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  • Permissions issue Mac OS X Client -> Mac OS X Server

    - by Meltemi
    I can't get access to a folder on our server and can't understand why. Perhaps someone will see what I'm overlooking... Trouble accessing /Library/Subdirectory/NextDirectory/ User joe can ssh to the server just fine and cd to /Library/Subdirectory/ however trying to cd into the next folder, NextDirectory, gives this error: -bash: cd: NextDirectory/: Permission denied both username joe & bill are members of the group admin and both can get INTO Subdirectory without any trouble... hostname:Library joe$ ls -l | grep Subdirectory drwxrwxr-x 3 bill admin 102 Jun 1 14:51 Subdirectory and from w/in the Subversion folder hostname:Subdirectory joe$ ls -l drwxrwx--- 5 root admin 170 Jun 1 22:19 NextDirectory bill can cd into NextDirectory but joe cannot!?! What am I overlooking? What tools do we have to troubleshoot this? thanks!

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  • Show #14 DotNetNuke 5.6.1, Razor/Webmatrix and WebCamps

    - by Chris Hammond
    Once again, it’s been far too long since the last show, this time just over 4 months, For Show #14 I am joined by Joe Brinkman. Take a listen and see what has been going on in the DNN world. Length: 47:56 Size: 43.8mb MP3 Download Welcome back to DNNVoice Welcome to guest Joe Brinkman ( http://blog.theaccidentalgeek.com/ ) Introduction to Joe and Welcome back from Chris Hammond ( http://www.chrishammond.com ) and what he's been doing DotNetNuke Training Free Extensions Module Development Templates...(read more)

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  • Great Programmer Productivity - Accounting for 10,000 fold difference?

    - by TheImpact
    "A great lathe operator commands several times the wage of an average lathe operator, but a great writer of software code is worth 10,000 times the price of an average software writer." - Bill Gates Say there's a "great" software engineer and an "average" software engineer on the same team. How can you account for one engineer being 10,000 times more productive? I can't quite fathom this, given they're both taking on their share of features, bugs and investigations, and consistently deliver with quality. Would my description possibly justify them to be above "average"? "great"? In a corporation like Microsoft, what % of software engineers are "average"? What % "great"?

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  • Beginning Java (Working with Arrays; Class Assignment)

    - by Jason
    I am to the point where I feel as if I correctly wrote the code for this homework assignment. We were given a skeleton and 2 classes that we had to import (FileIOHelper and Student). /* * Created: *** put the date here *** * * Author: *** put your name here *** * * The program will read information about students and their * scores from a file, and output the name of each student with * all his/her scores and the total score, plus the average score * of the class, and the name and total score of the students with * the highest and lowest total score. */ // import java.util.Scanner; import java.io.*; // C:\Users\Adam\info.txt public class Lab6 { public static void main(String[] args) throws IOException { // Fill in the body according to the following comments Scanner key boardFile = new Scanner(System.in); // Input file name String filename = getFileName(keyboardFile); //Open the file // Input number of students int numStudents = FileIOHelper.getNumberOfStudents(filename); Student students[] = new Student[numStudents]; // Input all student records and create Student array and // integer array for total scores int totalScore[] = new int[students.length]; for (int i = 0; i < students.length; i++){ for(int j = 1; j < 4; j++){ totalScore[i] = totalScore[i] + students[i].getScore(j); } } // Compute total scores and find students with lowest and // highest total score int maxScore = 0; int minScore = 0; for(int i = 0; i < students.length; i++){ if(totalScore[i] >= totalScore[maxScore]){ maxScore = i; } else if(totalScore[i] <= totalScore[minScore]){ minScore = i; } } // Compute average total score int allScores = 0; int average = 0; for (int i = 0; i < totalScore.length; i++){ allScores = allScores + totalScore[i]; } average = allScores / totalScore.length; // Output results outputResults(students, totalScore, maxScore, minScore, average); } // Given a Scanner in, this method prompts the user to enter // a file name, inputs it, and returns it. private static String getFileName(Scanner in) { // Fill in the body System.out.print("Enter the name of a file: "); String filename = in.next(); return filename; // Do not declare the Scanner variable in this method. // You must use the value this method receives in the // argument (in). } // Given the number of students records n to input, this // method creates an array of Student of the appropriate size, // reads n student records using the FileIOHelper, and stores // them in the array, and finally returns the Student array. private static Student[] getStudents(int n) { Student[] myStudents = new Student[n]; for(int i = 0; i <= n; i++){ myStudents[i] = FileIOHelper.getNextStudent(); } return myStudents; } // Given an array of Student records, an array with the total scores, // the indices in the arrays of the students with the highest and // lowest total scores, and the average total score for the class, // this method outputs a table of all the students appropriately // formatted, plus the total number of students, the average score // of the class, and the name and total score of the students with // the highest and lowest total score. private static void outputResults( Student[] students, int[] totalScores, int maxIndex, int minIndex, int average ) { // Fill in the body System.out.println("\nName \t\tScore1 \tScore2 \tScore3 \tTotal"); System.out.println("--------------------------------------------------------"); for(int i = 0; i < students.length; i++){ outputStudent(students[i], totalScores[i], average); System.out.println(); } System.out.println("--------------------------------------------------------"); outputNumberOfStudents(students.length); outputAverage(average); outputMaxStudent(students[maxIndex], totalScores[maxIndex]); outputMinStudent(students[minIndex], totalScores[minIndex]); System.out.println("--------------------------------------------------------"); } // Given a Student record, the total score for the student, // and the average total score for all the students, this method // outputs one line in the result table appropriately formatted. private static void outputStudent(Student s, int total, int avg) { System.out.print(s.getName() + "\t"); for(int i = 1; i < 4; i++){ System.out.print(s.getScore(i) + "\t"); } System.out.print(total + "\t"); if(total < avg){ System.out.print("-"); }else if(total > avg){ System.out.print("+"); }else{ System.out.print("="); } } // Given the number of students, this method outputs a message // stating what the total number of students in the class is. private static void outputNumberOfStudents(int n) { System.out.println("The total number of students in this class is: \t" + n); } // Given the average total score of all students, this method // outputs a message stating what the average total score of // the class is. private static void outputAverage(int average) { System.out.println("The average total score of the class is: \t" + average); } // Given the Student with highest total score and the student's // total score, this method outputs a message stating the name // of the student and the highest score. private static void outputMaxStudent( Student student, int score ) { System.out.println(student.getName() + " got the maximum total score of: \t" + score); } // Given the Student with lowest total score and the student's // total score, this method outputs a message stating the name // of the student and the lowest score. private static void outputMinStudent( Student student, int score ) { System.out.println(student.getName() + " got the minimum total score of: \t" + score); } } But now I get an error at the line totalScore[i] = totalScore[i] + students[i].getScore(j); Exception in thread "main" java.lang.NullPointerException at Lab6.main(Lab6.java:42)

