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  • What's New with PeopleTools

    Michael Ni, Vice President of Product Strategy at Oracle, goes through with Cliff the PeopleTools Roadmap that was outlined at OOW, discusses what's in the latest release of PeopleTools and gives an overview of the new features that will be in the next few releases of PeopleTools. Michael also comments to Cliff about the next generation of applications, Fusion Architecture and the Total Ownership Experience (TOE) across the lifecycle of the applications.

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  • Create Ubuntu repository on CentOS server with debmirror

    - by Wilco Groothand
    I want to create a UBUNTU repo mirror on my CentOS reposerver. I read about it and came to the conclusion that for our purpose debmirror was the correct solution, because I then could mirror a subset of the total repository. The problem is that with debmirror I run into the gpg key errors. Already solved in Ubuntu, but the apt-key solutions are not valid within CentOS. The command does not exists. I am totally stuck.

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  • About Google Adwords Ads

    Google AdWords is the flagship advertising model of Google and also the driving force behind its revenue. The total advertising revenues that Google earns in every fiscal year is completely because o... [Author: John Anthony - Computers and Internet - May 03, 2010]

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  • Please recommend a good book on general IT for junior developer

    - by Rachel
    I have just got a job as a junior java developer and I am finding learning the code fine as I taught myself to pass the interview tests, trouble is I have a biology degree and know little about general IT and computer /network jargon. I don't want to be seen as a total idiot and non - techie so please can you recommend a book I can read / study to quickly increase my general background knowledge so that I can compete on a level with IT graduates!

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  • Indie Software Developers - How do I handle taxes?

    - by Connor
    I apologize if this is the wrong site to post on, perhaps someone could point me to the proper place if it is not. Hello, I am 17 years old and currently develop applications/games for Android and iPhone as well as develop internet websites and code a variety of my own projects. I have been very fortunate and have made a large amount of money and continue to make money online to the point where I do not need a stable job, though I'd like to get one after college. I've never held a job anywhere, and have never had to pay taxes. I'm coming into a lot of issues and I am quite confused. I get money from MANY sources- 15 different advertisement networks(!), 4 different payment processors, 5 different affiliate networks and a variety of other sources. All of them pay to different places and at different times (checking account, PayPal, reloadable debit card, ect.) I essentially have a list in a Notepad with names and login information for each source. I have also created a PHP script that uses cURL to grab all the revenue from each service, add it all up, then text me every few hours so I can keep track. It's a mess, but it's working OK, and I can create custom reports (for IRS?). But enough of that, my questions are about taxes in the US, and how indie developers handle it all. I'm at slightly over $250k so far this year, with negligible earnings last year. I have it all stockpiled in a bank account and haven't touched it, I'm a bit scared to. What do I file as? A sole proprietor, a business, just a regular person? How can I handle all of the different revenue sources? (AdSense, CJ, LinkShare) So far none of them have sent me any paperwork on taxes and I've read that I'm supposed to pay taxes quarterly? Do I need paperwork from EACH source to file? Or can I just say I got $x total and that'd be it? What percentage do you pay of total earnings? Average? Should I create an LLC? A corporation? Or stay as a developer? What would be the cheapest options? Could I go to jail? I haven't touched the money except a few dollars to help my parents pay the mortgage once. Any insight would be great. My parents have no idea what I should do, both have no forms of higher education and both have no high school diploma's. They just live day by day with simple jobs. I appreciate any help or experience with this.

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  • Laser Cutter Plays Portal Theme Song While Cutting Aperture Science Logo [Video]

    - by Jason Fitzpatrick
    We’re at a total loss to think of anything more geeky than a laser cutter programmed to simultaneously laser cut the Aperture Science logo while playing “Stile Alive” from Portal. If you’re unfamiliar with the original tune, check out this video. [via Boing Boing] The HTG Guide to Hiding Your Data in a TrueCrypt Hidden Volume Make Your Own Windows 8 Start Button with Zero Memory Usage Reader Request: How To Repair Blurry Photos

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  • Equal Gifts Algorithm Problem

