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  • Binary Search Tree can't delete the root

    - by Ali Zahr
    Everything is working fine in this function, but the problem is that I can't delete the root, I couldn't figure out what's the bug here.I've traced the "else part" it works fine until the return, it returns the old value I don't know why. Plz Help! node *removeNode(node *Root, int key) { node *tmp = new node; if(key > Root->value) Root->right = removeNode(Root->right,key); else if(key < Root->value) Root->left = removeNode(Root->left, key); else if(Root->left != NULL && Root->right != NULL) { node *minNode = findNode(Root->right); Root->value = minNode->value; Root->right = removeNode(Root->right,Root->value); } else { tmp = Root; if(Root->left == NULL) Root = Root->right; else if(Root->right == NULL) Root = Root->left; delete tmp; } return Root; }

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  • mysql category tree search

    - by ffffff
    I have the following schema on MySQL 5.1 CREATE TABLE `mytest` ( `category` varchar(32) , `item_name` varchar(255) KEY `key1` (`category`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1; category column is filled with like that [:parent_parent_cat_id][:parent_cat_id][:leaf_cat_id] "10000200003000" if you can search all of the under categories :parent_parent_category_id SELECT * FROM mytest WHERE category LIKE "10000%"; it's using index key1; but How to use index when I wanna search :parent_cat_id? SELECT * FROM mytest WHERE category LIKE "%20000%"; Do you have a better solutions?

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  • Express XPath as an expression tree

    - by 47d_
    If I have an XPath query like NodeA/NodeB[@WIDTH and not(@WIDTH="20")] | NodeC[@WIDTH and not(@WIDTH="20")]/NodeD Is there any API available to visualize this XPath query as a stack of atomic expressions, something like (following is generic) Get results of NodeA, call it "first set" Get results of NodeB from "first set" Filter where [@WIDTH and not(@WIDTH="20")] Filter NodeD, call this "node d for B" Get results of NodeC from "first set" Filter where [@WIDTH and not(@WIDTH="20")] Filter NodeD, call this "node d for C" Combine "node d for B" and "node d for C" I am trying to see if we can convert the XPath expression into custom expression which is close to english and vice versa. If no API is available, what would be the best approach? Thanks in advance.

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  • transform file/directory structure into 'tree' in javascript

    - by dave
    I have an array of objects that looks like this: [{ name: 'test', size: 0, type: 'directory', path: '/storage/test' }, { name: 'asdf', size: 170, type: 'directory', path: '/storage/test/asdf' }, { name: '2.txt', size: 0, type: 'file', path: '/storage/test/asdf/2.txt' }] There could be any number of arbitrary path's, this is the result of iterating through files and folders within a directory. What I'm trying to do is determine the 'root' node of these. Ultimately, this will be stored in mongodb and use materialized path to determine it's relationships. In this example, /storage/test is a root with no parent. /storage/test/asdf has the parent of /storage/test which is the parent to /storage/test/asdf/2.txt. My question is, how would you go about iterating through this array, to determine the parent's and associated children? Any help in the right direction would be great! Thank you

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  • Javascript / Jquery Tree Travesal question

    - by Copper
    Suppose I have the following <ul> <li>Item 1</li> <li>Item 2 <ul> <li>Sub Item</li> </ul> </li> <li>Item 3</li> </ul> This list is auto-generated by some other code (so adding exclusive id's/class' is out of the question. Suppose I have some jquery code that states that if I mouseover an li, it gets a background color. However, if I mouseover the "Sub Item" list item, "Item 2" will be highlighted as well. How can I make it so that if the user mouses over "Sub Item" it only puts a background color on that and not on "Item 2" as well?

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  • How can I troubleshoot flash player/hardware conflict?

