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  • What does this red icon on my panel mean?

    - by Ceil
    This red icon.... It showed up after I upgraded to Ubuntu 12.04LTS; I can't figure out how to ditch it. It seems to be a notification thing. I am at a loss. The icon is here: I clicked it and read this: An error occurred, please run Package Manager from the right-click menu or apt-get in a terminal to see what is wrong. The error message was: 'Error: BrokenCount0'. This usually means that your installed packages have unset dependencies.

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  • Is a company order to switch to a certain IDE a red flag?

    - by Justin Alexander
    I recently joined a rapidly growing startup. In the past 3 months the development team has grown from 4 to 12. Until now they were very laissez-faire about what developers used to do their work. In fact one of the things I initially found attractive about the company is that most programmers used Linux, or whatever OS they felt best suited their efforts. Now orders, without discussion, have come down that everyone is to switch to Eclipse. A fine editor. I prefer SublimeText2, but it's just my personal taste. Is this a red flag? It seems capricious and unreasonably controlling to tell developers (non-MS) what IDE or tool-sets to use if they are already settled in and productive.

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  • Generate colors between red and green for a power meter?

    - by Simucal
    I'm writing a java game and I want to implement a power meter for how hard you are going to shoot something. I need to write a function that takes a int between 0 - 100, and based on how high that number is, it will return a color between Green (0 on the power scale) and Red (100 on the power scale). Similar to how volume controls work: What operation do I need to do on the Red, Green, and Blue components of a color to generate the colors between Green and Red? So, I could run say, getColor(80) and it will return an orangish color (its values in R, G, B) or getColor(10) which will return a more Green/Yellow rgb value. I know I need to increase components of the R, G, B values for a new color, but I don't know specifically what goes up or down as the colors shift from Green-Red. Progress: I ended up using HSV/HSB color space because I liked the gradiant better (no dark browns in the middle). The function I used was (in java): public Color getColor(double power) { double H = power * 0.4; // Hue (note 0.4 = Green, see huge chart below) double S = 0.9; // Saturation double B = 0.9; // Brightness return Color.getHSBColor((float)H, (float)S, (float)B); } Where "power" is a number between 0.0 and 1.0. 0.0 will return a bright red, 1.0 will return a bright green. Java Hue Chart: Thanks everyone for helping me with this!

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  • Does constantly checking the documentation make you a bad coder?

    - by cdburgess
    When writing PHP code for any given project, do you find you can write code off the top of your head? Or do you make multiple round trips to php.net? If it is the later, can you still be considered a good coder. This is a legitimate question as I find I have difficulty always remembering all of the functions that are available to me so I find I use php.net as a crutch. Is there anyway to improve this?

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  • nth-child doesn't respond to class

    - by Arne Stephensson
    Is it possible to get the nth-child pseudo-selector to work with a specific class? See this example: http://jsfiddle.net/fZGvH/ I want to have the second DIV.red turn red, but it doesn't apply the color as expected. Not only that, but when you specify this, it changes the 5th DIV to red: div.red:nth-child(6) When you specify this, it changes the 8th DIV to red: div.red:nth-child(9) It seems to be one DIV behind. There are only 8 DIV tags in the example so I don't know why nth-child(9) even works. Testing using Firefox 3.6, but in my actual production code the same problem occurs in Chrome. I'm not understanding something about how this is supposed to work, would appreciate clarification. Also, this will change the 6th DIV to red, but what I actually want is for it to change the second DIV.red to red: div.red:nth-of-type(6) And I don't understand why nth-child() and nth-of-type() respond differently, since there are only eight tags of the same type in the document.

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  • How do i pass a number from a list as a parameter in scheme?

