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  • Ways to ensure unique instances of a class?

    - by Peanut
    I'm looking for different ways to ensure that each instance of a given class is a uniquely identifiable instance. For example, I have a Name class with the field name. Once I have a Name object with name initialised to John Smith I don't want to be able to instantiate a different Name object also with the name as John Smith, or if instantiation does take place I want a reference to the orginal object to be passed back rather than a new object. I'm aware that one way of doing this is to have a static factory that holds a Map of all the current Name objects and the factory checks that an object with John Smith as the name doesn't already exist before passing back a reference to a Name object. Another way I could think of off the top of my head is having a static Map in the Name class and when the constructor is called throwing an exception if the value passed in for name is already in use in another object, however I'm aware throwing exceptions in a constructor is generally a bad idea. Are there other ways of achieving this?

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  • Podcast Show Notes: The Red Room Interview &ndash; Part 2

    - by Bob Rhubart
    Room bloggers Sean Boiling, Richard Ward, and Mervin Chaing bring their in-the-trenches perspective to the conversation once again in this week’s edition of the OTN ArchBeat Podcast. Listen. (Missed last week? No problemo: Listen to Part 1) In this segment the conversation turns to SOA governance and balancing the need for reuse against the need for speed.  It’s no mystery that many people react to the term “SOA Governance” in much the same way as they would to the sound of Darth Vader’s respirator. But Mervin explains how a simple change in terminology can go a long way toward lowering blood pressure. Those interested in connecting with Sean, Richard, or Mervin can do so via the links listed below: Sean Boiling - Sales Consulting Manager for Oracle Fusion Middleware LinkedIn | Twitter | Blog Richard Ward - SOA Channel Development Manager at Oracle LinkedIn | Blog Mervin Chiang - Consulting Principal at Leonardo Consulting LinkedIn | Twitter | Blog And you’ll find the complete list of the Red Room SOA Best Practice Posts in last week’s show notes. The third and final segment of the Red Room series runs next week.  I have enough material from the original interview for a fourth program,  but it’ll have to wait. Also, as mentioned last week, the podcast name change is now complete, from Arch2Arch, to ArchBeat. As WPBH-TV9 weatherman Phil Connors says, “Anything different is good.”   Technorati Tags: archbeat,podcast. arch2arch,soa,soa governance,oracle,otn Flickr Tags: archbeat,podcast. arch2arch,soa,soa governance,oracle,otn

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  • What is the effect of this order_by clause?