    - by sparthikas
    OBJECTIVE: Have a web browser on my Ubuntu install that can play youtube and hulu videos. Also would like to understand problem so that I can fix it again if I change software. Workarounds welcome, technical understanding and solution preferable. SYMPTOMS: Flash does not run - cannot make the right-click menu appear, an empty box is where video should be, changes to black box when hovering over other links. The "The Adobe Flash plugin has crashed" message does not appear with its sad lego face. cannot activate proprietary graphics driver - causes system to reboot to a prompt. SOLUTIONS TRIED: Replaced OS (tried slackware 13.37, fedora 17, linuxmint 13 maya, gentoo, lubuntu, and even winXP. lubuntu confirmed to work, don't remember how much tweaking, if any, this required. Slack, fedora, mint, and gentoo all failed to run flash just like ubuntu) many reinstalls of flash player via different methods, including cleaning up old installs first, also tried gnash and lightspark. replaced graphics card (replaced HIS IceQ Radeon HD 4670 AGP with older GeForce 5700 LE no change in problem) flash does successfully work on winXP installation with Catalyst AGP hotfix driver applied, however I consider windows wholly unacceptable for web browsing due to lack of security. Lubuntu install also works, however I do not want to be tied down to just using Lubuntu on this computer. SYSINFO: Have latest versions of Ubuntu, Firefox, and Flash on fresh Ubuntu install. Using Gigabyte 7s748 motherboard with Athlon XP 2800+ and 3 GB of RAM with Radeon HD 4670 AGP card, also a Dell soundblaster live series sound card (due to malfunction of onboard sound on motherboard) Wired internet connection, Maxtor 6Y120L3 HDD, Sony DVD RW AW-Q170A, Dell M993s monitor. NOTES: I do not know if the graphics driver issue and the flash troubles are linked. Substitution of older graphics card having same flash troubles seems to suggest they aren't. My troubleshooting method is rather reductionist, consisting mainly of "replace things until you find out what was causing the error by process of elimination" only it seems that this must be a conflict which arises when software decides how to configure itself on my hardware. That is, I know the hardware can run Flash, and I know that on other systems the same software can too, but somehow the combination fails. Consequently I feel out of my depth. I will keep trying things off and on, but I have spent probably about 30 man-hours in the last 4 months working on this problem with no joy other than the lubuntu workaround. Any help will be appreciated, I will be checking back and posting updates. Any pertinent questions regarding me or my computer will be answered, outputs from config files can be accessed and posted (IDK which ones or what parts to post).

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  • How to find Sub-trees in non-binary tree

    - by kenny
    I have a non-binary tree. I want to find all "sub-trees" that are connected to root. Sub-tree is a a link group of tree nodes. every group is colored in it's own color. What would be be the best approach? Run recursion down and up for every node? The data structure of every treenode is a list of children, list of parents. (the type of children and parents are treenodes) Clarification: Group defined if there is a kind of "closure" between nodes where root itself is not part of the closure. As you can see from the graph you can't travel from pink to other nodes (you CAN NOT use root). From brown node you can travel to it's child so this form another group. Finally you can travel from any cyan node to other cyan nodes so the form another group

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  • GUI question : representing large tree

    - by Peter
    I have a tree-like datastructure of some six levels deep, that I would like to represent on a single webpage (can be tabs, trees; ....) In each level both childnodes and content are possible. Presenting it like a real tree would be not very usable (too big). I was thinking in the lines of hiding parts of the tree when you drill down and presenting a breadcrumbs or the like to keep you informed as to where you are... I guess my question boils down to : any ideas / examples ? Tx!

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  • How to find siblings of a tree?

    - by smallB
    On my interview for an internship, I was asked following question: On a whiteboard write the simplest algorithm with use of recursion which would take a root of a so called binary tree (so called because it is not strictly speaking binary tree) and make every child in this tree connected with its sibling. So if I have: 1 / \ 2 3 / \ \ 4 5 6 / \ 7 8 then the sibling to 2 would be 3, to four five, to five six and to seven eight. I didn't do this, although I was heading in the right direction. Later (next day) at home I did it, but with the use of a debugger. It took me better part of two hours and 50 lines of code. I personally think that this was very difficult question, almost impossible to do correctly on a whiteboard. How would you solve it on a whiteboard? How to apprehend this question without using a debugger?