    - by wyatt
    I need to take a number from a list and convert it to a number so that i can pass it as a parameter. im trying to make a 1-bit adder in scheme. i've written the code for the or gate and the xor gate and also the half adder and now im trying to combine them all to make a full adder. im not sure if im going about it the right way. any input will be appreciated thank you. (define or-gate (lambda (a b) (if (= a 1) 1 (if (= b 1) 1 0)))) (define xor-gate (lambda (a b) (if (= a b) 0 1))) (define ha (lambda (a b) (list (xor-gate a b)(and-gate a b)))) (define fa (lambda (a b cin) (or-gate (cdr(ha cin (car (ha a b))))(cdr(ha a b))))) the issue i get when i run the program is that the half adder (ha) function outputs a list as a value and that makes the values incompatible with my other procedures because they require numbers and not lists. i feel like there is a simple solution but i dont know it.

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  • Exceptional DBA 2011 Jeff Moden on why you should enter in 2012

    - by Red and the Community
    My "reign" as the Red Gate Exceptional DBA is almost over and I was asked to say a few words about this wonderful award. Having been one of those folks that shied away from entering the contest during the first 3 years of the award, I thought I’d spend the time encouraging DBAs of all types to enter. Winning this award has some obvious benefits. You win a trip to PASS including money towards your flight, paid hotel stay, and, of course, paid admission. You win a wonderful bundle of software from Red Gate to make your job as a DBA a whole lot easier. You also win some pretty incredible notoriety for your resume. After all, it’s not everyone who wins a worldwide contest. To date, there are only 4 of us in the world who have won this award. You could be number 5! For me, all of that pales in comparison to what I found out during the entry process. I’m very confident in my skills, but I’m also humble. It was suggested to me that I enter the contest when it first started. I just couldn’t bring myself to nominate myself. When the 2011 nomination period opened up, several people again suggested that I enter, so I swallowed hard and asked several co-workers to have a look at the online nomination form and, if they thought me worthy, to write a nomination for me. I won’t bore you with the details, but what they wrote about me was one of the most incredible rewards that I could ever have hoped to receive. I had no idea of the impact that I’d made on my co-workers. Even if I hadn’t made it to the top 5 for the award, I had already won something very near and dear that no one can ever top. “Even if I hadn’t made it to the top 5 for the award, I had already won something very near and dear that no one can ever top.” There’s only one named winner and 4 "runners up" in this competition every year but don’t let that discourage you. Enter this competition. Even if you work in the proverbial "Mom’n'Pop" shop, get your boss and the people you work with directly to nominate you. Even if you don’t make it to the top 5, you might just find out that you’re more of a winner than you think. If you’re too proud to ask them, then take the time to nominate yourself instead of shying away like I did for the first 3 years. You work hard as a DBA and, as David Poole once said, if you’re the first person that people ask for help rather than one of the last, then you’re probably an Exceptional DBA. It’s time to stand up and be counted! Win or lose, the entry process can be a huge reward in itself. It was for me. Thank you, Red Gate, for giving me such a wonderful opportunity. Thanks for listening folks and for all that you do as DBAs. As ‘Red Green’ says, "We’re all in this together and I’m pullin’ for ya". –Jeff Moden Red Gate Exceptional DBA 2011

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  • What if I can't make my unit test fail in "Red, Green, Refactor" of TDD?

    - by Joshua Harris
    So let's say that I have a test: @Test public void MoveY_MoveZero_DoesNotMove() { Point p = new Point(50.0, 50.0); p.MoveY(0.0); Assert.assertAreEqual(50.0, p.Y); } This test then causes me to create the class Point: public class Point { double X; double Y; public void MoveY(double yDisplace) { throw new NotYetImplementedException(); } } Ok. It fails. Good. Then I remove the exception and I get green. Great, but of course I need to test if it changes value. So I write a test that calls p.MoveY(10.0) and checks if p.Y is equal to 60.0. It fails, so then I change the function to look like so: public void MoveY(double yDisplace) { Y += yDisplace; } Great, now I have green again and I can move on. I've tested not moving and moving in the positive direction, so naturally I should test a negative value. The only problem with this test is that if I wrote the test correctly, then it doesn't fail at first. That means that I didn't fit the principle of "Red, Green, Refactor." Of course, This is a first-world problem of TDD, but getting a fail at first is helpful in that it shows that your test can fail. Otherwise this seemingly innocent test that is just passing for incorrect reasons could fail later because it was written wrong. That might not be a problem if it happened 5 minutes later, but what if it happens to the poor-sap that inheirited your code two years later. What he knows is that MoveY does not work with negative values because that is what the test is telling him. But, it really could work and just be a bug in the test. I don't think that would happen in this particular case because the code sample is so simple, but if it were a large complicated system that might not be the case. It seems crazy to say that I want to fail my tests, but that is an important step in TDD, for good reasons.