    - by 7Aces
    Problem Link - http://opc.iarcs.org.in/index.php/problems/EQGIFTS It is Lavanya's birthday and several families have been invited for the birthday party. As is customary, all of them have brought gifts for Lavanya as well as her brother Nikhil. Since their friends are all of the erudite kind, everyone has brought a pair of books. Unfortunately, the gift givers did not clearly indicate which book in the pair is for Lavanya and which one is for Nikhil. Now it is up to their father to divide up these books between them. He has decided that from each of these pairs, one book will go to Lavanya and one to Nikhil. Moreover, since Nikhil is quite a keen observer of the value of gifts, the books have to be divided in such a manner that the total value of the books for Lavanya is as close as possible to total value of the books for Nikhil. Since Lavanya and Nikhil are kids, no book that has been gifted will have a value higher than 300 Rupees... For the problem, I couldn't think of anything except recursion. The code I wrote is given below. But the problem is that the code is time-inefficient and gives TLE (Time Limit Exceeded) for 9 out of 10 test cases! What would be a better approach to the problem? Code - #include<cstdio> #include<climits> #include<algorithm> using namespace std; int n,g[150][2]; int diff(int a,int b,int f) { ++f; if(f==n) { if(a>b) { return a-b; } else { return b-a; } } return min(diff(a+g[f][0],b+g[f][1],f),diff(a+g[f][1],b+g[f][0],f)); } int main() { int i; scanf("%d",&n); for(i=0;i<n;++i) { scanf("%d%d",&g[i][0],&g[i][1]); } printf("%d",diff(g[0][0],g[0][1],0)); } Note - It is just a practice question, & is not part of a competition.

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  • SCRIPT REFERENCE PROFILER TO GET JAVASCRIPT REFERENCE DETAILS

    Many of us came through a scenario like, you need the details of total JS files referred to a web page. It's not a matter if all JS are directly referring through physical file. But now a days for cache purose we are embedding JS to the assembly. In these situations how you will get the JS reference details? Here coming ScriptReferenceProfiler.

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  • Why no fortran standard library ?

    - by Stefano Borini
    To be a language focused on mathematics and scientific computing, I am always baffled by the total lack of useful mathematical routines in the Fortran standard library. One would expect it to be shipped at least with a routine to compute standard deviation and mean, but this is not the case. In particular with the introduction of Fortran 90 and the addition of modules (thus reducing namespace pollution), I don't see any reason why of this critical lack of services. I would like to hear your knowledge about why this is the case.

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  • Google Analytics - Showing multiple site stats at once

    - by John
    Is there a way in google analytics to add multiple sites to and show all the stats together? So like the graphs and total visits/unique hits all combined for all the sites added to the google analytics account? For example if I have: site1.com site2.com site3.com Under one google analytics account, is there a way in google analytics tool to merge them together so I can see a sum of all traffic in one report?

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  • Shopping Cart URL Structure

    - by Drew
    In regards to URL structure when it comes to guests and authenticated users, am I able to track traffic associated with both paths, but at the same time track total conversions going through the shopping cart? I have set up the following URL structure: Authenticated users follow this path: /cart /checkout /checkout-confirmation-ty Guests go like such: /cart /checkout-guest /checkout-confirmation-guest-ty can I track the authenticated and guest users separately? is this possible with Google Analytics?

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  • MySQL Enterprise Backup 3.8.2 - Overview

    - by Priya Jayakumar
      MySQL Enterprise Backup (MEB) is the ideal solution for backing up MySQL databases. MEB 3.8.2 is released in June 2013. MySQL Enterprise Backup 3.8.2 release’s main goal is to improve usability. With this release, users can know the progress of backup completed both in terms of size and as a percentage of the total. This release also offers options to be able to manage the behavior of MEB in case the space on the secondary storage is completely exhausted during backup. The progress indicator is a (short) string that indicates how far the execution of a time-consuming MEB command has progressed. It consists of one or more "meters" that measures the progress of the command. There are two options introduced to control the progress reporting function of mysqlbackup command (1) –show-progress (2) –progress-interval. The user can control the progress indicator by using “--show-progress” option in any of the MEB operations. This option instructs MEB to output periodically short reports on the progress of time-consuming commands. The argument of this option instructs where the output could be sent. For example it could be stderr, stdout, file, fifo and table. With the “--show-progress” option both the total size of the backup to be copied and the size that’s already copied will be shown. Along with this, the state of the operation for example data or meta-data being copied or tables being locked and other such operations will also be reported. This gives more clear information to the DBA on the progress of the backup that’s happening. Interval between progress report in seconds is controlled by “--progress-interval” option. For more information on this please refer progress-report-options. MEB can also be accessed through GUI from MySQL WorkBench’s next version. This can be used as the front end interface for MEB users to perform backup operations at the click of a button. This feature was highly requested by DBAs and will be very useful. Refer http://insidemysql.com/mysql-workbench-6-0-a-sneak-preview/ for WorkBench upcoming release info. Along with the progress report feature some of the important issues like below are also addressed in MEB 3.8.2. In MEB 3.8.2 a new command line option “--on-disk-full” is introduced to abort or warn the user when a backup process encounters a full disk condition. When no option is given, by default it would abort. A few issues related to “incremental-backup” are also addressed in this release. Please refer 3.8.2 documentation for more details. It would be good for MEB users to move to 3.8.2 to take incremental backups. Overall the added usability and the important defects fixed in this release makes MySQL Enterprise Backup 3.8.2 a promising release.  

