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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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  • Get list of duplicate rows in MySql

    - by user347033
    Hi, i have a table like this ID nachname vorname 1 john doe 2 john doe 3 jim doe 4 Michael Knight I need a query that will return all the fields (select *) from the records that have the same nachname and vorname (in this case, records 1 and 2). Can anyone help me with this? Thanks

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  • MySQL LEFT OUTER JOIN virtual table

    - by user1707323
    I am working on a pretty complicated query let me try to explain it to you. Here is the tables that I have in my MySQL database: students Table --- `students` --- student_id first_name last_name current_status status_change_date ------------ ------------ ----------- ---------------- -------------------- 1 John Doe Active NULL 2 Jane Doe Retread 2012-02-01 students_have_courses Table --- `students_have_courses` --- students_student_id courses_course_id s_date e_date int_date --------------------- ------------------- ---------- ---------- ----------- 1 1 2012-01-01 2012-01-04 2012-01-05 1 2 2012-01-05 NULL NULL 2 1 2012-01-10 2012-01-11 NULL students_have_optional_courses Table --- `students_have_optional_courses` --- students_student_id optional_courses_opcourse_id s_date e_date --------------------- ------------------------------ ---------- ---------- 1 1 2012-01-02 2012-01-03 1 1 2012-01-06 NULL 1 5 2012-01-07 NULL Here is my query so far SELECT `students_and_courses`.student_id, `students_and_courses`.first_name, `students_and_courses`.last_name, `students_and_courses`.courses_course_id, `students_and_courses`.s_date, `students_and_courses`.e_date, `students_and_courses`.int_date, `students_have_optional_courses`.optional_courses_opcourse_id, `students_have_optional_courses`.s_date, `students_have_optional_courses`.e_date FROM ( SELECT `c_s_a_s`.student_id, `c_s_a_s`.first_name, `c_s_a_s`.last_name, `c_s_a_s`.courses_course_id, `c_s_a_s`.s_date, `c_s_a_s`.e_date, `c_s_a_s`.int_date FROM ( SELECT `students`.student_id, `students`.first_name, `students`.last_name, `students_have_courses`.courses_course_id, `students_have_courses`.s_date, `students_have_courses`.e_date, `students_have_courses`.int_date FROM `students` LEFT OUTER JOIN `students_have_courses` ON ( `students_have_courses`.`students_student_id` = `students`.`student_id` AND (( `students_have_courses`.`s_date` >= `students`.`status_change_date` AND `students`.current_status = 'Retread' ) OR `students`.current_status = 'Active') ) WHERE `students`.current_status = 'Active' OR `students`.current_status = 'Retread' ) `c_s_a_s` ORDER BY `c_s_a_s`.`courses_course_id` DESC ) `students_and_courses` LEFT OUTER JOIN `students_have_optional_courses` ON ( `students_have_optional_courses`.students_student_id = `students_and_courses`.student_id AND `students_have_optional_courses`.s_date >= `students_and_courses`.s_date AND `students_have_optional_courses`.e_date IS NULL ) GROUP BY `students_and_courses`.student_id; What I want to be returned is the student_id, first_name, and last_name for all Active or Retread students and then LEFT JOIN the highest course_id, s_date, e_date, and int_date for the those students where the s_date is since the status_change_date if status is 'Retread'. Then LEFT JOIN the highest optional_courses_opcourse_id, s_date, and e_date from the students_have_optional_courses TABLE where the students_have_optional_courses.s_date is greater or equal to the students_have_courses.s_date and the students_have_optional_courses.e_date IS NULL Here is what is being returned: student_id first_name last_name courses_course_id s_date e_date int_date optional_courses_opcourse_id s_date_1 e_date_1 ------------ ------------ ----------- ------------------- ---------- ---------- ------------ ------------------------------ ---------- ---------- 1 John Doe 2 2012-01-05 NULL NULL 1 2012-01-06 NULL 2 Jane Doe NULL NULL NULL NULL NULL NULL NULL Here is what I want being returned: student_id first_name last_name courses_course_id s_date e_date int_date optional_courses_opcourse_id s_date_1 e_date_1 ------------ ------------ ----------- ------------------- ---------- ---------- ------------ ------------------------------ ---------- ---------- 1 John Doe 2 2012-01-05 NULL NULL 5 2012-01-07 NULL 2 Jane Doe NULL NULL NULL NULL NULL NULL NULL Everything is working except one thing, I cannot seem to get the highest students_have_optional_courses.optional_courses_opcourse_id no matter how I form the query Sorry, I just solved this myself after writing this all out I think it helped me think of the solution. Here is the solution query: SELECT `students_and_courses`.student_id, `students_and_courses`.first_name, `students_and_courses`.last_name, `students_and_courses`.courses_course_id, `students_and_courses`.s_date, `students_and_courses`.e_date, `students_and_courses`.int_date, `students_optional_courses`.optional_courses_opcourse_id, `students_optional_courses`.s_date, `students_optional_courses`.e_date FROM ( SELECT `c_s_a_s`.student_id, `c_s_a_s`.first_name, `c_s_a_s`.last_name, `c_s_a_s`.courses_course_id, `c_s_a_s`.s_date, `c_s_a_s`.e_date, `c_s_a_s`.int_date FROM ( SELECT `students`.student_id, `students`.first_name, `students`.last_name, `students_have_courses`.courses_course_id, `students_have_courses`.s_date, `students_have_courses`.e_date, `students_have_courses`.int_date FROM `students` LEFT OUTER JOIN `students_have_courses` ON ( `students_have_courses`.`students_student_id` = `students`.`student_id` AND (( `students_have_courses`.`s_date` >= `students`.`status_change_date` AND `students`.current_status = 'Retread' ) OR `students`.current_status = 'Active') ) WHERE `students`.current_status = 'Active' OR `students`.current_status = 'Retread' ) `c_s_a_s` ORDER BY `c_s_a_s`.`courses_course_id` DESC ) `students_and_courses` LEFT OUTER JOIN ( SELECT * FROM `students_have_optional_courses` ORDER BY `students_have_optional_courses`.optional_courses_opcourse_id DESC ) `students_optional_courses` ON ( `students_optional_courses`.students_student_id = `students_and_courses`.student_id AND `students_optional_courses`.s_date >= `students_and_courses`.s_date AND `students_optional_courses`.e_date IS NULL ) GROUP BY `students_and_courses`.student_id;