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  • VirtualBox 3.2 is released! A Red Letter Day?

    - by Fat Bloke
    Big news today! A new release of VirtualBox packed full of innovation and improvements. Over the next few weeks we'll take a closer look at some of these new features in a lot more depth, but today we'll whet your appetite with the headline descriptions. To start with, we should point out that this is the first Oracle-branded version which makes today a real Red-letter day ;-)  Oracle VM VirtualBox 3.2 Version 3.2 moves VirtualBox forward in 3 main areas ( handily, all beginning with "P" ) : performance, power and supported guest operating system platforms.  Let's take a look: Performance New Latest Intel hardware support - Harnessing the latest in chip-level support for virtualization, VirtualBox 3.2 supports new Intel Core i5 and i7 processor and Intel Xeon processor 5600 Series support for Unrestricted Guest Execution bringing faster boot times for everything from Windows to Solaris guests; New Large Page support - Reducing the size and overhead of key system resources, Large Page support delivers increased performance by enabling faster lookups and shorter table creation times. New In-hypervisor Networking - Significant optimization of the networking subsystem has reduced context switching between guests and host, increasing network throughput by up to 25%. New New Storage I/O subsystem - VirtualBox 3.2 offers a completely re-worked virtual disk subsystem which utilizes asynchronous I/O to achieve high-performance whilst maintaining high data integrity; New Remote Video Acceleration - The unique built-in VirtualBox Remote Display Protocol (VRDP), which is primarily used in virtual desktop infrastructure deployments, has been enhanced to deliver video acceleration. This delivers a rich user experience coupled with reduced computational expense, which is vital when servers are running hundreds of virtual machines; Power New Page Fusion - Traditional Page Sharing techniques have suffered from long and expensive cache construction as pages are scrutinized as candidates for de-duplication. Taking a smarter approach, VirtualBox Page Fusion uses intelligence in the guest virtual machine to determine much more rapidly and accurately those pages which can be eliminated thereby increasing the capacity or vm density of the system; New Memory Ballooning- Ballooning provides another method to increase vm density by allowing the memory of one guest to be recouped and made available to others; New Multiple Virtual Monitors - VirtualBox 3.2 now supports multi-headed virtual machines with up to 8 virtual monitors attached to a guest. Each virtual monitor can be a host window, or be mapped to the hosts physical monitors; New Hot-plug CPU's - Modern operating systems such Windows Server 2008 x64 Data Center Edition or the latest Linux server platforms allow CPUs to be dynamically inserted into a system to provide incremental computing power while the system is running. Version 3.2 introduces support for Hot-plug vCPUs, allowing VirtualBox virtual machines to be given more power, with zero-downtime of the guest; New Virtual SAS Controller - VirtualBox 3.2 now offers a virtual SAS controller, enabling it to run the most demanding of high-end guests; New Online Snapshot Merging - Snapshots are powerful but can eat up disk space and need to be pruned from time to time. Historically, machines have needed to be turned off to delete or merge snapshots but with VirtualBox 3.2 this operation can be done whilst the machines are running. This allows sophisticated system management with minimal interruption of operations; New OVF Enhancements - VirtualBox has supported the OVF standard for virtual machine portability for some time. Now with 3.2, VirtualBox specific configuration data is also stored in the standard allowing richer virtual machine definitions without compromising portability; New Guest Automation - The Guest Automation APIs allow host-based logic to drive operations in the guest; Platforms New USB Keyboard and Mouse - Support more guests that require USB input devices; New Oracle Enterprise Linux 5.5 - Support for the latest version of Oracle's flagship Linux platform; New Ubuntu 10.04 ("Lucid Lynx") - Support for both the desktop and server version of the popular Ubuntu Linux distribution; And as a man once said, "just one more thing" ... New Mac OS X (experimental) - On Apple hardware only, support for creating virtual machines run Mac OS X. All in all this is a pretty powerful release packed full of innovation and speedups. So what are you waiting for?  -FB 