    - by bread
    I don't understand what this order_by clause is doing and whether I need it or not: select c.customerid, c.firstname, c.lastname, i.order_date, i.item, i.price from items_ordered i, customers c where i.customerid = c.customerid group by c.customerid, i.item, i.order_date order by i.order_date desc; This produces this data: 10330 Shawn Dalton 30-Jun-1999 Pogo stick 28.00 10101 John Gray 30-Jun-1999 Raft 58.00 10410 Mary Ann Howell 30-Jan-2000 Unicycle 192.50 10101 John Gray 30-Dec-1999 Hoola Hoop 14.75 10449 Isabela Moore 29-Feb-2000 Flashlight 4.50 10410 Mary Ann Howell 28-Oct-1999 Sleeping Bag 89.22 10339 Anthony Sanchez 27-Jul-1999 Umbrella 4.50 10449 Isabela Moore 22-Dec-1999 Canoe 280.00 10298 Leroy Brown 19-Sep-1999 Lantern 29.00 10449 Isabela Moore 19-Mar-2000 Canoe paddle 40.00 10413 Donald Davids 19-Jan-2000 Lawnchair 32.00 10330 Shawn Dalton 19-Apr-2000 Shovel 16.75 10439 Conrad Giles 18-Sep-1999 Tent 88.00 10298 Leroy Brown 18-Mar-2000 Pocket Knife 22.38 10299 Elroy Keller 18-Jan-2000 Inflatable Mattress 38.00 10438 Kevin Smith 18-Jan-2000 Tent 79.99 10101 John Gray 18-Aug-1999 Rain Coat 18.30 10449 Isabela Moore 15-Dec-1999 Bicycle 380.50 10439 Conrad Giles 14-Aug-1999 Ski Poles 25.50 10449 Isabela Moore 13-Aug-1999 Unicycle 180.79 10101 John Gray 08-Mar-2000 Sleeping Bag 88.70 10299 Elroy Keller 06-Jul-1999 Parachute 1250.00 10438 Kevin Smith 02-Nov-1999 Pillow 8.50 10101 John Gray 02-Jan-2000 Lantern 16.00 10315 Lisa Jones 02-Feb-2000 Compass 8.00 10449 Isabela Moore 01-Sep-1999 Snow Shoes 45.00 10438 Kevin Smith 01-Nov-1999 Umbrella 6.75 10298 Leroy Brown 01-Jul-1999 Skateboard 33.00 10101 John Gray 01-Jul-1999 Life Vest 125.00 10330 Shawn Dalton 01-Jan-2000 Flashlight 28.00 10298 Leroy Brown 01-Dec-1999 Helmet 22.00 10298 Leroy Brown 01-Apr-2000 Ear Muffs 12.50 While if I remove the order_by clause completely, as in this query: select c.customerid, c.firstname, c.lastname, i.order_date, i.item, i.price from items_ordered i, customers c where i.customerid = c.customerid group by c.customerid, i.item, i.order_date; I get these results: 10101 John Gray 30-Dec-1999 Hoola Hoop 14.75 10101 John Gray 02-Jan-2000 Lantern 16.00 10101 John Gray 01-Jul-1999 Life Vest 125.00 10101 John Gray 30-Jun-1999 Raft 58.00 10101 John Gray 18-Aug-1999 Rain Coat 18.30 10101 John Gray 08-Mar-2000 Sleeping Bag 88.70 10298 Leroy Brown 01-Apr-2000 Ear Muffs 12.50 10298 Leroy Brown 01-Dec-1999 Helmet 22.00 10298 Leroy Brown 19-Sep-1999 Lantern 29.00 10298 Leroy Brown 18-Mar-2000 Pocket Knife 22.38 10298 Leroy Brown 01-Jul-1999 Skateboard 33.00 10299 Elroy Keller 18-Jan-2000 Inflatable Mattress 38.00 10299 Elroy Keller 06-Jul-1999 Parachute 1250.00 10315 Lisa Jones 02-Feb-2000 Compass 8.00 10330 Shawn Dalton 01-Jan-2000 Flashlight 28.00 10330 Shawn Dalton 30-Jun-1999 Pogo stick 28.00 10330 Shawn Dalton 19-Apr-2000 Shovel 16.75 10339 Anthony Sanchez 27-Jul-1999 Umbrella 4.50 10410 Mary Ann Howell 28-Oct-1999 Sleeping Bag 89.22 10410 Mary Ann Howell 30-Jan-2000 Unicycle 192.50 10413 Donald Davids 19-Jan-2000 Lawnchair 32.00 10438 Kevin Smith 02-Nov-1999 Pillow 8.50 10438 Kevin Smith 18-Jan-2000 Tent 79.99 10438 Kevin Smith 01-Nov-1999 Umbrella 6.75 10439 Conrad Giles 14-Aug-1999 Ski Poles 25.50 10439 Conrad Giles 18-Sep-1999 Tent 88.00 10449 Isabela Moore 15-Dec-1999 Bicycle 380.50 10449 Isabela Moore 22-Dec-1999 Canoe 280.00 10449 Isabela Moore 19-Mar-2000 Canoe paddle 40.00 10449 Isabela Moore 29-Feb-2000 Flashlight 4.50 10449 Isabela Moore 01-Sep-1999 Snow Shoes 45.00 10449 Isabela Moore 13-Aug-1999 Unicycle 180.79 I'm not sure what the order_by is doing here and if it's having the intended effects.