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  • GuestPost: Announcing gmStudio V9.85 for VB6/ASP/COM re-engineering

    - by Eric Nelson
    Mark Juras of GreatMigrations.com kindly sent me an article on gmStudio which I have posted on my old VB focused goto100 site. gmStudio is a programmable VB6/ASP/COM re-engineering tool that enables an agile tool-assisted rewrite methodology and helps teams dramatically lower the total cost, risk, and disruption of ambitious migration projects without sacrificing quality, control, or time to market. You can find the rest of the article over on goto100. Figure 1: the gmStudio Main Form

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  • PowerPivot, Parent/Child and Unary Operators

    - by AlbertoFerrari
    Following my last post about parent/child hierarchies in PowerPivot, I worked a bit more to implement a very useful feature of Parent/Child hierarchies in SSAS which is obviously missing in PowerPivot, i.e. unary operators. A unary operator is simply the aggregation function that needs to be used to aggregate values of children over their parent. Unary operators are very useful in accountings where you might have incomes and expenses in the same hierarchy and, at the total level, you want to subtract...(read more)

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  • How to Map a CSV or Tab Delimited File to MySQL Multi-Table Database [migrated]

    - by Keefer
    I've got a pretty substantial XLS file a client provided 830 total tabs/sheets. I've designed a multi table database with PHPMyAdmin (MySQL obviously) to house the information that's in there, and have populated about 5 of those sheets by hand to ensure the data will fit into the designed database. Is there a piece of software or some sort of tool that will help me format this XLS document and map it to the right places in the database?

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  • Ideas for attack damage algorithm (language irrelevant)

    - by Dillon
    I am working on a game and I need ideas for the damage that will be done to the enemy when your player attacks. The total amount of health that the enemy has is called enemyHealth, and has a value of 1000. You start off with a weapon that does 40 points of damage (may be changed.) The player has an attack stat that you can increase, called playerAttack. This value starts off at 1, and has a possible max value of 100 after you level it up many times and make it farther into the game. The amount of damage that the weapon does is cut and dry, and subtracts 40 points from the total 1000 points of health every time the enemy is hit. But what the playerAttack does is add to that value with a percentage. Here is the algorithm I have now. (I've taken out all of the gui, classes, etc. and given the variables very forward names) double totalDamage = weaponDamage + (weaponDamage*(playerAttack*.05)) enemyHealth -= (int)totalDamage; This seemed to work great for the most part. So I statrted testing some values... //enemyHealth ALWAYS starts at 1000 weaponDamage = 50; playerAttack = 30; If I set these values, the amount of damage done on the enemy is 125. Seemed like a good number, so I wanted to see what would happen if the players attack was maxed out, but with the weakest starting weapon. weaponDamage = 50; playerAttack = 100; the totalDamage ends up being 300, which would kill an enemy in just a few hits. Even with your attack that high, I wouldn't want the weakest weapon to be able to kill the enemy that fast. I thought about adding defense, but I feel the game will lose consistency and become unbalanced in the long run. Possibly a well designed algorithm for a weapon decrease modifier would work for lower level weapons or something like that. Just need a break from trying to figure out the best way to go about this, and maybe someone that has experience with games and keeping the leveling consistent could give me some ideas/pointers.

