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  • Login failed for user 'sa' because the account is currently locked out. The system administrator can

    - by cabhilash
    Login failed for user 'sa' because the account is currently locked out. The system administrator can unlock it. (Microsoft SQL Server, Error: 18486) SQL server has local password policies. If policy is enabled which locks down the account after X number of failed attempts then the account is automatically locked down.This error with 'sa' account is very common. sa is default administartor login available with SQL server. So there are chances that an ousider has tried to bruteforce your system. (This can cause even if a legitimate tries to access the account with wrong password.Sometimes a user would have changed the password without informing others. So the other users would try to lo) You can unlock the account with the following options (use another admin account or connect via windows authentication) Alter account & unlock ALTER LOGIN sa WITH PASSWORD='password' UNLOCK Use another account Almost everyone is aware of the sa account. This can be the potential security risk. Even if you provide strong password hackers can lock the account by providing the wrong password. ( You can provide extra security by installing firewall or changing the default port but these measures are not always practical). As a best practice you can disable the sa account and use another account with same privileges.ALTER LOGIN sa DISABLE You can edit the lock-ot options using gpedit.msc( in command prompt type gpedit.msc and press enter). Navigate to Account Lokout policy as shown in the figure The Following options are available Account lockout threshold This security setting determines the number of failed logon attempts that causes a user account to be locked out. A locked-out account cannot be used until it is reset by an administrator or until the lockout duration for the account has expired. You can set a value between 0 and 999 failed logon attempts. If you set the value to 0, the account will never be locked out. Failed password attempts against workstations or member servers that have been locked using either CTRL+ALT+DELETE or password-protected screen savers count as failed logon attempts. Account lockout duration This security setting determines the number of minutes a locked-out account remains locked out before automatically becoming unlocked. The available range is from 0 minutes through 99,999 minutes. If you set the account lockout duration to 0, the account will be locked out until an administrator explicitly unlocks it. If an account lockout threshold is defined, the account lockout duration must be greater than or equal to the reset time. Default: None, because this policy setting only has meaning when an Account lockout threshold is specified. Reset account lockout counter after This security setting determines the number of minutes that must elapse after a failed logon attempt before the failed logon attempt counter is reset to 0 bad logon attempts. The available range is 1 minute to 99,999 minutes. If an account lockout threshold is defined, this reset time must be less than or equal to the Account lockout duration. Default: None, because this policy setting only has meaning when an Account lockout threshold is specified.When creating SQL user you can set CHECK_POLICY=on which will enforce the windows password policy on the account. The following policies will be applied Define the Enforce password history policy setting so that several previous passwords are remembered. With this policy setting, users cannot use the same password when their password expires.  Define the Maximum password age policy setting so that passwords expire as often as necessary for your environment, typically, every 30 to 90 days. With this policy setting, if an attacker cracks a password, the attacker only has access to the network until the password expires.  Define the Minimum password age policy setting so that passwords cannot be changed until they are more than a certain number of days old. This policy setting works in combination with the Enforce password historypolicy setting. If a minimum password age is defined, users cannot repeatedly change their passwords to get around the Enforce password history policy setting and then use their original password. Users must wait the specified number of days to change their passwords.  Define a Minimum password length policy setting so that passwords must consist of at least a specified number of characters. Long passwords--seven or more characters--are usually stronger than short ones. With this policy setting, users cannot use blank passwords, and they have to create passwords that are a certain number of characters long.  Enable the Password must meet complexity requirements policy setting. This policy setting checks all new passwords to ensure that they meet basic strong password requirements.  Password must meet the following complexity requirement, when they are changed or created: Not contain the user's entire Account Name or entire Full Name. The Account Name and Full Name are parsed for delimiters: commas, periods, dashes or hyphens, underscores, spaces, pound signs, and tabs. If any of these delimiters are found, the Account Name or Full Name are split and all sections are verified not to be included in the password. There is no check for any character or any three characters in succession. Contain characters from three of the following five categories:  English uppercase characters (A through Z) English lowercase characters (a through z) Base 10 digits (0 through 9) Non-alphabetic characters (for example, !, $, #, %) A catch-all category of any Unicode character that does not fall under the previous four categories. This fifth category can be regionally specific.

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  • What's wrong with relative imports in Python?

    - by Oddthinking
    I recently upgraded versions of pylint, a popular Python style-checker. It has gone ballistic throughout my code, pointing out places where I import modules in the same package, without specifying the full package path. The new error message is W0403. W0403: Relative import %r, should be %r Used when an import relative to the package directory is detected. Example For example, if my packages are structured like this: /cake /__init__.py /icing.py /sponge.py /drink and in the sponge package I write: import icing instead of import cake.icing I will get this error. While I understand that not all Pylint messages are of equal importance, and I am not afraid to dismiss them, I don't understand why such a practice is considered a poor idea. I was hoping someone could explain the pitfalls, so I could improve my coding style rather than (as I currently plan to do) turning off this apparently spurious warning.

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  • Liskov Substitution Principle and the Oft Forgot Third Wheel