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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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  • Odd SQL Results

    - by Ryan Burnham
    So i have the following query Select id, [First], [Last] , [Business] as contactbusiness, (Case When ([Business] != '' or [Business] is not null) Then [Business] Else 'No Phone Number' END) from contacts The results look like id First Last contactbusiness (No column name) 2 John Smith 3 Sarah Jane 0411 111 222 0411 111 222 6 John Smith 0411 111 111 0411 111 111 8 NULL No Phone Number 11 Ryan B 08 9999 9999 08 9999 9999 14 David F NULL No Phone Number I'd expect record 2 to also show No Phone Number If i change the "[Business] is not null" to [Business] != null then i get the correct results id First Last contactbusiness (No column name) 2 John Smith No Phone Number 3 Sarah Jane 0411 111 222 0411 111 222 6 John Smith 0411 111 111 0411 111 111 8 NULL No Phone Number 11 Ryan B 08 9999 9999 08 9999 9999 14 David F NULL No Phone Number Normally you need to use is not null rather than != null. whats going on here?

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  • Creating an excel macro to sum lines with duplicate values

    - by john
    I need a macro to look at the list of data below, provide a number of instances it appears and sum the value of each of them. I know a pivot table or series of forumlas could work but i'm doing this for a coworker and it has to be a 'one click here' kinda deal. The data is as follows. A B Smith 200.00 Dean 100.00 Smith 100.00 Smith 50.00 Wilson 25.00 Dean 25.00 Barry 100.00 The end result would look like this Smith 3 350.00 Dean 2 125.00 Wilson 1 25.00 Barry 1 100.00 Thanks in advance for any help you can offer!

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  • Simple Select Statement on MySQL Database Hanging

    - by AlishahNovin
    I have a very simple sql select statement on a very large table, that is non-normalized. (Not my design at all, I'm just trying to optimize while simultaneously trying to convince the owners of a redesign) Basically, the statement is like this: SELECT FirstName, LastName, FullName, State FROM Activity Where (FirstName=@name OR LastName=@name OR FullName=@name) AND State=@state; Now, FirstName, LastName, FullName and State are all indexed as BTrees, but without prefix - the whole column is indexed. State column is a 2 letter state code. What I'm finding is this: When @name = 'John Smith', and @state = '%' the search is really fast and yields results immediately. When @name = 'John Smith', and @state = 'FL' the search takes 5 minutes (and usually this means the web service times out...) When I remove the FirstName and LastName comparisons, and only use the FullName and State, both cases above work very quickly. When I replace FirstName, LastName, FullName, and State searches, but use LIKE for each search, it works fast for @name='John Smith%' and @state='%', but slow for @name='John Smith%' and @state='FL' When I search against 'John Sm%' and @state='FL' the search finds results immediately When I search against 'John Smi%' and @state='FL' the search takes 5 minutes. Now, just to reiterate - the table is not normalized. The John Smith appears many many times, as do many other users, because there is no reference to some form of users/people table. I'm not sure how many times a single user may appear, but the table itself has 90 Million records. Again, not my design... What I'm wondering is - though there are many many problems with this design, what is causing this specific problem. My guess is that the index trees are just too large that it just takes a very long time traversing the them. (FirstName, LastName, FullName) Anyway, I appreciate anyone's help with this. Like I said, I'm working on convincing them of a redesign, but in the meantime, if I someone could help me figure out what the exact problem is, that'd be fantastic.

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  • Is it possible to use ContainsTable to get results for more than one column?

    - by LockeCJ
    Consider the following table: People FirstName nvarchar(50) LastName nvarchar(50) Let's assume for the moment that this table has a full-text index on it for both columns. Let's suppose that I wanted to find all of the people named "John Smith" in this table. The following query seems like a perfectly rational way to accomplish this: SELECT * from People p INNER JOIN CONTAINSTABLE(People,*,'"John*" AND "Smith*"') Unfortunately, this will return no results, assuming that there is no record in the People table that contains both "John" and "Smith" in either the FirstName or LastName columns. It will not match a record with "John" in the FirstName column, and "Smith" in the LastName column, or vice-versa. My question is this: How does one accomplish what I'm trying to do above? Please consider that the example above is simplified. The real table I'm working with has ten columns and the input I'm receiving is a single string which is split up based on standard word breakers (space, dash, etc.)