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  • SSAS Compare version 1.0 released

    - by Red Gate Software BI Tools Team
    We’re pleased to announce that SSAS Compare version 1.0 has been released as a free tool. Version 1.0 includes: Comparisons of live databases and XMLA or Analysis Services Project files MDX syntax diffs and highlighting Server comparisons Deployment wizard with summaries of scripted actions Bug fixes and engine and UI refinements We’ve tested it on as many cube configurations as we could find (not just good old AdventureWorks!), but we can’t provide support for free tools — so if you’re reliant on SSAS Compare for your cube deployment, use it at your own risk. See the user license agreement in the installer for more details. SSAS Compare’s come a long way from its humble beginnings as an internal tool first developed for Red Gate’s own BI developers. Today’s SSAS Compare is now much more stable — not to mention much easier to use — and something the team is proud to have released with Red Gate’s name on. Next: Deployment Manager We’re working on integrating SSAS Compare cube deployment with our new Deployment Manager tool, so you’ll be able to create cube deployment scripts and automate the deployment process, too.  We’re documenting the process in a white paper we’ll publish online in the next week. Thank you! Thanks to all the SSAS Compare users out there. Without your feedback, we could never have produced such a stable product so quickly. We hope you continue to find useful. See you in Deployment Manager!  

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  • Is the Leptonica implementation of 'Modified Median Cut' not using the median at all?

    - by TheCodeJunkie
    I'm playing around a bit with image processing and decided to read up on how color quantization worked and after a bit of reading I found the Modified Median Cut Quantization algorithm. I've been reading the code of the C implementation in Leptonica library and came across something I thought was a bit odd. Now I want to stress that I am far from an expert in this area, not am I a math-head, so I am predicting that this all comes down to me not understanding all of it and not that the implementation of the algorithm is wrong at all. The algorithm states that the vbox should be split along the lagest axis and that it should be split using the following logic The largest axis is divided by locating the bin with the median pixel (by population), selecting the longer side, and dividing in the center of that side. We could have simply put the bin with the median pixel in the shorter side, but in the early stages of subdivision, this tends to put low density clusters (that are not considered in the subdivision) in the same vbox as part of a high density cluster that will outvote it in median vbox color, even with future median-based subdivisions. The algorithm used here is particularly important in early subdivisions, and 3is useful for giving visible but low population color clusters their own vbox. This has little effect on the subdivision of high density clusters, which ultimately will have roughly equal population in their vboxes. For the sake of the argument, let's assume that we have a vbox that we are in the process of splitting and that the red axis is the largest. In the Leptonica algorithm, on line 01297, the code appears to do the following Iterate over all the possible green and blue variations of the red color For each iteration it adds to the total number of pixels (population) it's found along the red axis For each red color it sum up the population of the current red and the previous ones, thus storing an accumulated value, for each red note: when I say 'red' I mean each point along the axis that is covered by the iteration, the actual color may not be red but contains a certain amount of red So for the sake of illustration, assume we have 9 "bins" along the red axis and that they have the following populations 4 8 20 16 1 9 12 8 8 After the iteration of all red bins, the partialsum array will contain the following count for the bins mentioned above 4 12 32 48 49 58 70 78 86 And total would have a value of 86 Once that's done it's time to perform the actual median cut and for the red axis this is performed on line 01346 It iterates over bins and check they accumulated sum. And here's the part that throws me of from the description of the algorithm. It looks for the first bin that has a value that is greater than total/2 Wouldn't total/2 mean that it is looking for a bin that has a value that is greater than the average value and not the median ? The median for the above bins would be 49 The use of 43 or 49 could potentially have a huge impact on how the boxes are split, even though the algorithm then proceeds by moving to the center of the larger side of where the matched value was.. Another thing that puzzles me a bit is that the paper specified that the bin with the median value should be located, but does not mention how to proceed if there are an even number of bins.. the median would be the result of (a+b)/2 and it's not guaranteed that any of the bins contains that population count. So this is what makes me thing that there are some approximations going on that are negligible because of how the split actually takes part at the center of the larger side of the selected bin. Sorry if it got a bit long winded, but I wanted to be as thoroughas I could because it's been driving me nuts for a couple of days now ;)

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  • jQuery, array form radio button name problem.