    - by Stacy Vicknair
    Liskov Substitution Principle (LSP) is a principle of object oriented programming that many might be familiar with from the SOLID principles mnemonic from Uncle Bob Martin. The principle highlights the relationship between a type and its subtypes, and, according to Wikipedia, is defined by Barbara Liskov and Jeanette Wing as the following principle:   Let be a property provable about objects of type . Then should be provable for objects of type where is a subtype of .   Rectangles gonna rectangulate The iconic example of this principle is illustrated with the relationship between a rectangle and a square. Let’s say we have a class named Rectangle that had a property to set width and a property to set its height. 1: Public Class Rectangle 2: Overridable Property Width As Integer 3: Overridable Property Height As Integer 4: End Class   We all at some point here that inheritance mocks an “IS A” relationship, and by gosh we all know square IS A rectangle. So let’s make a square class that inherits from rectangle. However, squares do maintain the same length on every side, so let’s override and add that behavior. 1: Public Class Square 2: Inherits Rectangle 3:  4: Private _sideLength As Integer 5:  6: Public Overrides Property Width As Integer 7: Get 8: Return _sideLength 9: End Get 10: Set(value As Integer) 11: _sideLength = value 12: End Set 13: End Property 14:  15: Public Overrides Property Height As Integer 16: Get 17: Return _sideLength 18: End Get 19: Set(value As Integer) 20: _sideLength = value 21: End Set 22: End Property 23: End Class   Now, say we had the following test: 1: Public Sub SetHeight_DoesNotAffectWidth(rectangle As Rectangle) 2: 'arrange 3: Dim expectedWidth = 4 4: rectangle.Width = 4 5:  6: 'act 7: rectangle.Height = 7 8:  9: 'assert 10: Assert.AreEqual(expectedWidth, rectangle.Width) 11: End Sub   If we pass in a rectangle, this test passes just fine. What if we pass in a square?   This is where we see the violation of Liskov’s Principle! A square might "IS A” to a rectangle, but we have differing expectations on how a rectangle should function than how a square should! Great expectations Here’s where we pat ourselves on the back and take a victory lap around the office and tell everyone about how we understand LSP like a boss. And all is good… until we start trying to apply it to our work. If I can’t even change functionality on a simple setter without breaking the expectations on a parent class, what can I do with subtyping? Did Liskov just tell me to never touch subtyping again? The short answer: NO, SHE DIDN’T. When I first learned LSP, and from those I’ve talked with as well, I overlooked a very important but not appropriately stressed quality of the principle: our expectations. Our inclination is to want a logical catch-all, where we can easily apply this principle and wipe our hands, drop the mic and exit stage left. That’s not the case because in every different programming scenario, our expectations of the parent class or type will be different. We have to set reasonable expectations on the behaviors that we expect out of the parent, then make sure that those expectations are met by the child. Any expectations not explicitly expected of the parent aren’t expected of the child either, and don’t register as a violation of LSP that prevents implementation. You can see the flexibility mentioned in the Wikipedia article itself: A typical example that violates LSP is a Square class that derives from a Rectangle class, assuming getter and setter methods exist for both width and height. The Square class always assumes that the width is equal with the height. If a Square object is used in a context where a Rectangle is expected, unexpected behavior may occur because the dimensions of a Square cannot (or rather should not) be modified independently. This problem cannot be easily fixed: if we can modify the setter methods in the Square class so that they preserve the Square invariant (i.e., keep the dimensions equal), then these methods will weaken (violate) the postconditions for the Rectangle setters, which state that dimensions can be modified independently. Violations of LSP, like this one, may or may not be a problem in practice, depending on the postconditions or invariants that are actually expected by the code that uses classes violating LSP. Mutability is a key issue here. If Square and Rectangle had only getter methods (i.e., they were immutable objects), then no violation of LSP could occur. What this means is that the above situation with a rectangle and a square can be acceptable if we do not have the expectation for width to leave height unaffected, or vice-versa, in our application. Conclusion – the oft forgot third wheel Liskov Substitution Principle is meant to act as a guidance and warn us against unexpected behaviors. Objects can be stateful and as a result we can end up with unexpected situations if we don’t code carefully. Specifically when subclassing, make sure that the subclass meets the expectations held to its parent. Don’t let LSP think you cannot deviate from the behaviors of the parent, but understand that LSP is meant to highlight the importance of not only the parent and the child class, but also of the expectations WE set for the parent class and the necessity of meeting those expectations in order to help prevent sticky situations.   Code examples, in both VB and C# Technorati Tags: LSV,Liskov Substitution Principle,Uncle Bob,Robert Martin,Barbara Liskov,Liskov

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  • Why not Green Threads?

    - by redjamjar
    Whilst I know questions on this have been covered already (e.g. http://stackoverflow.com/questions/5713142/green-threads-vs-non-green-threads), I don't feel like I've got a satisfactory answer. The question is: why don't JVM's support green threads anymore? It says this on the code-style Java FAQ: A green thread refers to a mode of operation for the Java Virtual Machine (JVM) in which all code is executed in a single operating system thread. And this over on java.sun.com: The downside is that using green threads means system threads on Linux are not taken advantage of and so the Java virtual machine is not scalable when additional CPUs are added. It seems to me that the JVM could have a pool of system processes equal to the number of cores, and then run green threads on top of that. This could offer some big advantages when you have a very number large of threads which block often (mostly because current JVM's cap the number of threads). Thoughts?

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  • Use Drive Mirroring for Instant Backup in Windows 7