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  • Is adding users to the group www-data safe on Debian?

    - by John
    Many PHP applications do self-configuration and self-updating. This requires apache to have write access to the PHP files. While chgrp'ing them all to www-data appears like a good practice to avoid making them world writable, I also wish to allow users to create new files and edit existing one. Is adding users to the group www-data safe on Debian? For example: 775 root www-data /var/www 644 john www-data /var/www/johns_php_application.php 660 john www-data /var/www/johns_php_applications_configuration_file

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  • Handwritten linked list is segfaulting and I don't understand why

    - by Born2Smile
    Hi I was working on a bit of fun, making an interface to run gnuplot from within c++, and for some reason the my linked list implementation fails. The code below fails on the line plots-append(&plot). Stepping through the code I discovered that for some reason the destructor ~John() is called immediately after the constructor John(), and I cannot seem to figure out why. The code included below is a stripped down version operating only on Plot*. Originally I made the linked list as a template class. And it worked fine as ll<int and ll<char* but for some reason it fails as ll<Plot*. Could youp please help me figure out why it fails? and perhaps help me understand how to make it work? In advance: Thanks a heap! //B2S #include <string.h class Plot{ char title[80]; public: Plot(){ } }; class Link{ Plot* element; Link* next; Link* prev; friend class ll; }; class ll{ Link* head; Link* tail; public: ll(){ head = tail = new Link(); head-prev = tail-prev = head-next = tail-next = head; } ~ll(){ while (head!=tail){ tail = tail-prev; delete tail-next; } delete head; } void append(Plot* element){ tail-element = element; tail-next = new Link(); tail-next-prev = tail; tail-next = tail; } }; class John{ ll* plots; public: John(){ plots= new ll(); } ~John(){ delete plots; } John(Plot* plot){ John(); plots-append(plot); } }; int main(){ Plot p; John k(&p); }

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  • SQL Server Distinct Question

    - by RPS
    I need to be able to select only the first row for each name that has the greatest value. I have a table with the following: id name value 0 JOHN 123 1 STEVE 125 2 JOHN 127 3 JOHN 126 So I am looking to return: id name value 1 STEVE 125 2 JOHN 127 Any idea on the MSSQL Syntax on how to perform this operation?