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  • Setup MSSQL replication with peer to peer topology: problem setting up Conflict Detection

    - by Roel
    Hi, I'm setting up a SQL Replication strategy, using MSSQL2008 with peer-to-peer publications (2 servers, each one subscribes to the other). I followed this HOWTO from MSDN, and the setup seems to be working fine: add a record to one table on server A, query on server B shows the new record. So far, so good. So far I only have one table 'Templates': Id PK (calculated field) NodeId int default 1/2 (Server A = 1, Server B = 2) LocalId int autoid Name nvarchar(100) Now, I would like to enable 'Conflict detection', which should be enabled by default. But every time I try to save the 'Conflict Detection' feature in the Publication Properties I get the following error: Cannot save Peer conflict detection properties. An exception occurred while executing a Transact-SQL statement or batch.(Microsoft.SqlServer.ConnectionInfo) Program Location: at Microsoft.SqlServer.Management.Common.ServerConnection.ExecuteNonQuery(String sqlCommand, ExecutionTypes executionType) at Microsoft.SqlServer.Management.Common.ServerConnection.ExecuteNonQuery(String sqlCommand) at Microsoft.SqlServer.Replication.ReplicationObject.ExecCommand(String commandIn) at Microsoft.SqlServer.Replication.TransPublication.SetPeerConflictDetection(Boolean enablePeerConflictDetection, Int32 peerOriginatorID) at Microsoft.SqlServer.Management.UI.PubPropSubscriptionOptions.SaveP2PConflictDetection() at Microsoft.SqlServer.Management.UI.PubPropSubscriptionOptions.SaveProperties(ExecutionMode& executionResult) Column name 'Id' does not exist in the target table or view. Changed database context to 'TestDB'. (.Net SqlClient Data Provider) For help, click: http://go.microsoft.com/fwlink?ProdName=Microsoft+SQL+Server&ProdVer=10.00.2531&EvtSrc=MSSQLServer&EvtID=1911&LinkId=20476 Server Name: SERVER_A Error Number: 1911 Severity: 16 State: 1 Line Number: 2 Program Location: at System.Data.SqlClient.SqlConnection.OnError(SqlException exception, Boolean breakConnection) at System.Data.SqlClient.SqlInternalConnection.OnError(SqlException exception, Boolean breakConnection) at System.Data.SqlClient.TdsParser.ThrowExceptionAndWarning(TdsParserStateObject stateObj) at System.Data.SqlClient.TdsParser.Run(RunBehavior runBehavior, SqlCommand cmdHandler, SqlDataReader dataStream, BulkCopySimpleResultSet bulkCopyHandler, TdsParserStateObject stateObj) at System.Data.SqlClient.SqlCommand.RunExecuteNonQueryTds(String methodName, Boolean async) at System.Data.SqlClient.SqlCommand.InternalExecuteNonQuery(DbAsyncResult result, String methodName, Boolean sendToPipe) at System.Data.SqlClient.SqlCommand.ExecuteNonQuery() at Microsoft.SqlServer.Management.Common.ServerConnection.ExecuteNonQuery(String sqlCommand, ExecutionTypes executionType) Now, I googled the hell out of this error, and nothing shows up. I also can't seem to find out what the exact target table of the error "Column name 'Id' does not exist..." is. Has anyone every done this successfully? Am I missing something? Having this setup without conflict detection feels pretty useless... EDIT OK, so after some more research and setting up with different databases etc, I found out that the calculated 'Id' column of the Templates table is the culprit. I don't know why, but the replication doesn't seem to allow calculated columns (which are also primary key). It works now too, without the 'Id' column, and using the NodeId and LocalId as a combined PK. So now the question is, why isn't it allowed to have a calculated column as PK for replication with conflict detection?

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  • How to configure Beyond Compare 3 for Eclipse conflict resolution?

