backend.ldif file: # Load dynamic backend modules dn: cn=module,cn=config objectClass: olcModuleList cn: module olcModulepath: /usr/lib/ldap olcModuleload: back_hdb # Database settings dn: olcDatabase=hdb,cn=config objectClass: olcDatabaseConfig objectClass: olcHdbConfig olcDatabase: {1}hdb olcSuffix: dc=don,dc=com olcDbDirectory: /var/lib/ldap olcRootDN: cn=admin,dc=don,dc=com olcRootPW: secret olcDbConfig: set_cachesize 0 2097152 0 olcDbConfig: set_lk_max_objects 1500 olcDbConfig: set_lk_max_locks 1500 olcDbConfig: set_lk_max_lockers 1500 olcDbIndex: objectClass eq olcLastMod: TRUE olcDbCheckpoint: 512 30 olcAccess: to attrs=userPassword by dn="cn=admin,dc=don,dc=com" write by anonymous auth by self write by * none olcAccess: to attrs=shadowLastChange by self write by * read olcAccess: to dn.base="" by * read olcAccess: to * by dn="cn=admin,dc=don,dc=com" write by * read frontend.ldif file: # Create top-level object in domain dn: dc=don,dc=com objectClass: top objectClass: dcObject objectclass: organization o: Example Organization dc: Example description: LDAP Example # Admin user. dn: cn=admin,dc=don,dc=com objectClass: simpleSecurityObject objectClass: organizationalRole cn: admin description: LDAP administrator userPassword: secret dn: ou=people,dc=don,dc=com objectClass: organizationalUnit ou: people dn: ou=groups,dc=don,dc=com objectClass: organizationalUnit ou: groups dn: uid=john,ou=people,dc=don,dc=com objectClass: inetOrgPerson objectClass: posixAccount objectClass: shadowAccount uid: john sn: Doe givenName: John cn: John Doe displayName: John Doe uidNumber: 1000 gidNumber: 10000 userPassword: password gecos: John Doe loginShell: /bin/bash homeDirectory: /home/john shadowExpire: -1 shadowFlag: 0 shadowWarning: 7 shadowMin: 8 shadowMax: 999999 shadowLastChange: 10877 mail: [email protected] postalCode: 31000 l: Toulouse o: Example mobile: +33 (0)6 xx xx xx xx homePhone: +33 (0)5 xx xx xx xx title: System Administrator postalAddress: initials: JD dn: cn=example,ou=groups,dc=don,dc=com objectClass: posixGroup cn: example gidNumber: 10000 Can anyone help me?

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• How to configure ldap on ubuntu 10.04 server