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  • string.Replace does not work for quote

    - by Azhar
    ((string)dt.Rows[i][1]).Replace("'", "\'") I want the result that if any string have quote it change it into slash quote e.g John's - John\'s but the above replace function is not working fine. it results like John\'s but if we change the code to ((string)dt.Rows[i][1]).Replace("'", "\'") it gives the Result like John's does change it anyway.

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  • PHP Find and replace after in a string?

    - by TheBlackBenzKid
    I have some strings which contain the words LIMIT 3, 199 or LIMIT 0, 100. Basically I want to replace the word LIMIT and everything after it in my string. How do I do this in PHP? str_replace only replaces an item and the LIMIT text after is dynamic/ So it could be WHEN JOHN WAS TRYING HIS SQL QUERY, HE FOUND THAT LIMIT, 121 // RETURN WOULD BE WHEN JOHN WAS TRYING HIS SQL QUERY, HE FOUND THAT WHEN JOHN TRIED LIMIT 343, 333 HE FOUND // RETURN WOULD BE WHEN JOHN TRIED

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  • Exchange 2010 OWA - a few questions about using multiple mailboxes

    - by Alexey Smolik
    We have an Exchange 2010 SP2 deployment and we need that our users could access multiple mailboxes in OWA. The problem is that a user (eg John Smith) needs to access not just somebody else's (eg Tom Anderson) mailboxes, but his OWN mailboxes, e.g. in different domains: john[email protected], john[email protected], john[email protected], etc. Of course it is preferable for the user to work with all of his mailboxes from a single window. Such mailboxes can be added as multiple Exchange accounts in Outlook, that works almost fine. But in OWA, there are problems: 1) In the left pane - as I've learned - we can open only Inbox folders from other mailboxes. No way to view all folders like in Outlook? 2) With Send-As permissions set, when trying to send a message from another address, that message is saved in the Sent Items folder of the mailbox that is opened in OWA, and not in the mailbox the message is sent from. The same thing with the trash can. Is there a way to fix that? Also, this problem exists in desktop Outlook when mailboxes are added automatically via the Auto Mapping feature, so that we need to turn it off and add the accounts manually. Is there a simpler workaround? 3) Okay, suppose we only open Inbox folders in the left pane. The problem is that the mailbox names shown there are formed from Display Name attributes. But those names are all identical! All the mailboxes are owned by John Smith, so they should be all named John Smith - so that letter recepient sees "John Smith" in the "from" field, no matter what mailbox it is sent from. Also, the user knows what's his name - no need to tell him. He wants to know what mailbox he works with. So we need a way to either: a) customize OWA to show mailbox email address instead of user Display Name, or b) make Exchange use another attribute to put in the "from" field when sending letters 4) Okay, we can switch between mailboxes using "Open Other Mailbox" in the upper-right corner menu. But: a) To select a mailbox we need to enter its name (or first letters). It there a way to show a list of links to mailboxes the user has full access to? Eg in the page header... b) If we start entering the first letters, we see a popup list with possible mailboxes to be opened. But there are all mailboxes (apparently from GAL), not only mailboxes the user has permission to open! How to filter that popup list? c) The same problem as in (3) with mailbox naming. We can see the opened mailbox email address ONLY in the page URL, which is insufficient for many users. In the left pane we see "John Smith" which is useless. 5) Each mailbox is tied with a separate user in AD. If one has several mailboxes, we need to have additional dummy AD accounts, create additional OUs to store them, etc. That's not very nice, is there any standartized, optimal way to build such a structure? We would really appreciate any answers or additional info for any of these questions. Thank you in advance.