    - by Peter Boughton
    What is the correct parameters to get Beyond Compare 3 working with Eclipse/Subclipse conflict resolution? In Preferences > Team > SVN > Diff/Merge there's the option to specify an external program to resolve conflicts. The default parameters are: "${yours}" "${theirs}" "${base}" "${merged}" And it suggests TortoiseMerge settings of this: /theirs:"${theirs}" /base:"${base}" /mine:"${yours}" /merged:"${merged}" But what is the appropriate config for Beyond Compare?

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  • Quick Outline: Navigating Your PL/SQL Packages in Oracle SQL Developer

    - by thatjeffsmith
    If you’re browsing your packages using the Connections panel, you have a nice tree navigator to click around your packages and your variable, procedure, and functions. Click, click, click all day long, click, click, click while I sing this song… But What if you drill into your PL/SQL source from the worksheet and don’t have the Tree expanded? Let’s say you’re working on your script, something like - Hmm, what goes next again? So I need to reacquaint myself with just what my beer package requires, so I’m going to drill into it by doing a DESCRIBE (via SHIFT+F4), and now I have the package open. The package is open but the tree hasn’t auto-expanded. Please don’t tell me I have to do the click-click-click thing in the tree!?! Just Open the Quick Outline Panel Do you see it? Just right click in the procedure editor – select the ‘Quick Outline’ in the context menu, and voila! The navigational power of the tree, without needing to drill down the tree itself. If I want to drill into my procedure declaration, just click on said procedure name in the Quick Outline panel. This works for both package specs and bodies. Technically you can use this for stand alone procedures and functions, but the real power is demonstrated for packages.

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  • Asymptotic runtime of list-to-tree function

    - by Deestan
    I have a merge function which takes time O(log n) to combine two trees into one, and a listToTree function which converts an initial list of elements to singleton trees and repeatedly calls merge on each successive pair of trees until only one tree remains. Function signatures and relevant implementations are as follows: merge :: Tree a -> Tree a -> Tree a --// O(log n) where n is size of input trees singleton :: a -> Tree a --// O(1) empty :: Tree a --// O(1) listToTree :: [a] -> Tree a --// Supposedly O(n) listToTree = listToTreeR . (map singleton) listToTreeR :: [Tree a] -> Tree a listToTreeR [] = empty listToTreeR (x:[]) = x listToTreeR xs = listToTreeR (mergePairs xs) mergePairs :: [Tree a] -> [Tree a] mergePairs [] = [] mergePairs (x:[]) = [x] mergePairs (x:y:xs) = merge x y : mergePairs xs This is a slightly simplified version of exercise 3.3 in Purely Functional Data Structures by Chris Okasaki. According to the exercise, I shall now show that listToTree takes O(n) time. Which I can't. :-( There are trivially ceil(log n) recursive calls to listToTreeR, meaning ceil(log n) calls to mergePairs. The running time of mergePairs is dependent on the length of the list, and the sizes of the trees. The length of the list is 2^h-1, and the sizes of the trees are log(n/(2^h)), where h=log n is the first recursive step, and h=1 is the last recursive step. Each call to mergePairs thus takes time (2^h-1) * log(n/(2^h)) I'm having trouble taking this analysis any further. Can anyone give me a hint in the right direction?

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  • Binary Search Tree, cannot do traversal

    - by ihm
    Please see BST codes below. It only outputs "5". what did I do wrong? #include <iostream> class bst { public: bst(const int& numb) : root(new node(numb)) {} void insert(const int& numb) { root->insert(new node(numb), root); } void inorder() { root->inorder(root); } private: class node { public: node(const int& numb) : left(NULL), right(NULL) { value = numb; } void insert(node* insertion, node* position) { if (position == NULL) position = insertion; else if (insertion->value > position->value) insert(insertion, position->right); else if (insertion->value < position->value) insert(insertion, position->left); } void inorder(node* tree) { if (tree == NULL) return; inorder(tree->left); std::cout << tree->value << std::endl; inorder(tree->right); } private: node* left; node* right; int value; }; node* root; }; int main() { bst tree(5); tree.insert(4); tree.insert(2); tree.insert(10); tree.insert(14); tree.inorder(); return 0; }