    - by Trevor Bekolay
    Even with the best backup solution, a hard drive crash means you’ll lose a few hours of work. By enabling drive mirroring in Windows 7, you’ll always have an up-to-date copy of your data. Windows 7’s mirroring – which is only available in Professional, Enterprise, and Ultimate editions – is a software implementation of RAID 1, which means that two or more disks are holding the exact same data. The files are constantly kept in sync, so that if one of the disks fails, you won’t lose any data. Note that mirroring is not technically a backup solution, because if you accidentally delete a file, it’s gone from both hard disks (though you may be able to recover the file). As an additional caveat, having mirrored disks requires changing them to “dynamic disks,” which can only be read within modern versions of Windows (you may have problems working with a dynamic disk in other operating systems or in older versions of Windows). See this Wikipedia page for more information. You will need at least one empty disk to set up disk mirroring. We’ll show you how to mirror an existing disk (of equal or lesser size) without losing any data on the mirrored drive, and how to set up two empty disks as mirrored copies from the get-go. Mirroring an Existing Drive Click on the start button and type partitions in the search box. Click on the Create and format hard disk partitions entry that shows up. Alternatively, if you’ve disabled the search box, press Win+R to open the Run window and type in: diskmgmt.msc The Disk Management window will appear. We’ve got a small disk, labeled OldData, that we want to mirror in a second disk of the same size. Note: The disk that you will use to mirror the existing disk must be unallocated. If it is not, then right-click on it and select Delete Volume… to mark it as unallocated. This will destroy any data on that drive. Right-click on the existing disk that you want to mirror. Select Add Mirror…. Select the disk that you want to use to mirror the existing disk’s data and press Add Mirror. You will be warned that this process will change the existing disk from basic to dynamic. Note that this process will not delete any data on the disk! The new disk will be marked as a mirror, and it will starting copying data from the existing drive to the new one. Eventually the drives will be synced up (it can take a while), and any data added to the E: drive will exist on both physical hard drives. Setting Up Two New Drives as Mirrored If you have two new equal-sized drives, you can format them to be mirrored copies of each other from the get-go. Open the Disk Management window as described above. Make sure that the drives are unallocated. If they’re not, and you don’t need the data on either of them, right-click and select Delete volume…. Right-click on one of the unallocated drives and select New Mirrored Volume…. A wizard will pop up. Click Next. Click on the drives you want to hold the mirrored data and click Add. Note that you can add any number of drives. Click Next. Assign it a drive letter that makes sense, and then click Next. You’re limited to using the NTFS file system for mirrored drives, so enter a volume label, enable compression if you want, and then click Next. Click Finish to start formatting the drives. You will be warned that the new drives will be converted to dynamic disks. And that’s it! You now have two mirrored drives. Any files added to E: will reside on both physical disks, in case something happens to one of them. Conclusion While the switch from basic to dynamic disks can be a problem for people who dual-boot into another operating system, setting up drive mirroring is an easy way to make sure that your data can be recovered in case of a hard drive crash. Of course, even with drive mirroring, we advocate regular backups to external drives or online backup services. Similar Articles Productive Geek Tips Rebit Backup Software [Review]Disabling Instant Search in Outlook 2007Restore Files from Backups on Windows Home ServerSecond Copy 7 [Review]Backup Windows Home Server Folders to an External Hard Drive TouchFreeze Alternative in AutoHotkey The Icy Undertow Desktop Windows Home Server – Backup to LAN The Clear & Clean Desktop Use This Bookmarklet to Easily Get Albums Use AutoHotkey to Assign a Hotkey to a Specific Window Latest Software Reviews Tinyhacker Random Tips CloudBerry Online Backup 1.5 for Windows Home Server Snagit 10 VMware Workstation 7 Acronis Online Backup Windows Firewall with Advanced Security – How To Guides Sculptris 1.0, 3D Drawing app AceStock, a Tiny Desktop Quote Monitor Gmail Button Addon (Firefox) Hyperwords addon (Firefox) Backup Outlook 2010

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  • Cannot resolve the collation conflict between "Latin1_General_CI_AS" and "SQL_Latin1_General_CP1_CI_

    - by Patrick Olurotimi Ige
    I was writing a store proc for a report and i needed some data from another server so i added a linked server to connect to this new db server. when i do a select like below its all fine select a,b,c from Server.DatabaseName.dbo.table But when i use the table in a join i get the error "Cannot resolve the collation conflict between "Latin1_General_CI_AS" and "SQL_Latin1_General_CP1_CI_AS" in the equal to operation." I did check the collation set on the 2 databases and it was actually the same and had mo idea why i'm getting the error. I later found out that you could specifically tell it to use a COLLATE Just rewrite your join like this on a.name COLLATE Latin1_General_CI_AS = eaobjname Hope that helps and saves your precious time Patrick

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  • How to load Xmir on Ubuntu Touch Saucy

    - by jaro123
    How to load Xmir on Ubuntu 13.10 Nexus 7 tablet, unity8 + lightdm + mir, unity-system-compositor + Xorg installed. Still getting message above after system reboot in the /var/log/Xorg.0.log : [ 81.542] (II) "glx" will be loaded by default. [ 81.542] (WW) "xmir" is not to be loaded by default. Skipping. [ 81.542] (II) LoadModule: "dri2" Thanks Note: No, this question is not duplicate with How to load Xmir on Ubuntu 13.10 because. My note that this do not work on Ubuntu Touch Saucy + Nexus7 tablet was removed by fossfreedom. I do feel it duplicate also, but this moderator issue. There is no chance to response on equal topic, even when it is not obviously solving the problem. I do consider this confusing and causing incomplete answers.

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  • Box2d too much for Circle/Circle collision detection?

    - by Joey Green
    I'm using cocos2d to program a game and am using box2d for collision detection. Everything in my game is a circle and for some reason I'm having a problem with some times things are not being detected as a collision when they should be. I'm thinking of rolling up my own collision detection since I don't think it would be too hard. Questions are: Would this approach work for collision detection between circles? a. get radius of circle A and circle B. b. get distance of the center of circle A and circle B c. if the distance is greater than or equal to the sum of circle A radius and circle B radius then we have a hit Should box2d be used for such simple collision detection? There are no physics in this game.

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  • Flixel Game Over Screen

    - by Jamie Read
    I am new to game development but familiar with programming languages. I have started using Flixel and have a working Breakout game with score and lives. I am just stuck on how I can create a new screen/game over screen if a player runs out of lives. I would like the process to be like following: Check IF lives are equal to 0 Pause the game and display a new screen (probably transparent) that says 'Game Over' When a user clicks or hits ENTER restart the level Here is the function I currently have to update the lives: private function loseLive(_ball:FlxObject, _bottomWall:FlxObject):void { // check for game over if (lives_count == 0) { } else { FlxG:lives_count -= 1; lives.text = 'Lives: ' + lives_count.toString() } } Here is my main game.as: package { import org.flixel.*; public class Game extends FlxGame { private const resolution:FlxPoint = new FlxPoint(640, 480); private const zoom:uint = 2; private const fps:uint = 60; public function Game() { super(resolution.x / zoom, resolution.y / zoom, PlayState, zoom); FlxG.flashFramerate = fps; } } }

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  • How to rotate a set of points on z = 0 plane in 3-D, preserving pairwise distances?