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  • Rhythmbox 2.99.1 crashes when playing any song on Ubuntu 13.10

    - by John H
    Yesterday rhythmbox was running smoothly but today it crashes only a few seconds after I hit the play button, regardless of track. I've tried disabling plugins, re-installing rhythmbox via synaptics, clearing the library and the rhythmdb.xml-file and just adding one album. Still, it crashes. If I run the rhythmbox from the command line i get the following before I have to force quit: :~$ rhythmbox Failed to create secure directory (/run/user/1000/pulse): Permission denied Killed :~$ However, if i run rhythmbox via the command line as superuser it does work. But i get the following errors: sudo rhythmbox (rhythmbox:8335): Gtk-WARNING **: Failed to register client: GDBus.Error:org.freedesktop.DBus.Error.ServiceUnknown: The name org.gnome.SessionManager was not provided by any .service files (rhythmbox:8335): IBUS-WARNING **: The owner of /home/john/.config/ibus/bus is not root! (rhythmbox:8335): Rhythmbox-WARNING **: Unable to grab media player keys: GDBus.Error:org.freedesktop.DBus.Error.ServiceUnknown: The name org.gnome.SettingsDaemon was not provided by any .service files Traceback (most recent call last): File "/usr/lib/rhythmbox/plugins/rb/Loader.py", line 47, in _contents_cb (ok, contents, etag) = file.load_contents_finish(result) gi._glib.GError: Operation not supported Traceback (most recent call last): File "/usr/lib/python2.7/dist-packages/gi/overrides/GLib.py", line 634, in return (lambda data: callback(*data), user_data) […] I'm running rhythmbox on a Lenovo e335 thikpad edge. I hope I've supplied enough information. Cheers -John

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  • Mexico leading in Business Transformation Strategies:

    - by [email protected]
    By john[email protected] on April 15, 2010 8:31 AM By John Burke Group Vice President Oracle Applications Business Unit I recently completed a business tour in Mexico, and was surprised by both the economic vibrancy of the country and the thought leadership expressed by many of the customers I met. An example of the economic vibrancy of the country: across the street from my hotel was the local Bentley dealership, Coach Store, Yves Saint Laurent and of course a Starbucks. I only made it to Starbucks. Both the Coach Store and YSL had a line of folks waiting to get in... As for thought leadership, there were several illustrations only on the first day. I had the opportunity to meet with a branch of the Mexican Federal Government. Their questions were not about clerical task automation, far from it! We discussed citizen on-line access to fees and services - for example looking up the duty on an international goods shipment, or tracking that my taxes have been received, or the status of my request for a certain service. Eligibility, policies and status. Having an integrated rules or policy automation system that would allow businesses and citizens to access accurate information and ensure the proper collection of fees and payment for 3rd party provided services. Then in the afternoon, I met with the owner of a roofing company (note: most roofs in Mexico are flat and made of cement). This CEO started discussing how he wanted to transform his business from a cement products company to a service company and market 5-10-15 year service contracts which would guarantee the structural integrity of the roof and of course that the roof would remain waterproof. Although his products were guaranteed, they required an annual inspection and most home owners never schedule that inspection until it is too late and water damage has occurred. These emergency calls reduce his margin and reduce customer satisfaction. This lead to a discussion of business models in general and why long term differentiation can only come from service, not just for the music or news industries, but also for roofing companies! I completely agreed with the transformational concepts described in both meetings and quickly understood why there is a Bentley dealership near my hotel.

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  • Ubuntu 12.10 unmet dependencies

    - by John
    I have Ubuntu 12.10 with 3.2.0.24-generic #39-Ubuntu running on a Dell Inspiron 700m laptop. I have unmet dependencies as follows: root@John-700m:/home/John# sudo apt-get remove --purge linux-image-3.5.0-{18,27,31,34}-generic Reading package lists... Done Building dependency tree Reading state information... Done Package 'linux-image-3.5.0-18-generic' is not installed, so not removed Package 'linux-image-3.5.0-27-generic' is not installed, so not removed Package 'linux-image-3.5.0-31-generic' is not installed, so not removed Package 'linux-image-3.5.0-34-generic' is not installed, so not removed You might want to run 'apt-get -f install' to correct these: The following packages have unmet dependencies: linux-generic : Depends: linux-image-generic but it is not going to be installed linux-image-extra-3.5.0-18-generic : Depends: linux-image-3.5.0-18-generic but it is not going to be installed linux-image-extra-3.5.0-27-generic : Depends: linux-image-3.5.0-27-generic but it is not going to be installed linux-image-extra-3.5.0-31-generic : Depends: linux-image-3.5.0-31-generic but it is not going to be installed linux-image-extra-3.5.0-34-generic : Depends: linux-image-3.5.0-34-generic but it is not going to be installed E: Unmet dependencies. Try 'apt-get -f install' with no packages (or specify a solution). I tried to remove the packages, but I still get the same error message. Any help would certainly be appreciated. @Eric, well noted. Here is the reults: model name : Intel(R) Pentium(R) M processor 1.60GHz flags : fpu vme de pse tsc msr mce cx8 mtrr pge mca cmov clflush dts acpi mmx fxsr sse sse2 ss tm pbe up bts est tm2 Thanks again. @Eric, If I need to downgrade to Ubuntu 12.04 from 12.10, can I do it without having to mess my current partition or files? I also have Windows XP running on the machine. Best. @Eric, if you are online, I could use your help (or anybody's else). I installed the package for non-pae unit and I still get the same error. Any suggestions? Thanks