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  • How to cleanly add after-the-fact commits from the same feature into git tree

    - by Dennis
    I am one of two developers on a system. I make most of the commits at this time period. My current git workflow is as such: there is master branch only (no develop/release) I make a new branch when I want to do a feature, do lots of commits, and then when I'm done, I merge that branch back into master, and usually push it to remote. ...except, I am usually not done. I often come back to alter one thing or another and every time I think it is done, but it can be 3-4 commits before I am really done and move onto something else. Problem The problem I have now is that .. my feature branch tree is merged and pushed into master and remote master, and then I realize that I am not really done with that feature, as in I have finishing touches I want to add, where finishing touches may be cosmetic only, or may be significant, but they still belong to that one feature I just worked on. What I do now Currently, when I have extra after-the-fact commits like this, I solve this problem by rolling back my merge, and re-merging my feature branch into master with my new commits, and I do that so that git tree looks clean. One clean feature branch branched out of master and merged back into it. I then push --force my changes to origin, since my origin doesn't see much traffic at the moment, so I can almost count that things will be safe, or I can even talk to other dev if I have to coordinate. But I know it is not a good way to do this in general, as it rewrites what others may have already pulled, causing potential issues. And it did happen even with my dev, where git had to do an extra weird merge when our trees diverged. Other ways to solve this which I deem to be not so great Next best way is to just make those extra commits to the master branch directly, be it fast-forward merge, or not. It doesn't make the tree look as pretty as in my current way I'm solving this, but then it's not rewriting history. Yet another way is to wait. Maybe wait 24 hours and not push things to origin. That way I can rewrite things as I see fit. The con of this approach is time wasted waiting, when people may be waiting for a fix now. Yet another way is to make a "new" feature branch every time I realize I need to fix something extra. I may end up with things like feature-branch feature-branch-html-fix, feature-branch-checkbox-fix, and so on, kind of polluting the git tree somewhat. Is there a way to manage what I am trying to do without the drawbacks I described? I'm going for clean-looking history here, but maybe I need to drop this goal, if technically it is not a possibility.

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  • How can I implement a splay tree that performs the zig operation last, not first?

    - by Jakob
    For my Algorithms & Data Structures class, I've been tasked with implementing a splay tree in Haskell. My algorithm for the splay operation is as follows: If the node to be splayed is the root, the unaltered tree is returned. If the node to be splayed is one level from the root, a zig operation is performed and the resulting tree is returned. If the node to be splayed is two or more levels from the root, a zig-zig or zig-zag operation is performed on the result of splaying the subtree starting at that node, and the resulting tree is returned. This is valid according to my teacher. However, the Wikipedia description of a splay tree says the zig step "will be done only as the last step in a splay operation" whereas in my algorithm it is the first step in a splay operation. I want to implement a splay tree that performs the zig operation last instead of first, but I'm not sure how it would best be done. It seems to me that such an algorithm would become more complex, seeing as how one needs to find the node to be splayed before it can be determined whether a zig operation should be performed or not. How can I implement this in Haskell (or some other functional language)?

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  • Sentence Tree v/s Words List