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  • 10 lines of code per day is the global average!? -- true?

    - by Earlz
    Ok so last year I participated in a high school curriculum contest thing at a college(I currently attend this college). I actually got 1st in it but was still a bit angry I didn't get every single one right. The most baffling of questions on there was How many lines of code does the average programmer write per day? A. 5 B. 10 C. 25 D. 30 Aside from being a subjective question which depended on language and everything else I was more baffled at what they had as the correct answer. 10. Even on my bad days at my job I touch more than 10 lines of code(either adding, modifying, or deleting) per day. And when I took this test I had only programmed as a hobby where it was common for me to write a few hundred lines for one of my new projects per day. Where are they getting this random number of ten!? Is this published somewhere? A quick googling found me nothing.

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  • How can I compute the average cost for this solution of the element uniqueness problem?

    - by Alceu Costa
    In the book Introduction to the Design & Analysis of Algorithms, the following solution is proposed to the element uniqueness problem: ALGORITHM UniqueElements(A[0 .. n-1]) // Determines whether all the elements in a given array are distinct // Input: An array A[0 .. n-1] // Output: Returns "true" if all the elements in A are distinct // and false otherwise. for i := 0 to n - 2 do for j := i + 1 to n - 1 do if A[i] = A[j] return false return true How can I compute the average cost (i.e. number of comparisons for a given n) for this algorithm? What is a reasonable assumption about the input?

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  • Which parallel sorting algorithm has the best average case performance?

    - by Craig P. Motlin
    Sorting takes O(n log n) in the serial case. If we have O(n) processors we would hope for a linear speedup. O(log n) parallel algorithms exist but they have a very high constant. They also aren't applicable on commodity hardware which doesn't have anywhere near O(n) processors. With p processors, reasonable algorithms should take O(n/p log n/p) time. In the serial case, quick sort has the best runtime complexity on average. A parallel quick sort algorithm is easy to implement (see here and here). However it doesn't perform well since the very first step is to partition the whole collection on a single core. I have found information on many parallel sort algorithms but so far I have not seen anything pointing to a clear winner. I'm looking to sort lists of 1 million to 100 million elements in a JVM language running on 8 to 32 cores.

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  • Improving terminal server performance for a specfic app

    - by Matt
    We have a windows 2003 terminal server running 2X application load balancign that is hosting a client's application that is accessed by around 50 users. Each user has there own database. The database is a file based database. The application is developed under Delphi so I think the database may be BDE based. As you can imagine, there is probably quite a lot of disk i/o. Here are some of the perfmon settings. Logged in users (average) 20 - 25 CPU Utilization (average) 80 - 100% Disk Queue Length (average) 1.6 % Disk time (average) 111 Page faults/sec (average) 1400 The application takes on average about a minute to load up. As usual, the budget is tight. Is there basic windows performance tuning tips that people can recommend to improve things before we fork out on more RAM etc. Server is a 2.8GHz Xeon with 3GB of RAM.

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  • What kinds of protections against viruses does Linux provide out of the box for the average user?