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  • Merge\Combine two datatables

    - by madlan
    I'm trying to merge\combine two datatables. I've looked at various examples and answers but they seem to create duplicate rows or require indexes (merge on datatable etc) I can't do this via SQL as one source is from a linked Oracle server accessed via MSSQL and the other from a different MSSQL Server that does not have linked access. The data is currently very simple: Name, Email, Phone DataTable1: "John Clark", "", "01522 55231" "Alex King", "[email protected]", "01522 55266" "Marcus Jones", "[email protected]", "01522 55461" DataTable2: "John Clark", "john[email protected]", "01522 55231" "Alex King", "[email protected]", "" "Marcus Jones", "[email protected]", "01522 55461" "Warren bean", "[email protected]", "01522 522311" Giving a datatable with the following: "John Clark", "john[email protected]", "01522 55231" "Alex King", "[email protected]", "01522 55266" "Marcus Jones", "[email protected]", "01522 55461" "Warren bean", "[email protected]", "01522 522311" Name is the field to match records on, with the first datatable taking priority.

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  • Thunderbird: how to move mails into correct thread? (mailing lists)

    - by unor
    I'm subscribed to some mailing lists and every day people reply to a wrong mail (or they don't reply at all), so that their mail lands in the wrong (or a new) thread. I set the mail display in "View ? Sort by" to "Threaded". Example: mailing list "Foobar": [Foobar] random topic Re: [Foobar] random topic Re: [Foobar] random topic Re: [Foobar] random topic Re: [Foobar] random topic [Foobar] I'm John Doe Re: [Foobar] I'm John Doe Re: [Foobar] Welcome, John Re: [Foobar] random topic There are two discussions, one about "random topic", one about "John Doe". The subject line changes in the discussion about John Doe, which is fine (no problem here). But the last mail should be pigeonholed in the first thread. Instead it is at the top-level. Now, how can I move that last mail into the correct thread? I tried to drag&drop it at the mail I think it should be a reply to, but this doesn't work. I think theoretically it should be possible by fiddling with the mail headers after receiving the mail, but this doesn't seem to be a comfortable way.

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  • SQL group and order

    - by John Lambert
    I have multiple users with multiple entries recording times they arrive at destinations Somehow, with my select query I would like to only show the most recent entries for each unique user name. Here is the code that doesn't work: SELECT * FROM $dbTable GROUP BY xNAME ORDER BY xDATETIME DESC This does the name grouping fine, but as far as showing ONLY their most recent entry, is just shows the first entry it sees in the SQL table. I guess my question is, is this possible? Here is my data sample: john 7:00 chris 7:30 greg 8:00 john 8:15 greg 8:30 chris 9:00 and my desired result should only be john 8:15 chris 9:00 greg 8:30

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  • C# - Determine if class initializaion causes infinite recursion?

    - by John M
    I am working on porting a VB6 application to C# (Winforms 3.5) and while doing so I'm trying to break up the functionality into various classes (ie database class, data validation class, string manipulation class). Right now when I attempt to run the program in Debug mode the program pauses and then crashes with a StackOverFlowException. VS 2008 suggests a infinite recursion cause. I have been trying to trace what might be causing this recursion and right now my only hypothesis is that class initializations (which I do in the header(?) of each class). My thought is this: mainForm initializes classA classA initializes classB classB initializes classA .... Does this make sense or should I be looking elsewhere? UPDATE1 (a code sample): mainForm namespace john { public partial class frmLogin : Form { stringCustom sc = new sc(); stringCustom namespace john { class stringCustom { retrieveValues rv = new retrieveValues(); retrieveValues namespace john { class retrieveValues { stringCustom sc = new stringCustom();

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  • struct assignment operator on arrays

    - by Django fan
    Suppose I defined a structure like this: struct person { char name [10]; int age; }; and declared two person variables: person Bob; person John; where Bob.name = "Bob", Bob.age = 30 and John.name = "John",John.age = 25. and I called Bob = John; struct person would do a Memberwise assignment and assign Johns's member values to Bob's. But arrays can't assign to arrays, so how does the assignment of the "name" array work?