    - by borayeris
    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>click div to select hidden options</title> <script type="text/javascript" src="jquery-1.4.4.js"></script> <style type="text/css"> .clickDiv { width:50px; height:50px; cursor:crosshair; } .red {border:1px #000 solid;} .green {border:1px #000 solid;} .redBG {background:#F00;} .greenBG {background:#0F0;} </style> <script type="text/javascript"> $(function() { $('div.clickDiv.red').click(function(){ var secilenMadde=$(this).attr('madde'); $('div#write').text(secilenMadde); $('input[name='+secilenMadde+'][value=red]').attr('checked', 'checked'); $('div.clickDiv.red[madde='+secilenMadde+']').addClass('redBG'); $('div.clickDiv.green[madde='+secilenMadde+']').removeClass('greenBG'); }); $('div.clickDiv.green').click(function(){ var secilenMadde=$(this).attr('madde'); $('div#write').text(secilenMadde); $('input[name='+secilenMadde+'][value=green]').attr('checked', 'checked'); $('div.clickDiv.green[madde='+secilenMadde+']').addClass('greenBG'); $('div.clickDiv.red[madde='+secilenMadde+']').removeClass('redBG'); }); }); </script> </head> <body> <div id="write"></div> <form id="formId" name="formName" method="post"> <table> <tr> <td><div class="clickDiv red" madde="line1"></div></td> <td><div class="clickDiv green" madde="line1"></div></td> </tr> <tr> <td><div class="clickDiv red" madde="line2"></div></td> <td><div class="clickDiv green" madde="line2"></div></td> </tr> </table> <label for="line1red"><input id="line1red" type="radio" name="line1" value="red" /> Red</label> <label for="line1green"><input id="line1green" type="radio" name="line1" value="green" /> Green</label><br /> <label for="line2red"><input type="radio" name="line2" value="red" /> Red</label> <label for="line2green"><input type="radio" name="line2" value="green" /> Green</label> </form> </body> </html> This works. <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>click div to select hidden options</title> <script type="text/javascript" src="jquery-1.4.4.js"></script> <style type="text/css"> .clickDiv { width:50px; height:50px; cursor:crosshair; } .red {border:1px #000 solid;} .green {border:1px #000 solid;} .redBG {background:#F00;} .greenBG {background:#0F0;} </style> <script type="text/javascript"> $(function() { $('div.clickDiv.red').click(function(){ var secilenMadde=$(this).attr('madde'); $('div#write').text(secilenMadde); $('input[name='+secilenMadde+'][value=red]').attr('checked', 'checked'); $('div.clickDiv.red[madde='+secilenMadde+']').addClass('redBG'); $('div.clickDiv.green[madde='+secilenMadde+']').removeClass('greenBG'); }); $('div.clickDiv.green').click(function(){ var secilenMadde=$(this).attr('madde'); $('div#write').text(secilenMadde); $('input[name='+secilenMadde+'][value=green]').attr('checked', 'checked'); $('div.clickDiv.green[madde='+secilenMadde+']').addClass('greenBG'); $('div.clickDiv.red[madde='+secilenMadde+']').removeClass('redBG'); }); }); </script> </head> <body> <div id="write"></div> <form id="formId" name="formName" method="post"> <table> <tr> <td><div class="clickDiv red" madde="line[1]"></div></td> <td><div class="clickDiv green" madde="line[1]"></div></td> </tr> <tr> <td><div class="clickDiv red" madde="line[2]"></div></td> <td><div class="clickDiv green" madde="line[2]"></div></td> </tr> </table> <label for="line1red"><input id="line1red" type="radio" name="line[1]" value="red" /> Red</label> <label for="line1green"><input id="line1green" type="radio" name="line[1]" value="green" /> Green</label><br /> <label for="line2red"><input type="radio" name="line[2]" value="red" /> Red</label> <label for="line2green"><input type="radio" name="line[2]" value="green" /> Green</label> </form> </body> </html> This doesn't. I need input names as an array but it breaks my script. Why?