    - by cagirici
    I have a set of points double n[] on the plane z = 0. And I have another set of points double[] m on the plane ax + by + cz + d = 0. Length of n is equal to length of m. Also, euclidean distance between n[i] and n[j] is equal to euclidean distance between m[i] and m[j]. I want to rotate n[] in 3-D, such that for all i, n[i] = m[i] would be true. In other words, I want to turn a plane into another plane, preserving the pairwise distances. Here's my code in java. But it does not help so much: double[] rotate(double[] point, double[] currentEquation, double[] targetEquation) { double[] currentNormal = new double[]{currentEquation[0], currentEquation[1], currentEquation[2]}; double[] targetNormal = new double[]{targetEquation[0], targetEquation[1], targetEquation[2]}; targetNormal = normalize(targetNormal); double angle = angleBetween(currentNormal, targetNormal); double[] axis = cross(targetNormal, currentNormal); double[][] R = getRotationMatrix(axis, angle); return rotated; } double[][] getRotationMatrix(double[] axis, double angle) { axis = normalize(axis); double cA = (float)Math.cos(angle); double sA = (float)Math.sin(angle); Matrix I = Matrix.identity(3, 3); Matrix a = new Matrix(axis, 3); Matrix aT = a.transpose(); Matrix a2 = a.times(aT); double[][] B = { {0, axis[2], -1*axis[1]}, {-1*axis[2], 0, axis[0]}, {axis[1], -1*axis[0], 0} }; Matrix A = new Matrix(B); Matrix R = I.minus(a2); R = R.times(cA); R = R.plus(a2); R = R.plus(A.times(sA)); return R.getArray(); } This is what I get. The point set on the right side is actually part of a point set on the left side. But they are on another plane. Here's a 2-D representation of what I try to do: There are two lines. The line on the bottom is the line I have. The line on the top is the target line. The distances are preserved (a, b and c). Edit: I have tried both methods written in answers. They both fail (I guess). Method of Martijn Courteaux public static double[][] getRotationMatrix(double[] v0, double[] v1, double[] v2, double[] u0, double[] u1, double[] u2) { RealMatrix M1 = new Array2DRowRealMatrix(new double[][]{ {1,0,0,-1*v0[0]}, {0,1,0,-1*v0[1]}, {0,0,1,0}, {0,0,0,1} }); RealMatrix M2 = new Array2DRowRealMatrix(new double[][]{ {1,0,0,-1*u0[0]}, {0,1,0,-1*u0[1]}, {0,0,1,-1*u0[2]}, {0,0,0,1} }); Vector3D imX = new Vector3D((v0[1] - v1[1])*(u2[0] - u0[0]) - (v0[1] - v2[1])*(u1[0] - u0[0]), (v0[1] - v1[1])*(u2[1] - u0[1]) - (v0[1] - v2[1])*(u1[1] - u0[1]), (v0[1] - v1[1])*(u2[2] - u0[2]) - (v0[1] - v2[1])*(u1[2] - u0[2]) ).scalarMultiply(1/((v0[0]*v1[1])-(v0[0]*v2[1])-(v1[0]*v0[1])+(v1[0]*v2[1])+(v2[0]*v0[1])-(v2[0]*v1[1]))); Vector3D imZ = new Vector3D(findEquation(u0, u1, u2)); Vector3D imY = Vector3D.crossProduct(imZ, imX); double[] imXn = imX.normalize().toArray(); double[] imYn = imY.normalize().toArray(); double[] imZn = imZ.normalize().toArray(); RealMatrix M = new Array2DRowRealMatrix(new double[][]{ {imXn[0], imXn[1], imXn[2], 0}, {imYn[0], imYn[1], imYn[2], 0}, {imZn[0], imZn[1], imZn[2], 0}, {0, 0, 0, 1} }); RealMatrix rotationMatrix = MatrixUtils.inverse(M2).multiply(M).multiply(M1); return rotationMatrix.getData(); } Method of Sam Hocevar static double[][] makeMatrix(double[] p1, double[] p2, double[] p3) { double[] v1 = normalize(difference(p2,p1)); double[] v2 = normalize(cross(difference(p3,p1), difference(p2,p1))); double[] v3 = cross(v1, v2); double[][] M = { { v1[0], v2[0], v3[0], p1[0] }, { v1[1], v2[1], v3[1], p1[1] }, { v1[2], v2[2], v3[2], p1[2] }, { 0.0, 0.0, 0.0, 1.0 } }; return M; } static double[][] createTransform(double[] A, double[] B, double[] C, double[] P, double[] Q, double[] R) { RealMatrix c = new Array2DRowRealMatrix(makeMatrix(A,B,C)); RealMatrix t = new Array2DRowRealMatrix(makeMatrix(P,Q,R)); return MatrixUtils.inverse(c).multiply(t).getData(); } The blue points are the calculated points. The black lines indicate the offset from the real position.