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  • passwordless ssh not working

    - by kuurious
    I've tried to setup a password-less ssh b/w A to B and B to A as well. Generated the public and private key using ssh-keygen -trsa on both the machines. Used the ssh-copy-id utility to copy the public-keys from A to B as well as B to A. The passwordless ssh works from A to B but not from B to A. I've checked the permissions of the ~/ssh/ folder and seems to be normal. A's .ssh folder permissions: -rw------- 1 root root 13530 2011-07-26 23:00 known_hosts -rw------- 1 root root 403 2011-07-27 00:35 id_rsa.pub -rw------- 1 root root 1675 2011-07-27 00:35 id_rsa -rw------- 1 root root 799 2011-07-27 00:37 authorized_keys drwxrwx--- 70 root root 4096 2011-07-27 00:37 .. drwx------ 2 root root 4096 2011-07-27 00:38 . B's .ssh folder permissions: -rw------- 1 root root 884 2011-07-07 13:15 known_hosts -rw-r--r-- 1 root root 396 2011-07-27 00:15 id_rsa.pub -rw------- 1 root root 1675 2011-07-27 00:15 id_rsa -rw------- 1 root root 2545 2011-07-27 00:36 authorized_keys drwxr-xr-x 8 root root 4096 2011-07-06 19:44 .. drwx------ 2 root root 4096 2011-07-27 00:15 . A is an ubuntu 10.04 (OpenSSH_5.3p1 Debian-3ubuntu4, OpenSSL 0.9.8k 25 Mar 2009) B is a debian machine (OpenSSH_5.1p1 Debian-5, OpenSSL 0.9.8g 19 Oct 2007) From A: #ssh B works fine. From B: #ssh -vvv A ... ... debug1: SSH2_MSG_SERVICE_ACCEPT received debug2: key: /root/.ssh/identity ((nil)) debug2: key: /root/.ssh/id_rsa (0x7f1581f23a50) debug2: key: /root/.ssh/id_dsa ((nil)) debug3: Wrote 64 bytes for a total of 1127 debug1: Authentications that can continue: publickey,password debug3: start over, passed a different list publickey,password debug3: preferred gssapi-keyex,gssapi-with-mic,gssapi,publickey,keyboard-interactive,password debug3: authmethod_lookup publickey debug3: remaining preferred: keyboard-interactive,password debug3: authmethod_is_enabled publickey debug1: Next authentication method: publickey debug1: Trying private key: /root/.ssh/identity debug3: no such identity: /root/.ssh/identity debug1: Offering public key: /root/.ssh/id_rsa debug3: send_pubkey_test debug2: we sent a publickey packet, wait for reply debug3: Wrote 368 bytes for a total of 1495 debug1: Authentications that can continue: publickey,password debug1: Trying private key: /root/.ssh/id_dsa debug3: no such identity: /root/.ssh/id_dsa debug2: we did not send a packet, disable method debug3: authmethod_lookup password debug3: remaining preferred: ,password debug3: authmethod_is_enabled password debug1: Next authentication method: password [email protected]'s password: Which essentially means it's not authenticating using the file /root/id_rsa. I ran the ssh-add command in both the machines as well. The authentication part of /etc/ssh/sshd_config file is # Authentication: LoginGraceTime 120 PermitRootLogin yes StrictModes yes RSAAuthentication yes PubkeyAuthentication yes #AuthorizedKeysFile %h/.ssh/authorized_keys # Don't read the user's ~/.rhosts and ~/.shosts files I'm running out of ideas. Any help would be appreciated.

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  • When is a Seek not a Seek?