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  • Before the Summit of 2012

    - by Ajarn Mark Caldwell
    Today, Monday, was the first day of the PASS Summit Preconference training events, but instead I spent the day at the free SQL in the City event put on by Red Gate. For me this was not a financial decision (pre-con sessions cost extra above the general Summit registration) but rather a matter of interest.  I had already included money for pre-cons in this year’s training budget, but none of them really stood out to me, so even if the Red-Gate event were not going on at the same time, I probably would not have gone to any pre-cons this year.  However, the topics being presented at the SQL in the City event were of great interest to me.  There promised to be good information on Continuous Integration and automated deployment of database changes, which lately has been a real hot topic at my work.  And indeed, Red-Gate announced the release of a new tool (still in Early Access Program…a.k.a. Beta) which is called the Deployment Manager.  Since we are in the middle of a TFS implementation project, it will be interesting to see how this plays out and compares to what we put together with the automated builds in TFS.  But, as I understand it, the primary focus of Deployment Manager is not to be the Build process (Red Gate uses JetBrains’ Team City for that in their shop) but rather to aid in the deployment of those build packages, as well as providing easy rollback and a good visualization of which versions of software are in which environments.  It looks promising and I’ve already downloaded the installer package to play with it later. Overall, I was quite impressed with the SQL in the City event.  Having heard many current and past members of the PASS Board of Directors describe the challenges of putting on a large conference, and the growing pains that the PASS Summit has gone through, I am even more impressed that the Red Gate event ran as smoothly as it did.  And it is quite impressive the amount of money that Red Gate must have spent given that this was a no-charge event to attend, they had a very nice hot lunch, and the after-event drinks celebration.  Well done, folks! Of course it was great to hear from a variety of speakers.  Today I listened to some folks from Red Gate like Grant Fritchey (blog | @GFritchey) and David Atkinson (Product Manager for SQL Source Control and now the Deployment Manager tool set); and also Brent Ozar (blog | @BrentO) and Buck Woody (blog | @BuckWoody).  By the way, if you have never seen either Brent or Buck speak, you really should.  Different styles, but both are very entertaining and educational at the same time.  I love Buck’s sense of humor (here’s a tip…don’t be late to Buck’s session or you’ll become part of the presentation) and I praise Brent’s slides.  Brent’s style very much reminds me of that espoused by Garr Reynolds on his Presentation Zen blog (and book) and I am impressed that he can make a technical presentation so engaging. It was a great day, a great way to kick off the week, and I am excited to get into the full Summit!

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  • When did Red Hat start shipping PHP 5.3 with 5.x?!?

    - by Jason
    Okay this is a PSA more than a question because I know the answer: January 13, 2011. See: https://rhn.redhat.com/errata/RHEA-2011-0069.html Colour me surprised though, didn't hear anything about in the blogosphere until I got a Security Errata notice today. I have been using the REMI repo for this in the past but will switch over to the Red Hat blessed PHP 5.3. Don't down-vote me bro! I'll select as the best answer the source that broke the news first (other than Red Hat of course). People have wanted this for so long I'm just amazed that it's finally happened!

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  • What does a red icon in XP's "Unlock Computer" dialog mean?

    - by wikiti
    A user was working from home and had a colleague turn on her computer so she could remote desktop to it. All worked fine, but when she came into the office and used her computer for a while then locked it the computer icon had a red screen, instead of blue. Like in the following mockup: Mockup of red computer screen. It didn't cause any problems and it went away when she rebooted, but I was intrigued to find out whether there was something that caused it or if it was just a windows oddity. I believe she just closed the remote desktop session (without really logging off) from home and then disconnected from the VPN before coming to the office. Any ideas?