  - by user3215

  I am following the link to configure ldap on ubuntu 10.04 server but could not. when I try to use sudo ldapadd -x -D cn=admin,dc=don,dc=com -W -f frontend.ldif I'm getting the following error: Enter LDAP Password: <entered 'secret' as password> adding new entry "dc=don,dc=com" ldap_add: Naming violation (64) additional info: value of single-valued naming attribute 'dc' conflicts with value present in entry Again when I try to do the same, I'm getting the following error: root@avy-desktop:/home/avy# sudo ldapadd -x -D cn=admin,dc=don,dc=com -W -f frontend.ldif Enter LDAP Password: ldap_bind: Invalid credentials (49) Here is the backend.ldif file # Load dynamic backend modules dn: cn=module,cn=config objectClass: olcModuleList cn: module olcModulepath: /usr/lib/ldap olcModuleload: back_hdb # Database settings dn: olcDatabase=hdb,cn=config objectClass: olcDatabaseConfig objectClass: olcHdbConfig olcDatabase: {1}hdb olcSuffix: dc=don,dc=com olcDbDirectory: /var/lib/ldap olcRootDN: cn=admin,dc=don,dc=com olcRootPW: secret olcDbConfig: set_cachesize 0 2097152 0 olcDbConfig: set_lk_max_objects 1500 olcDbConfig: set_lk_max_locks 1500 olcDbConfig: set_lk_max_lockers 1500 olcDbIndex: objectClass eq olcLastMod: TRUE olcDbCheckpoint: 512 30 olcAccess: to attrs=userPassword by dn="cn=admin,dc=don,dc=com" write by anonymous auth by self write by * none olcAccess: to attrs=shadowLastChange by self write by * read olcAccess: to dn.base="" by * read olcAccess: to * by dn="cn=admin,dc=don,dc=com" write by * read frontend.ldif file: # Create top-level object in domain dn: dc=don,dc=com objectClass: top objectClass: dcObject objectclass: organization o: Example Organization dc: Example description: LDAP Example # Admin user. dn: cn=admin,dc=don,dc=com objectClass: simpleSecurityObject objectClass: organizationalRole cn: admin description: LDAP administrator userPassword: secret dn: ou=people,dc=don,dc=com objectClass: organizationalUnit ou: people dn: ou=groups,dc=don,dc=com objectClass: organizationalUnit ou: groups dn: uid=john,ou=people,dc=don,dc=com objectClass: inetOrgPerson objectClass: posixAccount objectClass: shadowAccount uid: john sn: Doe givenName: John cn: John Doe displayName: John Doe uidNumber: 1000 gidNumber: 10000 userPassword: password gecos: John Doe loginShell: /bin/bash homeDirectory: /home/john shadowExpire: -1 shadowFlag: 0 shadowWarning: 7 shadowMin: 8 shadowMax: 999999 shadowLastChange: 10877 mail: [email protected] postalCode: 31000 l: Toulouse o: Example mobile: +33 (0)6 xx xx xx xx homePhone: +33 (0)5 xx xx xx xx title: System Administrator postalAddress: initials: JD dn: cn=example,ou=groups,dc=don,dc=com objectClass: posixGroup cn: example gidNumber: 10000 Anybody could help me?

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• Excel VBA Text To Column

  - by Pat

  This is what I currently have: H101 John Doe Jane Doe Jack Doe H102 John Smith Jane Smith Katie Smith Jack Smith And here is what I want: H101 John Doe H101 Jane Doe H101 Jack Doe H102 John Smith H102 Jane Smith H102 Katie Smith H102 Jack Smith Obviously I want to do this on a bigger scale. The number of columns is between 1 & 6, so I cant limit it that way. I was able to get a script that allows me to put each individual on one row. However, I am having a hard time getting the first column to copy over to each row. Sub ToOneColumn() Dim i As Long, k As Long, j As Integer Application.ScreenUpdating = False Columns(2).Insert i = 0 k = 1 While Not IsEmpty(Cells(k, 3)) j = 3 While Not IsEmpty(Cells(k, j)) i = i + 1 Cells(i, 1) = Cells(k, 1) //CODE IN QUESTION Cells(i, 2) = Cells(k, j) Cells(k, j).Clear j = j + 1 Wend k = k + 1 Wend Application.ScreenUpdating = True End Sub Like I said, it was working fine to get everyone each on their own row, but can't figure out how to get that first column. It seems like it should be so simple, but it's bugging me. Any help is greatly appreciated.

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• Multiple contacts with shared information

  - by Keith Thompson

  Background: I currently have several hundred contacts, synchronized between a Microsoft Exchange server and several mobile devices. I also save exported copies of the contacts in .vcf format. Is there a good way (application, file format, whatever) to maintain contacts with shared information? A very common scenario is that I have contacts for two or more people who live in the same house, for example: John Doe 123 Main Street, Anytown USA Home: 555-555-1111 Work: 555-555-2222 Mobile: 555-555-3333 E-mail: [email protected] Jane Doe 123 Main Street, Anytown USA Home: 555-555-1111 Work: 555-555-4444 Mobile: 555-555-5555 E-mail: [email protected] As you can see, both contacts have the same home address and phone number, but distinct names and work and mobile phone numbers. (Other information might also be either shared or distinct.) The applications and file formats I'm familiar with don't seem to have a good way to deal with this. If I use a single "John & Jane Doe" contact for both, it's difficult to distinguish the distinct information (if I want to call Jane's mobile phone rather than John's). If I use a separate contact for each, I have to remember to update both of them (or all of them for N 2) when they move or change their home phone number. An ideal solution would let me create a record containing information for their household, and have each of their contact records contain a reference to the household record, so that when I view John's contact record I see both shared and distinct information. Is there anything out there that has good support this kind of thing? (I would think there would be, since it's a very common scenario.) (I suppose I could roll my own system that generates merged .vcf files from some extended format, but that wouldn't play well with synchronizing across multiple devices.)