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  • Detecting End of Animation

    - by Will
    So I am making a death animation for a game. enemy1 is a UIImageView, and what I'm doing is when an integer is less than or equal to zero, it calls this deathAnimation which only happens once. What I want to do is use a CGPointMake right when the animation is finished being called. Note that before the deathAnimation is called, there is another animation that is constantly being called 30 times a second. I'm not using anything like cocos2d. if (enemy1health <= 0) { [self slime1DeathAnimation]; //How can i detect the end of this animation } This is how the animation is done -(void)slime1DeathAnimation{ enemy1.animationImages = [[NSArray alloc] initWithObjects: [UIImage imageNamed:@"Slime Death 1.png"], [UIImage imageNamed:@"Slime Death 2.png"], [UIImage imageNamed:@"Slime Death 3.png"], [UIImage imageNamed:@"Slime Death 4.png"], [UIImage imageNamed:@"Slime Death 5.png"], nil]; enemy1.animationDuration = 0.5; enemy1.animationRepeatCount = 1; [enemy1 startAnimating]; } If you need more code just ask

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  • RDA Health Checks for SOA

    - by ShawnBailey
    What is a health check in RDA? A health check evaluates something in your environment to determine whether a change needs to be considered in order to avoid a problem or optimize fuctionality. Examples of what this 'something' might be are: Configuration Parameters JVM Options Runtime Statistics What have we done for SOA? In the latest release of RDA, 4.30, we have added a Rule Set for SOA called 'Oracle SOA 11g (11.1.1) Post Installation (Generic)'. This Rule Set contains 14 SOA related health checks. These checks were all derived from common issues / solutions we see in support of the SOA product. Many of the recommendations come from the product documentation while others are covered in the SOA Knowledge Base. Our goal is that you will be able to easily identify the areas of concern and understand the guidance available from the output of the Rule Set. Running the health checks for SOA The rules that the checks use are installed with RDA and bundled by product or functional area into what are called 'Rule Sets'. To view the available Rule Sets simply run the command from the RDA home location: rda.cmd (or .sh) -dT hcve This will bring up a list of the available HCVE (Health Check / Verification Engine) Rule Sets. Each Rule Set contains a group of related rules that are used for evalutation and display of results. A rule can be considered synonymous with a single health check and they are assigned an ID, Name and Description that can be seen when they are executed. The Rule Set for SOA is option number 11 and you just enter this selection at the prompt. The Rule Set will then execute to completion. After running an HCVE Rule Set the tool will write the output to the RDA_HOME/output folder. The simplest way to view the output is to drag the .htm file to a browser but of course it can also be uploaded to a Service Request for evaluation by Oracle Support. Many of the Rule Sets will prompt you for information before they can execute their rules but the SOA Rule Set will identify the SOA domains configured in your RDA setup.cfg file. This means that you don't need to answer all of the questions again about where stuff is but it also means that you must have configured RDA for SOA. To run the Rule Set: Download the latest version of RDA from MOS Doc ID 314422.1 Configure RDA for your SOA domains. Detailed steps can be found here In it's simplest form the command is 'rda.cmd (.sh) -S SOA' Go to the RDA home location and enter the command 'rda.cmd (or .sh) -dT hcve' Select option '11' It should be noted that this our first release of a SOA Rule Set so there will probably be some things we need to clean up or fix. None of these rules will actually modify anything on your system as they are read only and do the evaluations internally. Please let us know if you have any issues with the rules or ideas for new ones so we can make them as useful as possible. The Checks Here is a list of the SOA health checks by ID, Name and Description. ID Name Description A00100 SOA Domain Homes Lists the SOA domains that were indentified from the RDA setup.cfg file A00200 Coherence Protocol Conflict Checks to see if you have both Unicast and Multicast configured in the same domain. Checks both the setDomainEnv and config.xml entries (if it exists). We recommend Unicast with fully qualified host names or IP addresses. A00210 Coherence Fully Qualified Host Checks that the host names are fully qualified or that IP addresses are used. Will fail if unqualified host names are detected. A00220 Unicast Local Host Checks that the Coherence localhost is specified for use with Unicast A00300 JTA Timeout Checks that the JTA timeout is configured for the domain and lists the value. The bundled rule will only list the current values of the JTA timeout for each SOA Domain. In the future the rule with fail with a warning if the value is 300 seconds or lower. It is recommended that timeouts follow the pattern 'syncMaxWaitTime' < EJB Timeouts < JTA Timeout. The 300 second value is important because the EJB Timeouts default to 300 seconds. Additional information can be found in MOS Doc ID 880313.1. A00310 XA Max Time Checks that the JTA Maximum XA call time is set for the domain. Fails if it is not explicitly set or if the value is less than or equal to the default of 12000 ms. A00320 XA Timeout Checks that the XA timeout is enabled and that the value is '0' for the SOA Data Source (SOADataSource-jdbc.xml) A00330 JDBC Statement Timeout Checks that the Statement Timeout is set for all SOA Data Sources. Fails if the value is not set or if it is set to the default of -1. A00400 XA Driver Checks that the SOA Data Source is configured to use an XA driver. Fails if it is not. A00410 JDBC Capacity Settings Checks that the minimum and maximum capacity are equal for all SOA Data Sources. Fails if they are not and lists specifically which data sources failed. A00500 SOA Roles Checks that the default SOA roles 'SOAAdmin' and 'SOAOperator' are configured for the soa-infra application in the file sytem-jazn-data.xml. Fails if they are not. A00700 SOA-INFRA Deployment Checks that the soa-infra application is deployed to either a cluster, all members of a cluster or a stand alone server. A00710 SOA Deployments Checks that the SOA related applications are deployed to the same domain members as soa-infra. A00720 SOA Library Deployments Checks that the SOA related libraries are deployed to the same domain members as soa-infra. A00730 Data Source Deployments Checks that the SOA Data Sources are all targeted to the same domain members as soa-infra

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  • LSP vs OCP / Liskov Substitution VS Open Close

    - by Kolyunya
    I am trying to understand the SOLID principles of OOP and I've come to the conclusion that LSP and OCP have some similarities (if not to say more). the open/closed principle states "software entities (classes, modules, functions, etc.) should be open for extension, but closed for modification". LSP in simple words states that any instance of Foo can be replaced with any instance of Bar which is derived from Foo and the program will work the same very way. I'm not a pro OOP programmer, but it seems to me that LSP is only possible if Bar, derived from Foo does not change anything in it but only extends it. That means that in particular program LSP is true only when OCP is true and OCP is true only if LSP is true. That means that they are equal. Correct me if I'm wrong. I really want to understand these ideas. Great thanks for an answer.