    - by ChocoDeveloper
    I know others have asked this, but I have other questions related to this. In particular, I'm concerned about the damage that the virus can do the user itself (his files), not the OS in general nor other users of the same machine. This question came to my mind because of that ransomware virus that is encrypting machines all over the world, and then asking the user to send a payment in Bitcoin if he wants to recover his files. I have already received and opened the email that is supposed to contain the virus, so I guess I didn't do that bad because nothing happened. But would I have survived if I opened the attachment and it was aimed at Linux users? I guess not. One of the advantages is that files are not executable by default right after downloading them. Is that just a bad default in Windows and could be fixed with a proper configuration? As a Linux user, I thought my machine was pretty secure by default, and I was even told that I shouldn't bother installing an antivirus. But I have read some people saying that the most important (or only?) difference is that Linux is just less popular, so almost no one writes viruses for it. Is that right? What else can I do to be safe from this kind of ransomware virus? Not automatically executing random files from unknown sources seems to be more than enough, but is it? I can't think of many other things a user can do to protect his own files (not the OS, not other users), because he has full permissions on them.

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  • What is a realistic average time difference between servers in the same LAN?

    - by monster
    Until recently, we had at work a small cluster of about 20 small Windows servers (which have now all been virtualized). They were all configured to synchronize with the local time server. It was on an 1Gb sub-network in our own DC. I never got them to be less than about 100ms away from each other, which I consider to be an incredibly big difference. Is that a normal value? What is a realistic expectation of time difference between machines running on a 1Gb network, and all connected to the same time server, and updating frequently, say every 5 minutes? I would like to know this as setting timeouts and other parameters in a distributed application requires to take that difference into consideration.

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  • How to select from tableA sum of grouped numbers from tableB above their sums average in Oracle?

    - by Nazgulled
    I have data like this: tableA.ID --------- 1 2 3 tableB.ID tableB.NUM -------------------- 1 10 1 15 2 18 3 12 2 15 3 13 1 12 I need to select tableA IDs where the sum of their NUMs in tableB is above the average of all tableA IDs sums. In other words: SUM ID=1 -> 10+15+12 = 37 SUM ID=2 -> 18+12+15 = 45 SUM ID=3 -> 12+13 = 25 AVG ALL IDs -> (37+45+25)/3 = 35 The SELECT must only show ID 1 and 2 because 37 35, 45 35 but 25 < 35. This is my current query which is working fine: SELECT tableA.ID FROM tableA, tableB WHERE tableA.ID = tableB.ID HAVING SUM(tableB.NUM) > ( SELECT AVG(MY_SUM) FROM ( SELECT SUM(tableB.NUM) MY_SUM FROM tableA, tableB WHERE tableA.ID = tableB.ID GROUP BY tableA.ID ) ) GROUP BY tableA.ID But I have a feeling there might be a better way without all those nested SELECTs. Perhaps 2, but 3 feels like too much. I'm probably wrong though. For instance, why can't I do something simple like this: SELECT tableA.ID FROM tableA, tableB WHERE tableA.ID = tableB.ID HAVING SUM(tableB.NUM) > AVG(SUM(tableB.NUM)) GROUP BY tableA.ID Or this: SELECT tableA.ID, SUM(tableB.NUM) MY_SUM FROM tableA, tableB WHERE tableA.ID = tableB.ID HAVING MY_SUM > AVG(MY_SUM) GROUP BY tableA.ID

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  • Cucumber failed scenarios not providing details (Ubuntu)

    - by user361646
    When I run cucumber from my Ubuntu server I don't get details on why the scenario is failing. For example here is what I get: ..... cucumber features/messaging.feature:6 # Scenario: Joe can view his inbox cucumber features/messaging.feature:14 # Scenario: Joe can send a message cucumber features/messaging.feature:26 # Scenario: Joe can view a message in his inbox cucumber features/messaging.feature:35 # Scenario: Joe can reply to a message ..... Is there something I need to configure or pass to the cucumber command to see the details of the failed scenarios??

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  • Using the apostrophe for PHP associative array element's name

    - by Haskella
    Hi I've been trying to figure out how to put something like Joe's Fruits into a PHP array something like this: <?php $arr = array( 'Fruitland' => '3ddlskdf3', 'Joe's Fruits' => 'dddfdfer3', ); ?> Using the above for example (stackoverflow's code color should tell you this by now), the array will take it as 'Joe' between the two apostrophes instead of the whole 'Joe's Fruits' is there any way I can do this without just calling it 'Joes Fruits'?

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  • can not connect through SCP, but SSH connections works