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• How to add LDAP user to existing local group in RHEL?

  - by Highway of Life

  I'm attempting to add some of our LDAP users to a locally defined group on our RHEL server, however I get an error stating that the LDAP user is not found in /etc/passwd. What would be the best way to allow LDAP users to be added to local groups? My feeling is that this must be done manually. I could edit: /etc/group and add the LDAP group to the list. Would that be ideal? [server]# id apache uid=409(apache) gid=409(apache) groups=409(apache) context=user_u:system_r:unconfined_t:s0 [server]# id john.doe uid=11389(john.doe) gid=6097(ABC_Corporate_US) groups=6097(ABC_Corporate_US) context=user_u:system_r:unconfined_t:s0 [server]# /usr/sbin/usermod -a -G apache john.doe usermod: john.doe not found in /etc/passwd OS: RHEL (Red Hat Enterprise Linux Server release 5.3 (Tikanga)) Note: Updating the OS on this machine is not an option.

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• Archiving mails with postfix: how to filter mails?

  - by Tronic

  i wanto to implement the following scenario: we use a postfix mailserver. to archive all old and new mails, i want to setup a second postfix on our fileserver and create a single mailbox "archive". then every mail gets forwarded as bcc to this mailbox automatically. now, i want to create different folders in a maildir structure and let the server move each mail to the right subfolder of the mailbox based on its sender or receiver. e.g. when we get a mail to one of our employees named "John Doe" at [email protected], the mail should be moved to "Inbox/John Doe Incoming". the same applies when john doe sends a mail, folder would be "Inbox/John Doe Outgoing". how can i implement this filter behaviour. i heard of Procmail and Maildrop. Which of the two would you prefer? Which is more easy to configure? Any out-of-box solutions here? thanks in advance!

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• Where to register third level .name domains ?

  - by Max Toro

  ... like john.doe.name

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• How does AuthzSVNAccessFile work?

  - by grigy

  I have set up an SVN repo with WebDAV access. For some reason it does not let checkout. Here is my httpd.conf part: <Location /svn> DAV svn SVNParentPath /home/svn/repositories AuthzSVNAccessFile /home/svn/dav_svn.authz Satisfy Any Require valid-user AuthType Basic AuthName "Subversion Repository" AuthUserFile /home/svn/dav_svn.passwd </Location> I have two repositories named "first" and "second" and the content of dav_svn.authz is: [first:/] doe = rw * = r [second:/] doe = rw grig = rw * = r When I'm trying to checkout the second with user doe, I get this in error_log: user doe: authentication failure for "/svn/second": Password Mismatch In order to understand what can be the problem I would like to better understand how the AuthzSVNAccessFile is supposed to work.

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• Code Trivia #4

  - by João Angelo

  Got the inspiration for this one in a recent stackoverflow question. What should the following code output and why? class Program { class Author { public string FirstName { get; set; } public string LastName { get; set; } public override string ToString() { return LastName + ", " + FirstName; } } static void Main() { Author[] authors = new[] { new Author { FirstName = "John", LastName = "Doe" }, new Author { FirstName = "Jane", LastName="Doe" } }; var line1 = String.Format("Authors: {0} and {1}", authors); Console.WriteLine(line1); string[] serial = new string[] { "AG27H", "6GHW9" }; var line2 = String.Format("Serial: {0}-{1}", serial); Console.WriteLine(line2); int[] version = new int[] { 1, 0 }; var line3 = String.Format("Version: {0}.{1}", version); Console.WriteLine(line3); } } Update: The code will print the first two lines // Authors: Doe, John and Doe, Jane // Serial: AG27H-6GHW9 and then throw an exception on the third call to String.Format because array covariance is not supported in value types. Given this the third call of String.Format will not resolve to String.Format(string, params object[]), like the previous two, but to String.Format(string, object) which fails to provide the second argument for the specified format and will then cause the exception.