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  • Is it safe to resize root partition?

    - by binW
    My HDD is partitioned into two equal sized partitions. First is being used for Windows and Second for Ubuntu. Everything is working fine. But now I want to remove Windows and use the disk completely for Ubuntu. I can easily boot from live cd and use GParted to delete Windows partition and then expand Ubuntu partition to use the whole hard disk. But I want to know if its safe i.e Will resizing Ubuntu partition change any thing else like the partition UUID or any thing else? Do I need to reinstall grub after resizing the root partition? It would be great if some one who has already done this can give their advice here.

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  • Is case after case in a switch efficient?

    - by RandomGuy
    Just a random question regarding switch case efficiency in case after case; is the following code (assume pseudo code): function bool isValid(String myString){ switch(myString){ case "stringA": case "stringB": case "stringC": return true; default: return false; } more efficient than this: function bool isValid(String myString){ switch(myString){ case "stringA": return true; case "stringB": return true; case "stringC": return true; default: return false; } Or is the performance equal? I'm not thinking in a specific language but if needed let's assume it's Java or C (for this case would be needed to use chars instead of strings).

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  • Unexpected behaviour with glFramebufferTexture1D

    - by Roshan
    I am using render to texture concept with glFramebufferTexture1D. I am drawing a cube on non-default FBO with all the vertices as -1,1 (maximum) in X Y Z direction. Now i am setting viewport to X while rendering on non default FBO. My background is blue with white color of cube. For default FBO, i have created 1-D texture and attached this texture to above FBO with color attachment. I am setting width of texture equal to width*height of above FBO view-port. Now, when i render this texture to on another cube, i can see continuous white color on start or end of each face of the cube. That means part of the face is white and rest is blue. I am not sure whether this behavior is correct or not. I expect all the texels should be white as i am using -1 and 1 coordinates for cube rendered on non-default FBO. code: #define WIDTH 3 #define HEIGHT 3 GLfloat vertices8[]={ 1.0f,1.0f,1.0f, -1.0f,1.0f,1.0f, -1.0f,-1.0f,1.0f, 1.0f,-1.0f,1.0f,//face 1 1.0f,-1.0f,-1.0f, -1.0f,-1.0f,-1.0f, -1.0f,1.0f,-1.0f, 1.0f,1.0f,-1.0f,//face 2 1.0f,1.0f,1.0f, 1.0f,-1.0f,1.0f, 1.0f,-1.0f,-1.0f, 1.0f,1.0f,-1.0f,//face 3 -1.0f,1.0f,1.0f, -1.0f,1.0f,-1.0f, -1.0f,-1.0f,-1.0f, -1.0f,-1.0f,1.0f,//face 4 1.0f,1.0f,1.0f, 1.0f,1.0f,-1.0f, -1.0f,1.0f,-1.0f, -1.0f,1.0f,1.0f,//face 5 -1.0f,-1.0f,1.0f, -1.0f,-1.0f,-1.0f, 1.0f,-1.0f,-1.0f, 1.0f,-1.0f,1.0f//face 6 }; GLfloat vertices[]= { 0.5f,0.5f,0.5f, -0.5f,0.5f,0.5f, -0.5f,-0.5f,0.5f, 0.5f,-0.5f,0.5f,//face 1 0.5f,-0.5f,-0.5f, -0.5f,-0.5f,-0.5f, -0.5f,0.5f,-0.5f, 0.5f,0.5f,-0.5f,//face 2 0.5f,0.5f,0.5f, 0.5f,-0.5f,0.5f, 0.5f,-0.5f,-0.5f, 0.5f,0.5f,-0.5f,//face 3 -0.5f,0.5f,0.5f, -0.5f,0.5f,-0.5f, -0.5f,-0.5f,-0.5f, -0.5f,-0.5f,0.5f,//face 4 0.5f,0.5f,0.5f, 0.5f,0.5f,-0.5f, -0.5f,0.5f,-0.5f, -0.5f,0.5f,0.5f,//face 5 -0.5f,-0.5f,0.5f, -0.5f,-0.5f,-0.5f, 0.5f,-0.5f,-0.5f, 0.5f,-0.5f,0.5f//face 6 }; GLuint indices[] = { 0, 2, 1, 0, 3, 2, 4, 5, 6, 4, 6, 7, 8, 9, 10, 8, 10, 11, 12, 15, 14, 12, 14, 13, 16, 17, 18, 16, 18, 19, 20, 23, 22, 20, 22, 21 }; GLfloat texcoord[] = { 0.0, 0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0, 0.0, 0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0, 0.0, 0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0, 0.0, 0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0, 0.0, 0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0, 0.0, 0.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0 }; glGenTextures(1, &id1); glBindTexture(GL_TEXTURE_1D, id1); glGenFramebuffers(1, &Fboid); glTexParameterf(GL_TEXTURE_1D, GL_TEXTURE_MIN_FILTER, GL_NEAREST); glTexParameterf(GL_TEXTURE_1D, GL_TEXTURE_MAG_FILTER, GL_NEAREST); glTexParameterf(GL_TEXTURE_1D, GL_TEXTURE_WRAP_S, GL_CLAMP_TO_EDGE); glTexImage1D(GL_TEXTURE_1D, 0, GL_RGBA, WIDTH*HEIGHT , 0, GL_RGBA, GL_UNSIGNED_BYTE,0); glBindFramebuffer(GL_FRAMEBUFFER, Fboid); glFramebufferTexture1D(GL_DRAW_FRAMEBUFFER,GL_COLOR_ATTACHMENT0,GL_TEXTURE_1D,id1,0); draw_cube(); glBindFramebuffer(GL_FRAMEBUFFER, 0); draw(); } draw_cube() { glViewport(0, 0, WIDTH, HEIGHT); glClearColor(0.0f, 0.0f, 0.5f, 1.0f); glClear(GL_COLOR_BUFFER_BIT); glEnableVertexAttribArray(glGetAttribLocation(temp.psId,"position")); glVertexAttribPointer(glGetAttribLocation(temp.psId,"position"), 3, GL_FLOAT, GL_FALSE, 0,vertices8); glDrawArrays (GL_TRIANGLE_FAN, 0, 24); } draw() { glClearColor(1.0f, 0.0f, 0.0f, 1.0f); glClearDepth(1.0f); glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT); glEnableVertexAttribArray(glGetAttribLocation(shader_data.psId,"tk_position")); glVertexAttribPointer(glGetAttribLocation(shader_data.psId,"tk_position"), 3, GL_FLOAT, GL_FALSE, 0,vertices); nResult = GL_ERROR_CHECK((GL_NO_ERROR, "glVertexAttribPointer(position, 3, GL_FLOAT, GL_FALSE, 0,vertices);")); glEnableVertexAttribArray(glGetAttribLocation(shader_data.psId,"inputtexcoord")); glVertexAttribPointer(glGetAttribLocation(shader_data.psId,"inputtexcoord"), 2, GL_FLOAT, GL_FALSE, 0,texcoord); glBindTexture(*target11, id1); glDrawElements ( GL_TRIANGLES, 36,GL_UNSIGNED_INT, indices ); when i change WIDTH=HEIGHT=2, and call a glreadpixels with height, width equal to 4 in draw_cube() i can see first 2 pixels with white color, next two with blue(glclearcolor), next two white and then blue and so on.. Now when i change width parameter in glTeximage1D to 16 then ideally i should see alternate patches of white and blue right? But its not the case here. why so?