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  • Parsing complex string using regex

    - by wojtek_z
    My regex skills are not very good and recently a new data element has thrown my parser into a loop Take the following string "+USER=Bob Smith-GROUP=Admin+FUNCTION=Read/FUNCTION=Write" Previously I had the following for my regex : [+\\-/] Which would turn the result into USER=Bob Smith GROUP=Admin FUNCTION=Read FUNCTION=Write FUNCTION=Read But now I have values with dashes in them which is causing bad output New string looks like "+USER=Bob Smith-GROUP=Admin+FUNCTION=Read/FUNCTION=Write/FUNCTION=Read-Write" Which gives me the following result , and breaks the key = value structure. USER=Bob Smith GROUP=Admin FUNCTION=Read FUNCTION=Write FUNCTION=Read Write Can someone help me formulate a valid regex for handling this or point me to some key / value examples. Basically I need to be able to handle + - / signs in order to get combinations.

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  • Are there any well known algorithms to detect the presence of names?

    - by Rhubarb
    For example, given a string: "Bob went fishing with his friend Jim Smith." Bob and Jim Smith are both names, but bob and smith are both words. Weren't for them being uppercase, there would be less indication of this outside of our knowledge of the sentence. Without doing grammar analysis, are there any well known algorithms for detecting the presence of names, at least Western names?

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  • Need help joining tables...

    - by yuudachi
    I am a MySQL newbie, so sorry if this is a dumb question.. These are my tables. student table: SID (primary) student_name advisor (foreign key to faculty.facultyID) requested_advisor (foreign key to faculty.facultyID) faculty table: facultyID (primary key) advisor_name I want to query a table that shows everything in the student table, but I want advisor and requested_advisor to show up as names, not the ID numbers. so like it displays like this on the webpage: Student Name: Jane Smith SID: 860123456 Current Advisor: John Smith Requested advisor: James Smith not like this Student Name: Jane Smith SID: 860123456 Current Advisor: 1 Requested advisor: 2 SELECT student.student_name, SID, student_email, faculty.advisor_name FROM student INNER JOIN faculty ON student.advisor = faculty.facultyID; this comes out close, but I don't know how to get the requested_advisor to show up as a name.

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  • Setting variables in shell script by running commands

    - by rajya vardhan
    >cat /tmp/list1 john jack >cat /tmp/list2 smith taylor It is guaranteed that list1 and list2 will have equal number of lines. f(){ i=1 while read line do var1 = `sed -n '$ip' /tmp/list1` var2 = `sed -n '$ip' /tmp/list2` echo $i,$var1,$var2 i=`expr $i+1` echo $i,$var1,$var2 done < $INFILE } So output of f() should be: 1,john,smith 2,jack,taylor But getting 1,p,p 1+1,p,p If i replace following: var1 = `sed -n '$ip' /tmp/list1` var2 = `sed -n '$ip' /tmp/list2` with this: var1=`head -$i /tmp/vip_list|tail -1` var2=`head -$i /tmp/lb_list|tail -1` Then output: 1,john,smith 1,john,smith Not an expert of shell, so please excuse if sounds childish :)

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  • Creating one row of information in excel using a unique value

    - by user1426513
    This is my first post. I am currently working on a project at work which requires that I work with several different worksheets in order to create one mail master worksheet, as it were, in order to do a mail merge. The worksheet contains information regarding different purchases, and each purchaser is identified with their own ID number. Below is an example of what my spreadsheet looks like now (however I do have more columns): ID Salutation Address ID Name Donation ID Name Tickets 9 Mr. John Doe 123 12 Ms. Jane Smith 100.00 12 Ms.Jane Smith 300.00 12 Ms. Jane Smith 456 22 Mr. Mike Man 500.00 84 Ms. Jo Smith 300.00 What I would like to do is somehow sort my data so that everythign with the same unique identifier (ID) lines up on the same row. For example ID 12 Jane Smith - all the information for her will show up under her name matched by her ID number, and ID 22 will match up with 22 etc... When I merged all of my spreadsheets together, I sorted them all by ID number, however my problem is, not everyone who made a donation bought a ticket or some people just bought tickets and nothing us, so sorting doesn't work. Hopefully this makes sense. Thanks in advance.