    - by Rohit Jose
    I was recently tasked with building a Name Entity Recognizer as part of a project. The objective was to parse a given sentence and come up with all the possible combinations of the entities. One approach that was suggested was to keep a lookup table for all the know connector words like articles and conjunctions, remove them from the words list after splitting the sentence on the basis of the spaces. This would leave out the Name Entities in the sentence. A lookup is then done for these identified entities on another lookup table that associates them to the entity type, for example if the sentence was: Remember the Titans was a movie directed by Boaz Yakin, the possible outputs would be: {Remember the Titans,Movie} was {a movie,Movie} directed by {Boaz Yakin,director} {Remember the Titans,Movie} was a movie directed by Boaz Yakin {Remember the Titans,Movie} was {a movie,Movie} directed by Boaz Yakin {Remember the Titans,Movie} was a movie directed by {Boaz Yakin,director} Remember the Titans was {a movie,Movie} directed by Boaz Yakin Remember the Titans was {a movie,Movie} directed by {Boaz Yakin,director} Remember the Titans was a movie directed by {Boaz Yakin,director} Remember the {the titans,Movie,Sports Team} was {a movie,Movie} directed by {Boaz Yakin,director} Remember the {the titans,Movie,Sports Team} was a movie directed by Boaz Yakin Remember the {the titans,Movie,Sports Team} was {a movie,Movie} directed by Boaz Yakin Remember the {the titans,Movie,Sports Team} was a movie directed by {Boaz Yakin,director} The entity lookup table here would contain the following data: Remember the Titans=Movie a movie=Movie Boaz Yakin=director the Titans=Movie the Titans=Sports Team Another alternative logic that was put forward was to build a crude sentence tree that would contain the connector words in the lookup table as parent nodes and do a lookup in the entity table for the leaf node that might contain the entities. The tree that was built for the sentence above would be: The question I am faced with is the benefits of the two approaches, should I be going for the tree approach to represent the sentence parsing, since it provides a more semantic structure? Is there a better approach I should be going for solving it?

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  • vector rotations for branches of a 3d tree

    - by freefallr
    I'm attempting to create a 3d tree procedurally. I'm hoping that someone can check my vector rotation maths, as I'm a bit confused. I'm using an l-system (a recursive algorithm for generating branches). The trunk of the tree is the root node. It's orientation is aligned to the y axis. In the next iteration of the tree (e.g. the first branches), I might create a branch that is oriented say by +10 degrees in the X axis and a similar amount in the Z axis, relative to the trunk. I know that I should keep a rotation matrix at each branch, so that it can be applied to child branches, along with any modifications to the child branch. My questions then: for the trunk, the rotation matrix - is that just the identity matrix * initial orientation vector ? for the first branch (and subsequent branches) - I'll "inherit" the rotation matrix of the parent branch, and apply x and z rotations to that also. e.g. using glm::normalize; using glm::rotateX; using glm::vec4; using glm::mat4; using glm::rotate; vec4 vYAxis = vec4(0.0f, 1.0f, 0.0f, 0.0f); vec4 vInitial = normalize( rotateX( vYAxis, 10.0f ) ); mat4 mRotation = mat4(1.0); // trunk rotation matrix = identity * initial orientation vector mRotation *= vInitial; // first branch = parent rotation matrix * this branches rotations mRotation *= rotate( 10.0f, 1.0f, 0.0f, 0.0f ); // x rotation mRotation *= rotate( 10.0f, 0.0f, 0.0f, 1.0f ); // z rotation Are my maths and approach correct, or am I completely wrong? Finally, I'm using the glm library with OpenGL / C++ for this. Is the order of x rotation and z rotation important?

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  • Cloning from a given point in the snapshot tree

    - by Fat Bloke
    Although we have just released VirtualBox 4.3, this quick blog entry is about a longer standing ability of VirtualBox when it comes to Snapshots and Cloning, and was prompted by a question posed internally, here in Oracle: "Is there a way I can create a new VM from a point in my snapshot tree?". Here's the scenario: Let's say you have your favourite work VM which is Oracle Linux based and as you installed different packages, such as database, middleware, and the apps, you took snapshots at each point like this: But you then need to create a new VM for some other testing or to share with a colleague who will be using the same Linux and Database layers but may want to reconfigure the Middleware tier, and may want to install his own Apps. All you have to do is right click on the snapshot that you're happy with and clone: Give the VM that you are about to create a name, and if you plan to use it on the same host machine as the original VM, it's a good idea to "Reinitialize the MAC address" so there's no clash on the same network: Now choose the Clone type. If you plan to use this new VM on the same host as the original, you can use Linked Cloning else choose Full.  At this point you now have a choice about what to do about your snapshot tree. In our example, we're happy with the Linux and Database layers, but we may want to allow our colleague to change the upper tiers, with the option of reverting back to our known-good state, so we'll retain the snapshot data in the new VM from this point on: The cloning process then chugs along and may take a while if you chose a Full Clone: Finally, the newly cloned VM is ready with the subset of the Snapshot tree that we wanted to retain: Pretty powerful, and very useful.  Cheers, -FB 