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  • Avoiding That Null Reference!

    - by TheJuice
    A coworker had this conversation with another of our developers. Names have been changed to protect the guilty. Clueless: hey! Clueless: I am using the ?? operator for null check below Nice Guy: hey Clueless: FundLoanRequestBoatCollateral boatCollateral = request.BoatCollateral ?? null; Nice Guy: that's not exactly how it works Clueless: I want to achive: FundLoanRequestBoatCollateral boatCollateral = request.BoatCollateral != null ? request.BoatCollateral : null; Clueless: isnt that equal to:  FundLoanRequestBoatCollateral boatCollateral = request.BoatCollateral ?? null; Nice Guy: that is functionally equivalent to FundLoanRequestBoatCollateral boatCollateral = request.BoatCollateral Nice Guy: you're checking if it's null and if so setting it to null Clueless: yeah Clueless: if its null I want to set it to null Nice Guy: if it's null then you're already going to set it to null, no special logic needed Clueless: I wanted to avoid a null reference if BoatCollateral is null   The sad part of all of this is that "Clueless" has been with our company for years and has a Master's in Computer Science.

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  • Adding a comma to a resource name in Microsoft Project

    - by John Paul Cook
    Microsoft Project does not allow a comma to be added to a resource name. In healthcare, the norm is to refer to people using the pattern of Name, Title which in my case is John Cook, RN. Not all commas are equal. By substituting a different comma for the one Project doesn’t like, it’s possible to add a comma to a resource name. Figure 1. Error message after trying to add a comma to a resource name in Microsoft Project 2013. The error message refers to “the list separator character” that is commonly...(read more)

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  • Loading Wavefront Data into VAO and Render It

    - by Jordan LaPrise
    I have successfully loaded a triangulated wavefront(.obj) into 6 vectors, the first 3 vectors contain the locations for vertices, uv coords, and normals. The last three have the indices stored for each of the faces. I have been looking into using VAO's and VBO's to render, and I'm not quite sure how to load and render the data. One of my biggest concerns is the fact that indexed rendering only allows you to have one array of indices, meaning I somehow have to make all of the first three vectors the same size, the only way I thought of doing this, is to make 3 new vertex's of equal size, and load in the data for each face, but that would completely defeat the purpose of indexing. Any help would be appreciated. Thanks in advance, Jordan

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  • Swap and hibernation

    - by maaartinus
    I saw a lot of recommendations claiming that for hibernation the swap partition/file must be at least as large as the main memory. This makes no sense to me. Lets assume I have 8 GB of main memory and 8 GB swap area and want to hibernate: case 1: I'm using 4 GB of virtual memory - 8 GB of swap is unnecessarily large. case 2: I'm using 8 GB of virtual memory - 8 GB of swap is just right. case 3: I'm using 12 GB of virtual memory - 8 GB of swap is too small. The outcome is: A swap area of size equal to the memory size is sufficient for hibernate IFF it doesn't get used for swapping at all. So what is the reason behind the claim that you need at least as much swap area as main memory for hibernate to work? I know that virtual memory gets used for caching too, and that the cache may be simply discarded, but what happens to hibernation if a program allocates 12 GB of virtual memory (given the above memory and swap sizes)?