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  • How do I average the difference between specific values in TSQL?

    - by jvenema
    Hey folks, sorry this is a bit of a longer question... I have a table with the following columns: [ChatID] [User] [LogID] [CreatedOn] [Text] What I need to find is the average response time for a given user id, to another specific user id. So, if my data looks like: [1] [john] [20] [1/1/11 3:00:00] [Hello] [1] [john] [21] [1/1/11 3:00:23] [Anyone there?] [1] [susan] [22] [1/1/11 3:00:43] [Hello!] [1] [susan] [23] [1/1/11 3:00:53] [What's up?] [1] [john] [24] [1/1/11 3:01:02] [Not much] [1] [susan] [25] [1/1/11 3:01:08] [Cool] ...then I need to see that Susan has an average response time of (20 + 6) / 2 = 13 seconds to John, and John has an average of (9 / 1) = 9 seconds to Susan. I'm not even sure this can be done in set-based logic, but if anyone has any ideas, they'd be much appreciated!

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  • Must I loop to search results for a specific value?

    - by tag
    I have a table in the database: name Opinion Tim Tim has an opinion John other random text Dan Dan's random text Al Al says something else I call this data and get it back in getRecords.lastResult To access John's opinion, I could use: getRecords.lastResult[1].opinion But that's only because I know that John is the second record (record 1), but this may change. So the right way is to search through the results to first find the record index for John, then access his opinion. My guess is I need some sort of a loop? Is there an easier way to search for John directly without a loop?

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  • Installing CUDA on Ubuntu 12.04 with nvidia driver 295.59