    - by Joe Cabezas
    i am trying to connect to my server to transfer file using scp: $ scp -v -r -P <port> <user>@<host>:~/dir/ dir/ this is the output: OpenSSH_5.2p1, OpenSSL 0.9.8r 8 Feb 2011 debug1: Reading configuration data /Users/joe/.ssh/config debug1: Reading configuration data /etc/ssh_config debug1: Connecting to <host> [<host>] port <port>. debug1: Connection established. debug1: identity file /Users/joe/.ssh/identity type -1 debug1: identity file /Users/joe/.ssh/id_rsa type -1 debug1: identity file /Users/joe/.ssh/id_dsa type -1 ssh_exchange_identification: Connection closed by remote host but connecting via SSH works fine: $ ssh <user>@<host> -p <port> <user>@<host>'s password: <user>@<host>:~$ OK what can be wrong with this? my /etc/ssh/sshd_config file on the host is: # Package generated configuration file # See the sshd_config(5) manpage for details # What ports, IPs and protocols we listen for Port <port> # Use these options to restrict which interfaces/protocols sshd will bind to #ListenAddress :: #ListenAddress 0.0.0.0 Protocol 2 # HostKeys for protocol version 2 HostKey /etc/ssh/ssh_host_rsa_key HostKey /etc/ssh/ssh_host_dsa_key HostKey /etc/ssh/ssh_host_ecdsa_key #Privilege Separation is turned on for security UsePrivilegeSeparation yes # Lifetime and size of ephemeral version 1 server key KeyRegenerationInterval 3600 ServerKeyBits 768 # Logging SyslogFacility AUTH LogLevel INFO # Authentication: LoginGraceTime 120 PermitRootLogin yes StrictModes yes RSAAuthentication yes PubkeyAuthentication no #AuthorizedKeysFile %h/.ssh/authorized_keys # Don't read the user's ~/.rhosts and ~/.shosts files IgnoreRhosts yes # For this to work you will also need host keys in /etc/ssh_known_hosts RhostsRSAAuthentication no # similar for protocol version 2 HostbasedAuthentication no # Uncomment if you don't trust ~/.ssh/known_hosts for RhostsRSAAuthentication #IgnoreUserKnownHosts yes # To enable empty passwords, change to yes (NOT RECOMMENDED) PermitEmptyPasswords no # Change to yes to enable challenge-response passwords (beware issues with # some PAM modules and threads) ChallengeResponseAuthentication no # Change to no to disable tunnelled clear text passwords #PasswordAuthentication yes # Kerberos options #KerberosAuthentication no #KerberosGetAFSToken no #KerberosOrLocalPasswd yes #KerberosTicketCleanup yes # GSSAPI options #GSSAPIAuthentication no #GSSAPICleanupCredentials yes X11Forwarding yes X11DisplayOffset 10 PrintMotd no PrintLastLog yes TCPKeepAlive yes #UseLogin no #MaxStartups 10:30:60 #Banner /etc/issue.net # Allow client to pass locale environment variables AcceptEnv LANG LC_* Subsystem sftp /usr/lib/openssh/sftp-server # Set this to 'yes' to enable PAM authentication, account processing, # and session processing. If this is enabled, PAM authentication will # be allowed through the ChallengeResponseAuthentication and # PasswordAuthentication. Depending on your PAM configuration, # PAM authentication via ChallengeResponseAuthentication may bypass # the setting of "PermitRootLogin without-password". # If you just want the PAM account and session checks to run without # PAM authentication, then enable this but set PasswordAuthentication # and ChallengeResponseAuthentication to 'no'. UsePAM yes

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  • DNSBL listed at zen.spamhaus.org - cant get outgoing mail working? Am I interpreting the response correctly?

    - by Joe Hopfgartner
    I have problem with a mailserver and there is something I kind of not understand! I can connect, authenticate, specify the sender address - but when specifying the reciever i get a error 550 which looks like so: RCPT TO:[email protected] 550-DNSBL listed at zen.spamhaus.org 550 http://www.spamhaus.org/query/bl?ip=62.178.15.161 Now the strange thing is that 62.178.15.161 is my local client address. Not the servers ip address. Also the error code 550 seems to be defined as so: 550 Requested action not taken: mailbox unavailable To me that makes totally no sense. Why this error code with this spamhaus message? Why the local ip adress and not the servers? There is exim running and there is nothing turning up in the logs mail.err mail.info mail.log mail.warn in /var/log I looked up both the servers and the clients ip adress on blacklists. The clients ip adress is listed on some (as expected), but the server is totally clean. Here is the complete telnet log when I reproduced the error. Mail clients like Evolution and Thunderbird give me the same spamhaus error message. joe@joe-desktop:~$ telnet mail.hunsynth.org 25 Trying 193.164.132.42... Connected to mail.hunsynth.org. Escape character is '^]'. 220 hunsynth.org ESMTP Exim 4.69 Sat, 01 Jan 2011 17:52:45 +0100 HELP 214-Commands supported: 214 AUTH STARTTLS HELO EHLO MAIL RCPT DATA NOOP QUIT RSET HELP EHLO AUTH 250-hunsynth.org Hello chello062178015161.6.11.univie.teleweb.at [62.178.15.161] 250-SIZE 52428800 250-PIPELINING 250-AUTH PLAIN LOGIN CRAM-MD5 250-STARTTLS 250 HELP AUTH LOGIN 334 VXNlcm5hbWU6 dGVzdEBodW5zeW50aC5vcmc= 334 UGFzc3dvcmQ6 ***** 235 Authentication succeeded MAIL FROM:[email protected] 250 OK RCPT TO:[email protected] 550-DNSBL listed at zen.spamhaus.org 550 http://www.spamhaus.org/query/bl?ip=62.178.15.161 quit 221 hunsynth.org closing connection Connection closed by foreign host. joe@joe-desktop:~$ Update: I tried the same thing from my other server and could successfully send an email. So it really looks like the server does check the IP wich establiches the connection is in some blacklist. This is theoretically a good thing - but - the authentication on the server should prevent that? Or shouldn't it? Well I just think it would be absurd if I couldn't send email over my smtp server from my dynamic ISP connection because the dynamic is listed, altough i have a clean server with login?