    - by Paul White
    The following script creates a single-column clustered table containing the integers from 1 to 1,000 inclusive. IF OBJECT_ID(N'tempdb..#Test', N'U') IS NOT NULL DROP TABLE #Test ; GO CREATE TABLE #Test ( id INTEGER PRIMARY KEY CLUSTERED ); ; INSERT #Test (id) SELECT V.number FROM master.dbo.spt_values AS V WHERE V.[type] = N'P' AND V.number BETWEEN 1 AND 1000 ; Let’s say we need to find the rows with values from 100 to 170, excluding any values that divide exactly by 10.  One way to write that query would be: SELECT T.id FROM #Test AS T WHERE T.id IN ( 101,102,103,104,105,106,107,108,109, 111,112,113,114,115,116,117,118,119, 121,122,123,124,125,126,127,128,129, 131,132,133,134,135,136,137,138,139, 141,142,143,144,145,146,147,148,149, 151,152,153,154,155,156,157,158,159, 161,162,163,164,165,166,167,168,169 ) ; That query produces a pretty efficient-looking query plan: Knowing that the source column is defined as an INTEGER, we could also express the query this way: SELECT T.id FROM #Test AS T WHERE T.id >= 101 AND T.id <= 169 AND T.id % 10 > 0 ; We get a similar-looking plan: If you look closely, you might notice that the line connecting the two icons is a little thinner than before.  The first query is estimated to produce 61.9167 rows – very close to the 63 rows we know the query will return.  The second query presents a tougher challenge for SQL Server because it doesn’t know how to predict the selectivity of the modulo expression (T.id % 10 > 0).  Without that last line, the second query is estimated to produce 68.1667 rows – a slight overestimate.  Adding the opaque modulo expression results in SQL Server guessing at the selectivity.  As you may know, the selectivity guess for a greater-than operation is 30%, so the final estimate is 30% of 68.1667, which comes to 20.45 rows. The second difference is that the Clustered Index Seek is costed at 99% of the estimated total for the statement.  For some reason, the final SELECT operator is assigned a small cost of 0.0000484 units; I have absolutely no idea why this is so, or what it models.  Nevertheless, we can compare the total cost for both queries: the first one comes in at 0.0033501 units, and the second at 0.0034054.  The important point is that the second query is costed very slightly higher than the first, even though it is expected to produce many fewer rows (20.45 versus 61.9167). If you run the two queries, they produce exactly the same results, and both complete so quickly that it is impossible to measure CPU usage for a single execution.  We can, however, compare the I/O statistics for a single run by running the queries with STATISTICS IO ON: Table '#Test'. Scan count 63, logical reads 126, physical reads 0. Table '#Test'. Scan count 01, logical reads 002, physical reads 0. The query with the IN list uses 126 logical reads (and has a ‘scan count’ of 63), while the second query form completes with just 2 logical reads (and a ‘scan count’ of 1).  It is no coincidence that 126 = 63 * 2, by the way.  It is almost as if the first query is doing 63 seeks, compared to one for the second query. In fact, that is exactly what it is doing.  There is no indication of this in the graphical plan, or the tool-tip that appears when you hover your mouse over the Clustered Index Seek icon.  To see the 63 seek operations, you have click on the Seek icon and look in the Properties window (press F4, or right-click and choose from the menu): The Seek Predicates list shows a total of 63 seek operations – one for each of the values from the IN list contained in the first query.  I have expanded the first seek node to show the details; it is seeking down the clustered index to find the entry with the value 101.  Each of the other 62 nodes expands similarly, and the same information is contained (even more verbosely) in the XML form of the plan. Each of the 63 seek operations starts at the root of the clustered index B-tree and navigates down to the leaf page that contains the sought key value.  Our table is just large enough to need a separate root page, so each seek incurs 2 logical reads (one for the root, and one for the leaf).  We can see the index depth using the INDEXPROPERTY function, or by using the a DMV: SELECT S.index_type_desc, S.index_depth FROM sys.dm_db_index_physical_stats ( DB_ID(N'tempdb'), OBJECT_ID(N'tempdb..#Test', N'U'), 1, 1, DEFAULT ) AS S ; Let’s look now at the Properties window when the Clustered Index Seek from the second query is selected: There is just one seek operation, which starts at the root of the index and navigates the B-tree looking for the first key that matches the Start range condition (id >= 101).  It then continues to read records at the leaf level of the index (following links between leaf-level pages if necessary) until it finds a row that does not meet the End range condition (id <= 169).  Every row that meets the seek range condition is also tested against the Residual Predicate highlighted above (id % 10 > 0), and is only returned if it matches that as well. You will not be surprised that the single seek (with a range scan and residual predicate) is much more efficient than 63 singleton seeks.  It is not 63 times more efficient (as the logical reads comparison would suggest), but it is around three times faster.  Let’s run both query forms 10,000 times and measure the elapsed time: DECLARE @i INTEGER, @n INTEGER = 10000, @s DATETIME = GETDATE() ; SET NOCOUNT ON; SET STATISTICS XML OFF; ; WHILE @n > 0 BEGIN SELECT @i = T.id FROM #Test AS T WHERE T.id IN ( 101,102,103,104,105,106,107,108,109, 111,112,113,114,115,116,117,118,119, 121,122,123,124,125,126,127,128,129, 131,132,133,134,135,136,137,138,139, 141,142,143,144,145,146,147,148,149, 151,152,153,154,155,156,157,158,159, 161,162,163,164,165,166,167,168,169 ) ; SET @n -= 1; END ; PRINT DATEDIFF(MILLISECOND, @s, GETDATE()) ; GO DECLARE @i INTEGER, @n INTEGER = 10000, @s DATETIME = GETDATE() ; SET NOCOUNT ON ; WHILE @n > 0 BEGIN SELECT @i = T.id FROM #Test AS T WHERE T.id >= 101 AND T.id <= 169 AND T.id % 10 > 0 ; SET @n -= 1; END ; PRINT DATEDIFF(MILLISECOND, @s, GETDATE()) ; On my laptop, running SQL Server 2008 build 4272 (SP2 CU2), the IN form of the query takes around 830ms and the range query about 300ms.  The main point of this post is not performance, however – it is meant as an introduction to the next few parts in this mini-series that will continue to explore scans and seeks in detail. When is a seek not a seek?  When it is 63 seeks © Paul White 2011 email: [email protected] twitter: @SQL_kiwi