    - by johnmcd
    I have been trying to get cuda to run on a nvidia gt 650m based laptop. I am running Ubuntu 12.04 with the nvidia 295.59 driver. Also, my laptop uses Optimus so I have install the driver via bumblebee. Bumblebee is not working correctly yet -- however I believe it is possible to install CUDA independently. To install CUDA I have followed the instructions detailed here: How can I get nVidia CUDA or OpenCL working on a laptop with nVidia discrete card/Intel Integrated Graphics? However I am still running into problem building the sdk. I made the changes specified at the above link in common.mk, but I got the following (snippet) from the build process: make[2]: Entering directory `/home/john/NVIDIA_GPU_Computing_SDK/C/src/fluidsGL' /usr/bin/ld: warning: libnvidia-tls.so.302.17, needed by /usr/lib/nvidia-current/libGL.so, not found (try using -rpath or -rpath-link) /usr/bin/ld: warning: libnvidia-glcore.so.302.17, needed by /usr/lib/nvidia-current/libGL.so, not found (try using -rpath or -rpath-link) /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv018tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv012glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv017glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv012tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv015tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv019tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv000glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv017tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv013tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv013glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv018glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv022tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv007tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv009tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv020tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv014glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv015glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv016tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv001glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv006tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv021tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv011tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv020glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv019glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv002glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv021glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv014tls' collect2: ld returned 1 exit status make[2]: *** [../../bin/linux/release/fluidsGL] Error 1 make[2]: Leaving directory `/home/john/NVIDIA_GPU_Computing_SDK/C/src/fluidsGL' make[1]: *** [src/fluidsGL/Makefile.ph_build] Error 2 make[1]: Leaving directory `/home/john/NVIDIA_GPU_Computing_SDK/C' make: *** [all] Error 2 The libraries that ld warns about are on my system and are installed on the system: $ locate libnvidia-tls.so.302.17 libnvidia-glcore.so.302.17 /usr/lib/nvidia-current/libnvidia-glcore.so.302.17 /usr/lib/nvidia-current/libnvidia-tls.so.302.17 /usr/lib/nvidia-current/tls/libnvidia-tls.so.302.17 /usr/lib32/nvidia-current/libnvidia-glcore.so.302.17 /usr/lib32/nvidia-current/libnvidia-tls.so.302.17 /usr/lib32/nvidia-current/tls/libnvidia-tls.so.302.17 however /usr/lib/nvidia-current and /usr/lib32/nvidia-current are not being picked up by ldconfig. I have tried adding them by adding a file to /etc/ld.so.conf.d/ which gets past this error, however now I am getting the following error: make[2]: Entering directory `/home/john/NVIDIA_GPU_Computing_SDK/C/src/deviceQueryDrv' cc1plus: warning: command line option ‘-Wimplicit’ is valid for C/ObjC but not for C++ [enabled by default] obj/x86_64/release/deviceQueryDrv.cpp.o: In function `main': deviceQueryDrv.cpp:(.text.startup+0x5f): undefined reference to `cuInit' deviceQueryDrv.cpp:(.text.startup+0x99): undefined reference to `cuDeviceGetCount' deviceQueryDrv.cpp:(.text.startup+0x10b): undefined reference to `cuDeviceComputeCapability' deviceQueryDrv.cpp:(.text.startup+0x127): undefined reference to `cuDeviceGetName' deviceQueryDrv.cpp:(.text.startup+0x16a): undefined reference to `cuDriverGetVersion' deviceQueryDrv.cpp:(.text.startup+0x1f0): undefined reference to `cuDeviceTotalMem_v2' deviceQueryDrv.cpp:(.text.startup+0x262): undefined reference to `cuDeviceGetAttribute' deviceQueryDrv.cpp:(.text.startup+0x457): undefined reference to `cuDeviceGetAttribute' deviceQueryDrv.cpp:(.text.startup+0x4bc): undefined reference to `cuDeviceGetAttribute' deviceQueryDrv.cpp:(.text.startup+0x502): undefined reference to `cuDeviceGetAttribute' deviceQueryDrv.cpp:(.text.startup+0x533): undefined reference to `cuDeviceGetAttribute' obj/x86_64/release/deviceQueryDrv.cpp.o:deviceQueryDrv.cpp:(.text.startup+0x55e): more undefined references to `cuDeviceGetAttribute' follow collect2: ld returned 1 exit status make[2]: *** [../../bin/linux/release/deviceQueryDrv] Error 1 make[2]: Leaving directory `/home/john/NVIDIA_GPU_Computing_SDK/C/src/deviceQueryDrv' make[1]: *** [src/deviceQueryDrv/Makefile.ph_build] Error 2 make[1]: Leaving directory `/home/john/NVIDIA_GPU_Computing_SDK/C' make: *** [all] Error 2 I would appreciate any help that anyone can provide me with. If I can provide any further information please let me know. Thanks.

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  • Checking data of all same class elements

    - by Tiffani
    I need the code to check the data-name value of all instances of .account-select. Right now it just checks the first .account-select element and not any subsequent ones. The function right now is on click of an element such as John Smith, it checks the data-name of the .account-select lis. If the data-names are the same, it does not create a new li with the John Smith data. If no data-names are equal to John Smith, then it adds an li with John Smith. This is the JS-Fiddle I made for it so you can see what I am referring to: http://jsfiddle.net/rsxavior/vDCNy/22/ Any help would be greatly appreciated. This is the Jquery Code I am using right now. $('.account').click(function () { var acc = $(this).data("name"); var sel = $('.account-select').data("name"); if (acc === sel) { } else { $('.account-hidden-li').append('<li class="account-select" data-name="'+ $(this).data("name") +'">' + $(this).data("name") + '<a class="close bcn-close" data-dismiss="alert" href="#">&times;</a></li>'); } }); And the HTML: <ul> <li><a class="account" data-name="All" href="#">All</a></li> <li><a class="account" data-name="John Smith" href="#">John Smith</a></li> </ul> <ul class="account-hidden-li"> <ul>

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  • Pattern/Matcher in Java?