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  • The Joy Of Hex

    - by Jim Giercyk
    While working on a mainframe integration project, it occurred to me that some basic computer concepts are slipping into obscurity. For example, just about anyone can tell you that a 64-bit processor is faster than a 32-bit processer. A grade school child could tell you that a computer “speaks” in ‘1’s and ‘0’s. Some people can even tell you that there are 8 bits in a byte. However, I have found that even the most seasoned developers often can’t explain the theory behind those statements. That is not a knock on programmers; in the age of IntelliSense, what reason do we have to work with data at the bit level? Many computer theory classes treat bit-level programming as a thing of the past, no longer necessary now that storage space is plentiful. The trouble with that mindset is that the world is full of legacy systems that run programs written in the 1970’s.  Today our jobs require us to extract data from those systems, regardless of the format, and that often involves low-level programming. Because it seems knowledge of the low-level concepts is waning in recent times, I thought a review would be in order.       CHARACTER: See Spot Run HEX: 53 65 65 20 53 70 6F 74 20 52 75 6E DECIMAL: 83 101 101 32 83 112 111 116 32 82 117 110 BINARY: 01010011 01100101 01100101 00100000 01010011 01110000 01101111 01110100 00100000 01010010 01110101 01101110 In this example, I have broken down the words “See Spot Run” to a level computers can understand – machine language.     CHARACTER:  The character level is what is rendered by the computer.  A “Character Set” or “Code Page” contains 256 characters, both printable and unprintable.  Each character represents 1 BYTE of data.  For example, the character string “See Spot Run” is 12 Bytes long, exclusive of the quotation marks.  Remember, a SPACE is an unprintable character, but it still requires a byte.  In the example I have used the default Windows character set, ASCII, which you can see here:  http://www.asciitable.com/ HEX:  Hex is short for hexadecimal, or Base 16.  Humans are comfortable thinking in base ten, perhaps because they have 10 fingers and 10 toes; fingers and toes are called digits, so it’s not much of a stretch.  Computers think in Base 16, with numeric values ranging from zero to fifteen, or 0 – F.  Each decimal place has a possible 16 values as opposed to a possible 10 values in base 10.  Therefore, the number 10 in Hex is equal to the number 16 in Decimal.  DECIMAL:  The Decimal conversion is strictly for us humans to use for calculations and conversions.  It is much easier for us humans to calculate that [30 – 10 = 20] in decimal than it is for us to calculate [1E – A = 14] in Hex.  In the old days, an error in a program could be found by determining the displacement from the entry point of a module.  Since those values were dumped from the computers head, they were in hex. A programmer needed to convert them to decimal, do the equation and convert back to hex.  This gets into relative and absolute addressing, a topic for another day.  BINARY:  Binary, or machine code, is where any value can be expressed in 1s and 0s.  It is really Base 2, because each decimal place can have a possibility of only 2 characters, a 1 or a 0.  In Binary, the number 10 is equal to the number 2 in decimal. Why only 1s and 0s?  Very simply, computers are made up of lots and lots of transistors which at any given moment can be ON ( 1 ) or OFF ( 0 ).  Each transistor is a bit, and the order that the transistors fire (or not fire) is what distinguishes one value from  another in the computers head (or CPU).  Consider 32 bit vs 64 bit processing…..a 64 bit processor has the capability to read 64 transistors at a time.  A 32 bit processor can only read half as many at a time, so in theory the 64 bit processor should be much faster.  There are many more factors involved in CPU performance, but that is the fundamental difference.    DECIMAL HEX BINARY 0 0 0000 1 1 0001 2 2 0010 3 3 0011 4 4 0100 5 5 0101 6 6 0110 7 7 0111 8 8 1000 9 9 1001 10 A 1010 11 B 1011 12 C 1100 13 D 1101 14 E 1110 15 F 1111   Remember that each character is a BYTE, there are 2 HEX characters in a byte (called nibbles) and 8 BITS in a byte.  I hope you enjoyed reading about the theory of data processing.  This is just a high-level explanation, and there is much more to be learned.  It is safe to say that, no matter how advanced our programming languages and visual studios become, they are nothing more than a way to interpret bits and bytes.  There is nothing like the joy of hex to get the mind racing.

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  • How to fill certain application design learning "gaps"?

    - by e4rthdog
    In life it doesnt matter if you do one thing for 15 years. You will end up waking one day and asking stuff that are equal to "how do i walk?" :) My specific question is that as a new entrant to C# and OOP i am stepping into many little "details" that need to be addressed. Written a lot of code in VB.NET / cobol / simple php e.t.c surely does not help much into the OOP world... So , even after reading entry level books for C# and watching some videos i recently found out about the "factory model design" for applications. I would appreciate if any of you guys recomment some reading on application design architecture for further reading...

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  • Open source framework quality [closed]

    - by Jonas Byström
    It's not hard to find snippets, components or tools/toolkits in the open source world which holds the quality bar really high. Myself I use git, python, linux, gcc, bash and a whole range of others on a daily basis, and I love them. But when it comes to bigger frameworks, which are intended for facilitating larger tasks of an application without much interference, I'm not as enthusiastic. I've tried a few commercial frameworks (game engines), which were okay, but all big open source frameworks which I've used myself, or which I have seen used in applications were decidedly worse than the commercial equivalent. But I'm not sure if my experience was typical. Where have bigger open source frameworks for facilitating larger tasks of an application been able to equal or exceed commercial frameworks, and how were they better?

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  • Recommended reading for (Object Oriented) application design architecture?

    - by e4rthdog
    In life it doesnt matter if you do one thing for 15 years. You will end up waking one day and asking stuff that are equal to "how do i walk?" :) My specific question is that as a new entrant to C# and OOP i am stepping into many little "details" that need to be addressed. Written a lot of code in VB.NET / cobol / simple php e.t.c surely does not help much into the OOP world... So , even after reading entry level books for C# and watching some videos i recently found out about the "factory model design" for applications. I would appreciate if any of you guys recomment some reading on application design architecture for further reading...

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  • I need help with algorithms, how do I improve?

    - by David Burr
    I usually do well at figuring out solutions to programming assignments but for some reason, I'm really struggling in my Algorithms class. I'm not failing but I know I can do better. When I'm confronted with problems like "Divide the array to 2 subarrays so that the sum of each subarray is equal to the other subarray," I feel like my brain won't cooperate and think and I end up not being able to solve it. Some of the things I'm doing right now to help myself: reading CLR (1st ed.) -- it takes a lot of time for stuff to sink in and I can't understand most of it solving some problems -- no matter how much I try, most of the time, I end up googling for the solution before I understand how to solve it I know that good algorithmic skills are very important because lots of good companies ask these sorts of questions in their interview process so I'm a bit worried right now. What else can can I do to improve my algorithmic/problem solving skills? Any advice on how to deal with this?

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