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  • Performance considerations for common SQL queries

    - by Jim Giercyk
    Originally posted on: http://geekswithblogs.net/NibblesAndBits/archive/2013/10/16/performance-considerations-for-common-sql-queries.aspxSQL offers many different methods to produce the same results.  There is a never-ending debate between SQL developers as to the “best way” or the “most efficient way” to render a result set.  Sometimes these disputes even come to blows….well, I am a lover, not a fighter, so I decided to collect some data that will prove which way is the best and most efficient.  For the queries below, I downloaded the test database from SQLSkills:  http://www.sqlskills.com/sql-server-resources/sql-server-demos/.  There isn’t a lot of data, but enough to prove my point: dbo.member has 10,000 records, and dbo.payment has 15,554.  Our result set contains 6,706 records. The following queries produce an identical result set; the result set contains aggregate payment information for each member who has made more than 1 payment from the dbo.payment table and the first and last name of the member from the dbo.member table.   /*************/ /* Sub Query  */ /*************/ SELECT  a.[Member Number] ,         m.lastname ,         m.firstname ,         a.[Number Of Payments] ,         a.[Average Payment] ,         a.[Total Paid] FROM    ( SELECT    member_no 'Member Number' ,                     AVG(payment_amt) 'Average Payment' ,                     SUM(payment_amt) 'Total Paid' ,                     COUNT(Payment_No) 'Number Of Payments'           FROM      dbo.payment           GROUP BY  member_no           HAVING    COUNT(Payment_No) > 1         ) a         JOIN dbo.member m ON a.[Member Number] = m.member_no         /***************/ /* Cross Apply  */ /***************/ SELECT  ca.[Member Number] ,         m.lastname ,         m.firstname ,         ca.[Number Of Payments] ,         ca.[Average Payment] ,         ca.[Total Paid] FROM    dbo.member m         CROSS APPLY ( SELECT    member_no 'Member Number' ,                                 AVG(payment_amt) 'Average Payment' ,                                 SUM(payment_amt) 'Total Paid' ,                                 COUNT(Payment_No) 'Number Of Payments'                       FROM      dbo.payment                       WHERE     member_no = m.member_no                       GROUP BY  member_no                       HAVING    COUNT(Payment_No) > 1                     ) ca /********/                    /* CTEs  */ /********/ ; WITH    Payments           AS ( SELECT   member_no 'Member Number' ,                         AVG(payment_amt) 'Average Payment' ,                         SUM(payment_amt) 'Total Paid' ,                         COUNT(Payment_No) 'Number Of Payments'                FROM     dbo.payment                GROUP BY member_no                HAVING   COUNT(Payment_No) > 1              ),         MemberInfo           AS ( SELECT   p.[Member Number] ,                         m.lastname ,                         m.firstname ,                         p.[Number Of Payments] ,                         p.[Average Payment] ,                         p.[Total Paid]                FROM     dbo.member m                         JOIN Payments p ON m.member_no = p.[Member Number]              )     SELECT  *     FROM    MemberInfo /************************/ /* SELECT with Grouping   */ /************************/ SELECT  p.member_no 'Member Number' ,         m.lastname ,         m.firstname ,         COUNT(Payment_No) 'Number Of Payments' ,         AVG(payment_amt) 'Average Payment' ,         SUM(payment_amt) 'Total Paid' FROM    dbo.payment p         JOIN dbo.member m ON m.member_no = p.member_no GROUP BY p.member_no ,         m.lastname ,         m.firstname HAVING  COUNT(Payment_No) > 1   We can see what is going on in SQL’s brain by looking at the execution plan.  The Execution Plan will demonstrate which steps and in what order SQL executes those steps, and what percentage of batch time each query takes.  SO….if I execute all 4 of these queries in a single batch, I will get an idea of the relative time SQL takes to execute them, and how it renders the Execution Plan.  We can settle this once and for all.  Here is what SQL did with these queries:   Not only did the queries take the same amount of time to execute, SQL generated the same Execution Plan for each of them.  Everybody is right…..I guess we can all finally go to lunch together!  But wait a second, I may not be a fighter, but I AM an instigator.     Let’s see how a table variable stacks up.  Here is the code I executed: /********************/ /*  Table Variable  */ /********************/ DECLARE @AggregateTable TABLE     (       member_no INT ,       AveragePayment MONEY ,       TotalPaid MONEY ,       NumberOfPayments MONEY     ) INSERT  @AggregateTable         SELECT  member_no 'Member Number' ,                 AVG(payment_amt) 'Average Payment' ,                 SUM(payment_amt) 'Total Paid' ,                 COUNT(Payment_No) 'Number Of Payments'         FROM    dbo.payment         GROUP BY member_no         HAVING  COUNT(Payment_No) > 1   SELECT  at.member_no 'Member Number' ,         m.lastname ,         m.firstname ,         at.NumberOfPayments 'Number Of Payments' ,         at.AveragePayment 'Average Payment' ,         at.TotalPaid 'Total Paid' FROM    @AggregateTable at         JOIN dbo.member m ON m.member_no = at.member_no In the interest of keeping things in groupings of 4, I removed the last query from the previous batch and added the table variable query.  Here’s what I got:     Since we first insert into the table variable, then we read from it, the Execution Plan renders 2 steps.  BUT, the combination of the 2 steps is only 22% of the batch.  It is actually faster than the other methods even though it is treated as 2 separate queries in the Execution Plan.  The argument I often hear against Table Variables is that SQL only estimates 1 row for the table size in the Execution Plan.  While this is true, the estimate does not come in to play until you read from the table variable.  In this case, the table variable had 6,706 rows, but it still outperformed the other queries.  People argue that table variables should only be used for hash or lookup tables.  The fact is, you have control of what you put IN to the variable, so as long as you keep it within reason, these results suggest that a table variable is a viable alternative to sub-queries. If anyone does volume testing on this theory, I would be interested in the results.  My suspicion is that there is a breaking point where efficiency goes down the tubes immediately, and it would be interesting to see where the threshold is. Coding SQL is a matter of style.  If you’ve been around since they introduced DB2, you were probably taught a little differently than a recent computer science graduate.  If you have a company standard, I strongly recommend you follow it.    If you do not have a standard, generally speaking, there is no right or wrong answer when talking about the efficiency of these types of queries, and certainly no hard-and-fast rule.  Volume and infrastructure will dictate a lot when it comes to performance, so your results may vary in your environment.  Download the database and try it!