    - by user1007059
    I have a certain text in Java, and I want to use pattern and matcher to extract something from it. This is my program: public String getItemsByType(String text, String start, String end) { String patternHolder; StringBuffer itemLines = new StringBuffer(); patternHolder = start + ".*" + end; Pattern pattern = Pattern.compile(patternHolder); Matcher matcher = pattern.matcher(text); while (matcher.find()) { itemLines.append(text.substring(matcher.start(), matcher.end()) + "\n"); } return itemLines.toString(); } This code works fully WHEN the searched text is on the same line, for instance: String text = "My name is John and I am 18 years Old"; getItemsByType(text, "My", "John"); immediately grabs the text "My name is John" out of the text. However, when my text looks like this: String text = "My name\nis John\nand I'm\n18 years\nold"; getItemsByType(text, "My", "John"); It doesn't grab anything, since "My" and "John" are on different lines. How do I solve this?

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  • Parsing back to 'messy' API strcuture

    - by Eric Fail
    I'm fetching data from an online database (REDcap) via API and the data gets delivered in as comma separated string like this, RAW.API <- structure("id,event_arm,name,dob,pushed_text,pushed_calc,complete\n\"01\",\"event_1_arm_1\",\"John\",\"1979-05-01\",\"\",\"\",2\n\"01\",\"event_2_arm_1\",\"John\",\"2012-09-02\",\"abc\",\"123\",1\n\"01\",\"event_3_arm_1\",\"John\",\"2012-09-10\",\"\",\"\",2\n\"02\",\"event_1_arm_1\",\"Mary\",\"1951-09-10\",\"def\",\"456\",2\n\"02\",\"event_2_arm_1\",\"Mary\",\"1978-09-12\",\"\",\"\",2\n", "`Content-Type`" = structure(c("text/html", "utf-8"), .Names = c("", "charset"))) I have this script that nicely parses it into a data frame, (df <- read.table(file = textConnection(RAW.API), header = TRUE, sep = ",", na.strings = "", stringsAsFactors = FALSE)) id event_arm name dob pushed_text pushed_calc complete 1 1 event_1_arm_1 John 1979-05-01 <NA> NA 2 2 1 event_2_arm_1 John 2012-09-02 abc 123 1 3 1 event_3_arm_1 John 2012-09-10 <NA> NA 2 4 2 event_1_arm_1 Mary 1951-09-10 def 456 2 5 2 event_2_arm_1 Mary 1978-09-12 <NA> NA 2 I then do some calculations and write them to pushed_text and pushed_calc whereafter I need to format the data back to the messy comma separated structure it came in. I imagine something like this, API.back <- `some magic command`(df, ...) identical(RAW.API, API.back) [1] TRUE Some command that can format my data from the data frame I made, df, back to the structure that the raw API-object came in, RAW.API. Any help would be very appreciated.

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  • Can MySQL Nested Select return list of results

    - by John
    Hi I want to write a mysql statement which will return a list of results from one table along with a comma separated list of field from another table. I think an example might better explain it Table 1 ======================== id First_Name Surname ---------------------- 1 Joe Bloggs 2 Mike Smith 3 Jane Doe Table 2 ======================== id Person_Id Job_id --------------------- 1 1 1 2 1 2 3 2 2 4 3 3 5 3 4 I want to return a list of people with a comma separated list of job_ids. So my result set would be id First_Name Surname job_id ------------------------------ 1 Joe Bloggs 1,2 2 Mike Smith 2 3 Jane Doe 3,4 I guess the sql would be something like select id, First_Name, Surname, (SELECT job_id FROM Table 2) as job_id from Table 1 but obviously this does not work so need to change the '(SELECT job_id FROM Table 2) as job_id' part. Hope this makes sense Thanks John

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  • Exchange 2010: Receiving "You can't send a message on behalf of this user..." error when trying to configure delegate access

    - by Beaming Mel-Bin
    I am trying to give someone (John Doe) delegate access to another account (Jane Doe) in our test environment. However, I receive the following error from Exchange: *Subject:* Undeliverable: You have been designated as a delegate for Jane Dow *To:* Jane Dow Delivery has failed to these recipients or groups: John Doe You can't send a message on behalf of this user unless you have permission to do so. Please make sure you're sending on behalf of the correct sender, or request the necessary permission. If the problem continues, please contact your helpdesk. Diagnostic information for administrators: Generating server: /O=UNIONCO/OU=EXCHANGE ADMINISTRATIVE GROUP (FYD132341234)/CN=RECIPIENTS/CN=jdoe #MSEXCH:MSExchangeIS:/DC=local/DC=unionco:MAILBOX-1[578:0x000004DC:0x0000001D] #EX# Can someone help me troubleshoot this?

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