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  • What should be done with class names that conflict (common) framework names

    - by Earlz
    What should be done exactly when the most obvious class name for a component is taken by a framework? In my case, I need to make a class that describes an HTTP request. Of course, the most common name is "taken" as System.Web.HttpRequest. What should I do? This project will be used in a web context, so I'd really rather not force people to not import the System.Web namespace, or type out all of my class names manually. What is the usual way of dealing with this? I can come up with this: Prefix class name with a project shortname Try to come up with a different name that means the same thing(I've tried and can't come up with anything) Force users to choose between namespaces

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  • Conflict between variable substitution and CJK characters in BASH

    - by AndreasT
    I encountered a problem with variable substitution in the BASH shell. Say you define a variable a. Then the command $> echo ${a//[0-4]/} prints its value with all the numbers ranged between 0 and 4 removed: $> a="Hello1265-3World" $> echo ${a//[0-4]/} Hello65-World This seems to work just fine, but let's take a look at the next example: $> b="?1265-3?" $> echo ${b//[0-4]/} ?1265-3? Substitution did not take place: I assume that is because b contains CJK characters. This issue extends to all cases in which square brackets are involved. Surprisingly enough, variable substitution without square brackets works fine in both cases: $> a="Hello1265-3World" $> echo ${a//2/} Hello165-3World $> b="?1265-3?" $> echo ${b//2/} ?165-3? Is it a bug or am I missing something? I use Lubuntu 12.04, terminal is lxterminal and echo $BASH_VERSION returns 4.2.24(1)-release. EDIT: Andrew Johnson in his comment stated that with gnome-terminal 4.2.37(1)-release the command works fine. I wonder whether it is a problem of lxterminal or of its specific 4.2.24(1)-release version.

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  • DNS - domain conflict?

    - by Stefanos.Ioannou
    I was given two domains: domain.com & domain.info (they are on GoDaddy). And I was also given two servers, 107.105.38.99 - Rails app and 107.107.90.17 - Wordpress platform, on Digital Ocean. At first, I was instructed to associate domain.com with the 107.107.38.99 (Rails app). Then I was instructed to de-associate this IP with domain.com and associated the 107.107.90.17 with the domain name domain.com. Then I was instructed to associated domain.info with the 107.107.38.99 (Rails app). Right now, when I go to domain.com the WordPress platform (107.107.90.17) loads fine and that is what is expected. But when I go to domain.info for the Rails app (107.107.38.99) I get redirected to domain.com. This is not expected and this is really weird for me. When I ping domain.info I get this: PING domain.info (107.107.38.99): 56 data bytes 64 bytes from 107.107.38.99: icmp_seq=0 ttl=50 time=74.601 ms Which is the expected result showing the correct IP but I don't understand why I get redirected to domain.com...(which when I ping is:) domain 64 bytes from 107.107.90.17: icmp_seq=0 ttl=50 time=75.057 ms The PTR Records on Digital Ocean are as follows: IP Address PTR Record 107.107.38.99 domain.info. 107.107.90.17 domain.com. and the DNS configurations on Digital Ocean are: domain.com A: @ 107.107.90.17 CNAME: * @ domain.info A: @ 107.107.38.99 CNAME: * @ I am not sure what the issue is, if you have any clue please let me know, I will be really grateful. If you need any other info let me know.

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