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• Class design issue

  - by user2865206

  I'm new to OOP and a lot of times I become stumped in situations similar to this example: Task: Generate an XML document that contains information about a person. Assume the information is readily available in a database. Here is an example of the structure: <Person> <Name>John Doe</Name> <Age>21</Age> <Address> <Street>100 Main St.</Street> <City>Sylvania</City> <State>OH</State> </Address> <Relatives> <Parents> <Mother> <Name>Jane Doe</Name> </Mother> <Father> <Name>John Doe Sr.</Name> </Father> </Parents> <Siblings> <Brother> <Name>Jeff Doe</Name> </Brother> <Brother> <Name>Steven Doe</Name> </Brother> </Siblings> </Relatives> </Person> Ok lets create a class for each tag (ie: Person, Name, Age, Address) Lets assume each class is only responsible for itself and the elements directly contained Each class will know (have defined by default) the classes that are directly contained within them Each class will have a process() function that will add itself and its childeren to the XML document we are creating When a child is drawn, as in the previous line, we will have them call process() as well Now we are in a recursive loop where each object draws their childeren until all are drawn But what if only some of the tags need to be drawn, and the rest are optional? Some are optional based on if the data exists (if we have it, we must draw it), and some are optional based on the preferences of the user generating the document How do we make sure each object has the data it needs to draw itself and it's childeren? We can pass down a massive array through every object, but that seems shitty doesnt it? We could have each object query the database for it, but thats a lot of queries, and how does it know what it's query is? What if we want to get rid of a tag later? There is no way to reference them. I've been thinking about this for 20 hours now. I feel like I am misunderstanding a design principle or am just approaching this all wrong. How would you go about programming something like this? I suppose this problem could apply to any senario where there are classes that create other classes, but the classes created need information to run. How do I get the information to them in a way that doesn't seem fucky? Thanks for all of your time, this has been kicking my ass.

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• Tricky SQL query involving consecutive values

  - by Gabriel

  I need to perform a relatively easy to explain but (given my somewhat limited skills) hard to write SQL query. Assume we have a table similar to this one: exam_no | name | surname | result | date ---------+------+---------+--------+------------ 1 | John | Doe | PASS | 2012-01-01 1 | Ryan | Smith | FAIL | 2012-01-02 <-- 1 | Ann | Evans | PASS | 2012-01-03 1 | Mary | Lee | FAIL | 2012-01-04 ... | ... | ... | ... | ... 2 | John | Doe | FAIL | 2012-02-01 <-- 2 | Ryan | Smith | FAIL | 2012-02-02 2 | Ann | Evans | FAIL | 2012-02-03 2 | Mary | Lee | PASS | 2012-02-04 ... | ... | ... | ... | ... 3 | John | Doe | FAIL | 2012-03-01 3 | Ryan | Smith | FAIL | 2012-03-02 3 | Ann | Evans | PASS | 2012-03-03 3 | Mary | Lee | FAIL | 2012-03-04 <-- Note that exam_no and date aren't necessarily related as one might expect from the kind of example I chose. Now, the query that I need to do is as follows: From the latest exam (exam_no = 3) find all the students that have failed (John Doe, Ryan Smith and Mary Lee). For each of these students find the date of the first of the batch of consecutively failing exams. Another way to put it would be: for each of these students find the date of the first failing exam that comes after their last passing exam. (Look at the arrows in the table). The resulting table should be something like this: name | surname | date_since_failing ------+---------+-------------------- John | Doe | 2012-02-01 Ryan | Smith | 2012-01-02 Mary | Lee | 2012-01-04 Ann | Evans | 2012-02-03 How can I perform such a query? Thank you for your time.

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