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  • Welcome to the Weblog on Oracle ADF Mobile!

    - by joe.huang
    Welcome to ADF Mobile team's weblog.  My name is Joe Huang - I am the product manager for ADF Mobile.  Oracle ADF Mobile is a part of Oracle's Application Development Framework (ADF) that support the development of enterprise/business applications that run on mobile devices.  The development tool for this framework is of course Oracle JDeveloper.  As some of you may know, we currently support the development of mobile browser-based application - this part of product is called ADF Mobile Browser.  Additionally, we are close to release a technology preview of ADF Mobile Client, which supports development of on-device, disconnect capable mobile applications.  What's truly unique about ADF Mobile development process is that it's a very visual and declarative experience, while still allow power Java developers to completely extend the framework to their liking.  The framework also provides a rich set of services needed by an enterprise-grade mobile application - these services would literally take years to implement if they are to be built from the ground up.  However, by using JDeveloper and ADF Mobile, you get the entire framework at your service!In the coming entries, the ADF Mobile product development team will publish any news, best practices, our observation on mobile technology trends, or just our experiences in playing with "gadgets".  Be sure to check back on this page!Sincerely,Joe HuangOracle

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  • ADF Mobile Client is now Generally Available!

    - by joe.huang
    ADF Mobile Client is now generally available!  The press release went out this morning, and the ADF Mobile Client extensions can now be downloaded in the JDeveloper Update Center.  There is also a new Oracle Mobile Computing Strategy White Paper and Data Sheet available, for a high level overview of ADF Mobile. To get started with ADF Mobile Client development, please leverage the following resources: Oracle Technology Network ADF Mobile Landing Page: Review this page for all available resources for ADF Mobile development. Getting Start with ADF Mobile Client Demo: Short demo of the end-to-end development process. Tutorial for Mobile Application Development using ADF Mobile Client ADF Mobile Client Developer Guide ADF Mobile Client Samples: available in the JDeveloper Extension itself.  Located in <JDeveloper Install Location>/jdev/extensions/oracle.adfnmc.core/Samples directory.  Blogs will follow, describing each of the sample applications in more detail. Oracle Database Mobile Server: If database synchronization is needed, please follow this link to download/install Mobile Server. Leverage JDeveloper Forum for any ADF Mobile related questions. You will need the latest (11g Patch Set 3, or 11.1.1.4.0) version of JDeveloper to use this extension.  To download the ADF Mobile Client extension in JDeveloper, you would go to Help Menu, select “Check For Update”, and look for ADF Mobile Client extension in the Official Oracle Extensions and Updates center.  You can also directly download the extension from Oracle Technology Network. Check it out!  For any issues with accessing any of the links above, please contact me directly. Thanks, Joe Huang ([email protected])

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  • How to avoid Cartesian product in an INNER JOIN query?

    - by flhe
    I have 6 tables, let's call them a,b,c,d,e,f. Now I want to search all the colums (except the ID columns) of all tables for a certain word, let's say 'Joe'. What I did was, I made INNER JOINS over all the tables and then used LIKE to search the columns. INNER JOIN ... ON INNER JOIN ... ON.......etc. WHERE a.firstname ~* 'Joe' OR a.lastname ~* 'Joe' OR b.favorite_food ~* 'Joe' OR c.job ~* 'Joe'.......etc. The results are correct, I get all the colums I was looking for. But I also get some kind of cartesian product, I get 2 or more lines with almost the same results. How can i avoid this? I want so have each line only once, since the results should appear on a web search.

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  • What container type provides better (average) performance than std::map?