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  • Using Url Rewrite to Block Page Requests

    - by The Official Microsoft IIS Site
    The other day I was checking the traffic stats for my WordPress blog to see which of my posts were the most popular. I was a little concerned to see that wp-login.php was in the Top 5 total requests almost every month. Since I’m the only author on my blog my logins could not possibly account for the traffic hitting that page. The only explanation could be that the additional traffic was coming from automated hacking attempts. Any server administrator concerned about security knows that “ footprinting...(read more)

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  • MPEG-2 playback inconsistent

    - by DustByte
    Many years ago I gave up on Linux because video playback was choppy. Now I'm back, and video playback is still playing up... I have two MPEG files: good.mpg bad.mpg. Here is some information about the two files, using avprobe: My machine is Intel Core 2 Duo E8400 @ 3.00GHz x 2, 64-bit. I do not know what graphics card I have. I run Ubuntu 12.04. So far I have had no problems with YouTube and playback of various video files, including playback of the file good.mpg, included in the avprobe snapshot above. However, the file bad.mpg gives me headache! The file bad.mpg is produced by a respectable "Old-video-tapes-to-DVD" company. I converted over 10 Video-8 tapes to MPEG through them, and today I collected my hard drive containing the MPEG files. Unfortunately I have problem watching them! Here are some details: Using Totem Movie Player 3.0.1 works well for several seconds, then it gets choppy and the playback is not at all smooth. Also the player easily freezes for a while when trying to jump to another position in the file. Most strangely though, the total time is shown as 0:42 (42 seconds) instead of the true 00:39:11: The VLC media player is doing a better job. It shows the correct total length, but as soon as I jump in the video to a new position, it stalls. Playback also stalls after 30 seconds if I press play and leave it. Using Handbrake and choosing bad.mpg as the source, gives me: There is only one title to choose, and it is 6 min 53 seconds. I would have guessed the full 39 minutes of the video should have shown. Lastly, putting the file bad.mpg in Dropbox and viewing it on my iPad with the Dropbox app seems fine (disregard the lack of easy jumping forward due to real-time encoding when streaming it). My question is simple: What is going on?! Why do I have problem to play the MPEG-2 files I just paid good money for (the issue with bad.mpg applies to all files I had encoded)? Is it an issue with my particular Linux machine? The graphics card? But why has everything worked fine so far, and why does not the good.mpg file cause any problems?