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  • How can I replace red in an image with a certain shade of gray?

    - by Malcolm Frexner
    In this image I have to change red to grey: I know I can just set the saturation to zero, but then the result is a grey that is to dark. I could just change the brightness, but that would also change the left lower part of the picture: Is there an adjustment that only works on the red parts of the image? I can't use a selection, because the setting has to be applied to lots of images. EDIT I tried to use "replace adjustment tool", but that did not work well for the shdows and bright parts of the image, even with the largest fuzzines. I used blue as the replacement colour, to have a better impression of what it does.

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  • Why am I seeing red dots on my LCD screen?

    - by mydoghasworms
    My laptop is about 2.5 years old. Now I am starting to see red dots on certain shades of colour (mainly dark colours, blues and blacks), and it is not limited to certain pixels, because when you move a window around, the red dots move with it, staying on the certain shades of colour. Is this a problem with the LCD screen, or is it the GPU? Is there a way to determine this? It is clearly not a driver issue, because it happens in Linux and Windows, and my Windows setup has not changed prior to the issue starting.

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  • initWithCoder works, but init seems to be overwriting my objects properties?

    - by Zigrivers
    Hi guys, I've been trying to teach myself how to use the Archiving/Unarchiving methods of NSCoder, but I'm stumped. I have a Singleton class that I have defined with 8 NSInteger properties. I am trying to save this object to disk and then load from disk as needed. I've got the save part down and I have the load part down as well (according to NSLogs), but after my "initWithCoder:" method loads the object's properties appropriately, the "init" method runs and resets my object's properties back to zero. I'm probably missing something basic here, but would appreciate any help! My class methods for the Singleton class: + (Actor *)shareActorState { static Actor *actorState; @synchronized(self) { if (!actorState) { actorState = [[Actor alloc] init]; } } return actorState; } -(id)init { if (self = [super init]) { NSLog(@"New Init for Actor started...\nStrength: %d", self.strength); } return self; } -(id)initWithCoder:(NSCoder *)coder { if (self = [super init]) { strength = [coder decodeIntegerForKey:@"strength"]; dexterity = [coder decodeIntegerForKey:@"dexterity"]; stamina = [coder decodeIntegerForKey:@"stamina"]; will = [coder decodeIntegerForKey:@"will"]; intelligence = [coder decodeIntegerForKey:@"intelligence"]; agility = [coder decodeIntegerForKey:@"agility"]; aura = [coder decodeIntegerForKey:@"aura"]; eyesight = [coder decodeIntegerForKey:@"eyesight"]; NSLog(@"InitWithCoder executed....\nStrength: %d\nDexterity: %d", self.strength, self.dexterity); [self retain]; } return self; } -(void) encodeWithCoder:(NSCoder *)encoder { [encoder encodeInteger:strength forKey:@"strength"]; [encoder encodeInteger:dexterity forKey:@"dexterity"]; [encoder encodeInteger:stamina forKey:@"stamina"]; [encoder encodeInteger:will forKey:@"will"]; [encoder encodeInteger:intelligence forKey:@"intelligence"]; [encoder encodeInteger:agility forKey:@"agility"]; [encoder encodeInteger:aura forKey:@"aura"]; [encoder encodeInteger:eyesight forKey:@"eyesight"]; NSLog(@"encodeWithCoder executed...."); } -(void)dealloc { //My dealloc stuff goes here [super dealloc]; } I'm a noob when it comes to this stuff and have been trying to teach myself for the last month, so forgive anything obvious. Thanks for the help!

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  • SQL Group By Modulo of Row Count

    - by Alex Czarto
    I have the following sample data: Id Name Quantity 1 Red 1 2 Red 3 3 Blue 1 4 Red 1 5 Yellow 3 So for this example, there are a total of 5 Red, 1 Blue, and 3 Yellow. I am looking for a way to group them by Color, but with a maximum of 2 items per group (sorting is not important). Like so: Name QuantityInPackage Red 2 Red 2 Red 1 Blue 1 Yellow 2 Yellow 1 Any suggestions on how to accomplish this using T-SQL on MS-SQL 2005?