    - by Truncheon
    In the following example a std::map structure is filled with 26 values from A - Z (for key) and 0 - 26 for value. The time taken (on my system) to lookup the last entry (10000000 times) is roughly 250 ms for the vector, and 125 ms for the map. (I compiled using release mode, with O3 option turned on for g++ 4.4) But if for some odd reason I wanted better performance than the std::map, what data structures and functions would I need to consider using? I apologize if the answer seems obvious to you, but I haven't had much experience in the performance critical aspects of C++ programming. UPDATE: This example is rather trivial and hides the true complexity of what I'm trying to achieve. My real world project is a simple scripting language that uses a parser, data tree, and interpreter (instead of a VM stack system). I need to use some kind of data structure (perhaps map) to store the variables names created by script programmers. These are likely to be pretty randomly named, so I need a lookup method that can quickly find a particular key within a (probably) fairly large list of names. #include <ctime> #include <map> #include <vector> #include <iostream> struct mystruct { char key; int value; mystruct(char k = 0, int v = 0) : key(k), value(v) { } }; int find(const std::vector<mystruct>& ref, char key) { for (std::vector<mystruct>::const_iterator i = ref.begin(); i != ref.end(); ++i) if (i->key == key) return i->value; return -1; } int main() { std::map<char, int> mymap; std::vector<mystruct> myvec; for (int i = 'a'; i < 'a' + 26; ++i) { mymap[i] = i - 'a'; myvec.push_back(mystruct(i, i - 'a')); } int pre = clock(); for (int i = 0; i < 10000000; ++i) { find(myvec, 'z'); } std::cout << "linear scan: milli " << clock() - pre << "\n"; pre = clock(); for (int i = 0; i < 10000000; ++i) { mymap['z']; } std::cout << "map scan: milli " << clock() - pre << "\n"; return 0; }

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  • Updating a specific key/value inside of an array field with MongoDB

    - by Jesta
    As a preface, I've been working with MongoDB for about a week now, so this may turn out to be a pretty simple answer. I have data already stored in my collection, we will call this collection content, as it contains articles, news, etc. Each of these articles contains another array called author which has all of the author's information (Address, Phone, Title, etc). The Goal - I am trying to create a query that will update the author's address on every article that the specific author exists in, and only the specified author block (not others that exist within the array). Sort of a "Global Update" to a specific article that affects his/her information on every piece of content that exists. Here is an example of what the content with the author looks like. { "_id" : ObjectId("4c1a5a948ead0e4d09010000"), "authors" : [ { "user_id" : null, "slug" : "joe-somebody", "display_name" : "Joe Somebody", "display_title" : "Contributing Writer", "display_company_name" : null, "email" : null, "phone" : null, "fax" : null, "address" : null, "address2" : null, "city" : null, "state" : null, "zip" : null, "country" : null, "image" : null, "url" : null, "blurb" : null }, { "user_id" : null, "slug" : "jane-somebody", "display_name" : "Jane Somebody", "display_title" : "Editor", "display_company_name" : null, "email" : null, "phone" : null, "fax" : null, "address" : null, "address2" : null, "city" : null, "state" : null, "zip" : null, "country" : null, "image" : null, "url" : null, "blurb" : null }, ], "tags" : [ "tag1", "tag2", "tag3" ], "title" : "Title of the Article" } I can find every article that this author has created by running the following command: db.content.find({authors: {$elemMatch: {slug: 'joe-somebody'}}}); So theoretically I should be able to update the authors record for the slug joe-somebody but not jane-somebody (the 2nd author), I am just unsure exactly how you reach in and update every record for that author. I thought I was on the right track, and here's what I've tried. b.content.update( {authors: {$elemMatch: {slug: 'joe-somebody'} } }, {$set: {address: '1234 Avenue Rd.'} }, false, true ); I just believe there's something I am missing in the $set statement to specify the correct author and point inside of the correct array. Any ideas? **Update** I've also tried this now: b.content.update( {authors: {$elemMatch: {slug: 'joe-somebody'} } }, {$set: {'authors.$.address': '1234 Avenue Rd.'} }, false, true );

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  • What are some good usability guidelines an average developer should follow?

    - by Allain Lalonde
    I'm not a usability specialist, and I really don't care to be one. I just want a small set of rules of thumb that I can follow while coding my User Interfaces so that my product has decent usability. At first I thought that this question would be easy to answer "Use your common sense", but if it's so common among us developers we wouldn't, as a group, have a reputation for our horrible interfaces. Any Suggestions?

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  • How to solve High Load average issue in Linux systems?