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  • Mix metrics for June 14, 2010

    - by tim.bonnemann
    We've been busy working on a few improvements to Mix which we plan to roll out over the coming weeks. In the meantime, here are our latest community metrics once again: Registered Mix users (weekly growth) 64,769 (+0.9%) Active users (percent of total) Last 30 days: 4,682 (7.2%) Last 60 days: 8,251 (12.7%) Last 90 days: 11,936 (18.4%) Traffic (30-day) Visits: 13,674 Page views: 77,808 Twitter Followers: 3,451 List mentions: 205 User-generated content (30-day) New ideas: 29 New questions: 38 New comments: 167 Groups There are currently 1,440 Mix groups (requires login).

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  • Friday Fun: Mad Virus

    - by Asian Angel
    In this week’s game infection of all cell-kind is the ultimate goal as you lead your virus army to victory. Will you succeed in infecting everything in your path or will you be stopped just short of total domination? HTG Explains: Learn How Websites Are Tracking You Online Here’s How to Download Windows 8 Release Preview Right Now HTG Explains: Why Linux Doesn’t Need Defragmenting

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  • Ubuntu 11.10 doesn't detect external usb hard drive

    - by Andrew
    I have been batting with this issue for a bit and cannot find the answer to it. So the Dmesg see's the device, being Symwave WDC WD64.... media@Media-pc:~$ dmesg | tail -n 20 [78678.719497] scsi 10:0:0:0: Direct-Access Generic- USB3.0 CRW -0 1.00 PQ: 0 ANSI: 0 CCS [78678.725621] scsi 10:0:0:1: Direct-Access Generic- USB3.0 CRW -1 1.00 PQ: 0 ANSI: 0 CCS [78684.073837] scsi 11:0:0:0: Direct-Access SYMWAVE WDC WD6400AAKS-0 3B01 PQ: 0 ANSI: 4 [78691.008126] scsi 11:0:0:0: uas_eh_abort_handler tag 0 [78691.008139] scsi 11:0:0:0: uas_eh_device_reset_handler tag 0 [78691.008147] scsi 11:0:0:0: uas_eh_target_reset_handler tag 0 [78691.008154] scsi 11:0:0:0: uas_eh_bus_reset_handler tag 0 [78691.080307] usb 2-2.4: reset high speed USB device number 9 using ehci_hcd [78691.221427] scsi 11:0:0:0: Device offlined - not ready after error recovery [78691.221498] scsi 11:0:0:0: rejecting I/O to offline device [78691.221519] scsi 11:0:0:0: rejecting I/O to offline device [78691.222952] scsi 11:0:0:1: Enclosure SYMWAVE SES 3B01 PQ: 0 ANSI: 4 [78691.223156] scsi 11:0:0:2: uas_sense_old: urb length 26 disagrees with IU sense data length 510, using 18 bytes of sense data [78691.225061] sd 11:0:0:0: Attached scsi generic sg3 type 0 [78691.225344] ses 11:0:0:1: Attached Enclosure device [78691.225495] ses 11:0:0:1: Attached scsi generic sg4 type 13 [78691.226266] sd 10:0:0:0: Attached scsi generic sg5 type 0 [78691.226653] sd 10:0:0:1: Attached scsi generic sg6 type 0 [78691.241647] sd 10:0:0:0: [sdd] Attached SCSI removable disk [78691.243832] sd 10:0:0:1: [sde] Attached SCSI removable disk It looks like it attaches sdd and sde. Now when i look in the disk utility it shows "Hard disk Symwave WD6400AAKS-0 device /dev/sdc doesn't show any other info then that, if i format, it says that it cannot open /dev/sdc no device or address error. Underneeth the device it does show two general usb3.0 CRW that are sdd and sde. Now if I do a fdisk -l it doesn't show the device: media@Media-pc:~$ sudo fdisk -l Disk /dev/sda: 80.0 GB, 80026361856 bytes 255 heads, 63 sectors/track, 9729 cylinders, total 156301488 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x000247de Device Boot Start End Blocks Id System /dev/sda1 * 2048 152176639 76087296 83 Linux /dev/sda2 152178686 156301311 2061313 5 Extended /dev/sda5 152178688 156301311 2061312 82 Linux swap / Solaris Disk /dev/sdb: 1000.2 GB, 1000204886016 bytes 255 heads, 63 sectors/track, 121601 cylinders, total 1953525168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x948fc822 Device Boot Start End Blocks Id System /dev/sdb1 63 1953520064 976760001 7 HPFS/NTFS/exFAT So now I am confused. Any ideas how I get fdisk to see the device?

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