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  • Mono Project: How to install Mono framework on Red Hat Linux which is compiled on centOS ?

    - by funwithcoding
    We have Red Hat Enterprise Linux servers at work place. However we dont have Red Hat Linux desktops. So I used CentOS 5.4 to compile the Mono sources and generated the Mono framework for CentOS and tested with some sample codes and I am satisfied. I want to transfer this compiled framework to Red Hat Enterprise Linux 5. How Can I do that? Do I have to compile the Mono framework statically or do I have to copy the linked libraries as well? I am not familiar with linux much. Any help is highly appreciated.

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  • Programmatically swap colors from a loaded bitmap to Red, Green, Blue or Gray, pixel by pixel.

    - by eyeClaxton
    Download source code here: http://www.eyeClaxton.com/download/delphi/ColorSwap.zip I would like to take a original bitmap (light blue) and change the colors (Pixel by Pixel) to the red, green, blue and gray equivalence relation. To get an idea of what I mean, I have include the source code and a screen shot. Any help would be greatly appreciated. If more information is needed, please feel free to ask. If you could take a look at the code below, I have three functions that I'm looking for help on. The functions "RGBToRed, RGBToGreen and RGBToRed" I can't seem to come up with the right formulas. unit MainUnit; interface uses Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms, Dialogs, ExtCtrls, StdCtrls; type TMainFrm = class(TForm) Panel1: TPanel; Label1: TLabel; Panel2: TPanel; Label2: TLabel; Button1: TButton; BeforeImage1: TImage; AfterImage1: TImage; RadioGroup1: TRadioGroup; procedure FormCreate(Sender: TObject); procedure Button1Click(Sender: TObject); private { Private declarations } public { Public declarations } end; var MainFrm: TMainFrm; implementation {$R *.DFM} function RGBToGray(RGBColor: TColor): TColor; var Gray: Byte; begin Gray := Round( (0.90 * GetRValue(RGBColor)) + (0.88 * GetGValue(RGBColor)) + (0.33 * GetBValue(RGBColor))); Result := RGB(Gray, Gray, Gray); end; function RGBToRed(RGBColor: TColor): TColor; var Red: Byte; begin // Not sure of the algorithm for this color Result := RGB(Red, Red, Red); end; function RGBToGreen(RGBColor: TColor): TColor; var Green: Byte; begin // Not sure of the algorithm for this color Result := RGB(Green, Green, Green); end; function RGBToBlue(RGBColor: TColor): TColor; var Blue: Byte; begin // Not sure of the algorithm for this color Result := RGB(Blue, Blue, Blue); end; procedure TMainFrm.FormCreate(Sender: TObject); begin BeforeImage1.Picture.LoadFromFile('Images\RightCenter.bmp'); end; procedure TMainFrm.Button1Click(Sender: TObject); var Bitmap: TBitmap; I, X: Integer; Color: Integer; begin Bitmap := TBitmap.Create; try Bitmap.LoadFromFile('Images\RightCenter.bmp'); for X := 0 to Bitmap.Height do begin for I := 0 to Bitmap.Width do begin Color := ColorToRGB(Bitmap.Canvas.Pixels[I, X]); case Color of $00000000: ; // Skip any Color Here! else case RadioGroup1.ItemIndex of 0: Bitmap.Canvas.Pixels[I, X] := RGBToBlue(Color); 1: Bitmap.Canvas.Pixels[I, X] := RGBToRed(Color); 2: Bitmap.Canvas.Pixels[I, X] := RGBToGreen(Color); 3: Bitmap.Canvas.Pixels[I, X] := RGBToGray(Color); end; end; end; end; AfterImage1.Picture.Graphic := Bitmap; finally Bitmap.Free; end; end; end. Okay, I apologize for not making it clearer. I'm trying to take a bitmap (blue in color) and swap the blue pixels with another color. Like the shots below.

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