    - by RoCkStUnNeRs
    The following is the different load with cpu time in different time limit . The below output has parsed from the top command. TIME LOAD US SY NICE ID WA HI SI ST 12:02:27 208.28 4.2%us 1.0%sy 0.2%ni 93.9%id 0.7%wa 0.0%hi 0.0%si 0.0%st 12:23:22 195.48 4.2%us 1.0%sy 0.2%ni 93.9%id 0.7%wa 0.0%hi 0.0%si 0.0%st 12:34:55 199.15 4.2%us 1.0%sy 0.2%ni 93.9%id 0.7%wa 0.0%hi 0.0%si 0.0%st 13:41:50 203.66 4.2%us 1.0%sy 0.2%ni 93.8%id 0.8%wa 0.0%hi 0.0%si 0.0%st 13:42:58 278.63 4.2%us 1.0%sy 0.2%ni 93.8%id 0.8%wa 0.0%hi 0.0%si 0.0%st Following is the additional Information of the system? cat /proc/cpuinfo processor : 0 vendor_id : GenuineIntel cpu family : 6 model : 23 model name : Intel(R) Xeon(R) CPU E5410 @ 2.33GHz stepping : 10 cpu MHz : 1992.000 cache size : 6144 KB physical id : 0 siblings : 4 core id : 0 cpu cores : 4 apicid : 0 initial apicid : 0 fdiv_bug : no hlt_bug : no f00f_bug : no coma_bug : no fpu : yes fpu_exception : yes cpuid level : 13 wp : yes flags : fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov pat pse36 clflush dts acpi mmx fxsr sse sse2 ss ht tm pbe lm constant_tsc arch_perfmon pebs bts pni monitor ds_cpl vmx est tm2 ssse3 cx16 xtpr dca sse4_1 lahf_lm bogomips : 4658.69 clflush size : 64 power management: processor : 1 vendor_id : GenuineIntel cpu family : 6 model : 23 model name : Intel(R) Xeon(R) CPU E5410 @ 2.33GHz stepping : 10 cpu MHz : 1992.000 cache size : 6144 KB physical id : 0 siblings : 4 core id : 1 cpu cores : 4 apicid : 1 initial apicid : 1 fdiv_bug : no hlt_bug : no f00f_bug : no coma_bug : no fpu : yes fpu_exception : yes cpuid level : 13 wp : yes flags : fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov pat pse36 clflush dts acpi mmx fxsr sse sse2 ss ht tm pbe lm constant_tsc arch_perfmon pebs bts pni monitor ds_cpl vmx est tm2 ssse3 cx16 xtpr dca sse4_1 lahf_lm bogomips : 4655.00 clflush size : 64 power management: processor : 2 vendor_id : GenuineIntel cpu family : 6 model : 23 model name : Intel(R) Xeon(R) CPU E5410 @ 2.33GHz stepping : 10 cpu MHz : 1992.000 cache size : 6144 KB physical id : 0 siblings : 4 core id : 2 cpu cores : 4 apicid : 2 initial apicid : 2 fdiv_bug : no hlt_bug : no f00f_bug : no coma_bug : no fpu : yes fpu_exception : yes cpuid level : 13 wp : yes flags : fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov pat pse36 clflush dts acpi mmx fxsr sse sse2 ss ht tm pbe lm constant_tsc arch_perfmon pebs bts pni monitor ds_cpl vmx est tm2 ssse3 cx16 xtpr dca sse4_1 lahf_lm bogomips : 4655.00 clflush size : 64 power management: processor : 3 vendor_id : GenuineIntel cpu family : 6 model : 23 model name : Intel(R) Xeon(R) CPU E5410 @ 2.33GHz stepping : 10 cpu MHz : 1992.000 cache size : 6144 KB physical id : 0 siblings : 4 core id : 3 cpu cores : 4 apicid : 3 initial apicid : 3 fdiv_bug : no hlt_bug : no f00f_bug : no coma_bug : no fpu : yes fpu_exception : yes cpuid level : 13 wp : yes flags : fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov pat pse36 clflush dts acpi mmx fxsr sse sse2 ss ht tm pbe lm constant_tsc arch_perfmon pebs bts pni monitor ds_cpl vmx est tm2 ssse3 cx16 xtpr dca sse4_1 lahf_lm bogomips : 4654.99 clflush size : 64 power management: Memory: total used free shared buffers cached Mem: 2 1 1 0 0 0 Swap: 5 0 5 let me know why the system is getting abnormally this much high load?

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  • getting userbase vote average and individual user's vote in the same query?

    - by Andrew Heath
    Here goes: T1 [id] [desc] 1 lovely 2 ugly 3 slender T2 [id] [userid] [vote] 1 1 3 1 2 5 1 3 2 2 1 1 2 2 4 2 3 4 In one query (if possible) I'd like to return: T1.id, T1.desc, AVG(T2.vote), T2.vote (for user viewing the page) I can get the first 3 items with: SELECT T1.id, T1.desc, AVG(T2.vote) FROM T1 LEFT JOIN T2 ON T1.id=T2.id GROUP BY T1.id and I can get the first, second, and fourth items with: SELECT T1.id, T1.desc, T2.vote FROM T1 LEFT JOIN T2 ON T1.id=T2.id WHERE T2.userid='1' GROUP BY T1.id but I'm at a loss as to how to get all four items in one query. I tried inserting a select as the fourth term: SELECT T1.id, T1.desc, AVG(T2.vote), (SELECT T2.vote FROM T2 WHERE T2.userid='1') AS userVote etc etc but I get an error that the select returns more than one row... Help? My reason for wanting to do this in one query instead of two is that I want to be able to sort the data within MySQL rather than one it's been split into a number of arrays.

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