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  • w2k3 chkdsk errors as vmware 1.8 guest

    - by Sean Kirkpatrick
    We have two Dell servers (CentOS 5.x) hosting a variety of VMs including 3 W2K3 and 1 W2K servers as guests as well as a handful of other Linux guests. Each Windows VM has 2 drives, C: and D:. on 2 of the W2K3 and the W2K boxes we have recurring errors appearing on a daily basis as reported by CHKDSK. We'll run CHKDSK /f and reboot all affected machines and the errors are reported as fixed. The next day CHKDSK will report the same errors. None of the linux guests nor either host report errors when the are rebooted. The RAID controllers are not reporting errors. We're beginning to think that these are phantom errors somehow, but I'm not willing to go to the bank on that just yet. Anybody have similar experiences or advice? Thanks! Sean

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  • How to Open Local File in AS3?

    - by Sean
    Hi, i have some code that looks like this function main9Click(event:MouseEvent):void { var main9URL:URLRequest = new URLRequest("N:\ICT\Nationals\Unit 2\Pages\Cars"); navigateToURL(main9URL, "_self"); } mainBtn6.addEventListener(MouseEvent.CLICK, main9Click); I need to make it open so when you click on that button it opens the file Cheers Sean

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  • - (void)alertViewCancel:(UIAlertView *)alertView is not called

    - by Sean
    Hi all I've got the problem that the UIAlertViewDelegate method - (void)alertViewCancel:(UIAlertView *)alertView is not called when I cancel a AlertView with it's cancel button. Weird is that the delegate method - (void)alertView:(UIAlertView *)alertView clickedButtonAtIndex:(NSInteger)buttonIndex works perfectly. Does anyone have an idea? Thanks in advance Sean

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  • Change URL of WebBrowser in C#

    - by Sean
    I am designing a simple application using c# and it has a web browser in it. I need to either (preferably) refresh the page that is currently inside the broswer, or navigate to a "new" url. I tried Browser.Url = new Uri("http://www.pandora.com/"); but I get an error when I try to compile it. Error 1 Cannot implicitly convert type 'string' to 'System.Uri' c:\users\sean\documents\visual studio 2010\Projects\Pandora\Pandora\Form1.cs 51 27 Pandora What am I doing wrong?

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  • XAML ToolTip + IsHitTestVisible="False"

    - by Sean
    We need to have mouse clicks and drags "ignored" by our View1 but the ToolTip must still function in that view. The reason is View1 is above View2 in Z-Order, so View1 can tint View2 a red color and show a warning via ToolTip; however the ToolTip accompanying View1 will not work if IsHitTestVisible="False". Anyone know a work around so the ToolTip will display on mouse move/over and the rest of mouse events are ignored by View1 and go to View2? Thanks, Sean

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  • uitableview delegate methods are not called

    - by Sean
    hi all i got the problem that the tableview methods are not called the first time the tableview is shown. if switch back to the previous view and then click the button to show the tableview again, the methods are called this time. i've to say that i show an actionsheet while the tableview is loading. the actionsheet i call in the ViewWillAppear method. thanks in advance sean

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  • Run script after switching user account "to the same account"

    - by Peter Sivák
    In Ubuntu, when I click on Switch User Account... and then choose the same account to log in (for example if my name is John Smith, I click on switch user account and then log into the John Smith account again), how can I run a script after that? (I know, that I can run a script after "first" login by putting it in /etc/profile file, but this script is not executed again when I choose switch user account and then immediately log in back to the same account.)

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  • GDL Presents: Women Techmakers with JESS3

    GDL Presents: Women Techmakers with JESS3 Join Leslie, COO and Co-founder of JESS3, in conversation with Megan Smith and Betsy Masiello, as they discuss Leslie's experience growing a design business from two employees to a transnational operation. Hosts: Megan Smith - Vice President, Google [x] | Betsy Masiello - Policy Manager Guest: Leslie Bradshaw - President, COO and Co-founder, JESS3 From: GoogleDevelopers Views: 0 3 ratings Time: 01:00:00 More in Science & Technology

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  • How to count the most recent value based on multiple criteria?

    - by Andrew
    I keep a log of phone calls like the following where the F column is LVM = Left Voice Mail, U = Unsuccessful, S = Successful. A1 1 B1 Smith C1 John D1 11/21/2012 E1 8:00 AM F1 LVM A2 2 B2 Smith C2 John D2 11/22/2012 E1 8:15 AM F2 U A3 3 B3 Harvey C3 Luke D3 11/22/2012 E1 8:30 AM F3 S A4 4 B4 Smith C4 John D4 11/22/2012 E1 9:00 AM F4 S A5 5 B5 Smith C5 John D5 11/23/2012 E5 8:00 AM F5 LVM This is a small sample. I actually have over 700 entries. In my line of work, it is important to know how many unsuccessful (LVM or U) calls I have made since the last Successful one (S). Since values in the F column can repeat, I need to take into consideration both the B and C column. Also, since I can make a successful call with a client and then be trying to contact them again, I need to be able to count from the last successful call. My G column is completely open which is where I would like to put a running total for each client (G5 would = 1 ideally while G4 = 0, G3 = 0, G2 = 2, G1 = 1 but I want these values calculated automatically so that I do not have scroll through 700 names).

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  • MS Word 2010: Hide citation title when 2 publications by same first author from different years are in one citation block

    - by srunni
    I'm trying to hide the display of the titles for two publications by the same first author from different years that are in the same citation block. By default, the title is shown in citations when there are two publications by the same author in a given document. The easiest way to get around this is to right click on the citation, click "Edit Citation", and then suppress the title. However, the issue with this is that if there are 2 citations in 1 citation block (i.e., "(Smith, J., et al. 2010, Smith, J., et al. 2011)" rather than "(Smith, J., et al. 2010) (Smith, J., et al. 2011)"), then using that suppress option only suppresses the title for the first citation (in this case, the 2010 publication). OTOH, if I try to initially insert the publications in separate citation blocks, I can suppress the title in both citations, but I can't cut and paste one into the other's citation block. I can click "Cut" and the citation that was just cut disappears, but the "Paste" option is not available when my cursor is in the second citation block. Any ideas? Thanks!

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  • LDAP change user pass on client

    - by Sean
    I am trying to allow ldap users to change their password on client machines. I have tried pam every which way I can think of /etc/ldap.conf & /etc/pam_ldap.conf, as well. At this point I'm stuck. Client: Ubuntu 11.04 Server: Debian 6.0 The current output is this: sobrien4@T-E700F-1:~$ passwd passwd: Authentication service cannot retrieve authentication info passwd: password unchanged /var/log/auth.log gives this during the command: May 9 10:49:06 T-E700F-1 passwd[18515]: pam_unix(passwd:chauthtok): user "sobrien4" does not exist in /etc/passwd May 9 10:49:06 T-E700F-1 passwd[18515]: pam_ldap: ldap_simple_bind Can't contact LDAP server May 9 10:49:06 T-E700F-1 passwd[18515]: pam_ldap: reconnecting to LDAP server... May 9 10:49:06 T-E700F-1 passwd[18515]: pam_ldap: ldap_simple_bind Can't contact LDAP server getent passwd |grep sobrien4 (note keeping short since testing with that account, however it outputs all ldap users): sobrien4:Ffm1oHzwnLz0U:10000:12001:Sean O'Brien:/home/sobrien4:/bin/bash getent group shows all ldap groups. /etc/pam.d/common-password (Note this is just the most current, I have tried a lot of different options): password required pam_cracklib.so retry=3 minlen=8 difok=3 password [success=1 default=ignore] pam_unix.so use_authtok md5 password required pam_ldap.so use_authtok password required pam_permit.so Popped open wireshark as well, the server & client are talking. I have the password changing working on the server. I.E. the server that runs slapd, I can log in with the ldap user and change the passwords. I tried copying the working configs from the server initially and no dice. I also tried cloning it, and just changing ip & host, and no go. My guess is that the client is not authorized by ip or hostname to change a pass. Pertaining to the slapd conf, I saw this in a guide and tried it: access to attrs=loginShell,gecos by dn="cn=admin,dc=cengineering,dc=etb" write by self write by * read access to * by dn="cn=admin,dc=cengineering,dc=etb" write by self write by * read So ldap seems to be working okay, just can't change the password.

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