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  • Reading Binary data from a Serial Port.

    - by rross
    I previously have been reading NMEA data from a GPS via a serial port using C#. Now I'm doing something similar, but instead of GPS from a serial. I'm attempting to read a KISS Statement from a TNC. I'm using this event handler. comport.DataReceived += new SerialDataReceivedEventHandler(port_DataReceived); Here is port_DataReceived. private void port_DataReceived(object sender, SerialDataReceivedEventArgs e) { string data = comport.ReadExisting(); sBuffer = data; try { this.Invoke(new EventHandler(delegate { ProcessBuffer(sBuffer); })); } catch { } } The problem I'm having is that the method is being called several times per statement. So the ProcessBuffer method is being called with only a partial statment. How can I read the whole statement?

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  • latest Xemacs on Windows binary download

    - by anjanb
    I'm trying to get an updated version for Windows vista. I previously got 21.4.22 but it's been a year since that release. The linux versions should be 22.x. I'm wondering if anyone else builds stable binaries for Windows ? 21.4.22 has several bugs and I cannot figure out how to fix them. I know Xemacs is not as active as GNU emacs but still aren't there any Xemacs users on windows who build their own copies even if the official site doesn't ? I would like to be able to compare buffers, files and directories apart from being able to edit any file : java, javascript, ruby, .bat, .sh, .xml, etc.

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  • Getting Image size of JPEG from its binary

    - by rajeshsr
    Hi I have a lot of jpeg files with varying image size. For instance, here is the first 640 bytes as given by hexdump of an image of size 256*384(pixels): 0000000: ffd8 ffe0 0010 4a46 4946 0001 0101 0048 ......JFIF.....H 0000010: 0048 0000 ffdb 0043 0003 0202 0302 0203 .H.....C........ 0000020: 0303 0304 0303 0405 0805 0504 0405 0a07 ................ 0000030: 0706 080c 0a0c 0c0b 0a0b 0b0d 0e12 100d ................ I guess the size information mus be within these lines. But am unable to see which bytes give the sizes correctly. Can anyone help me find the fields that contains the size information?

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  • Need algorithm to add Node in binary tree

    - by m.qayyum
    •if your new element is less or equal than the current node, you go to the left subtree, otherwise to the right subtree and continue traversing •if you arrived at a node, where you can not go any deeper, because there is no subtree, this is the place to insert your new element (5)Root (3)-------^--------(7) (2)---^----(5) ^-----(8) (5)--^ i want to add this last node with data 5...but i can't figure it out...I need a algorithm to do that or in java language

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  • Nicely printing/showing a binary tree in Haskell

    - by nicole
    I have a tree data type: data Tree a b = Branch b (Tree a b) (Tree a b) | Leaf a ...and I need to make it an instance of Show, without using deriving. I have found that nicely displaying a little branch with two leaves is easy: instance (Show a, Show b) => Show (Tree a b) where show (Leaf x) = show x show (Branch val l r) = " " ++ show val ++ "\n" ++ show l ++ " " ++ show r But how can I extend a nice structure to a tree of arbitrary size? It seems like determining the spacing would require me to know just how many leaves will be at the very bottom (or maybe just how many leaves there are in total) so that I can allocate all the space I need there and just work 'up.' I would probably need to call a size function. I can see this being workable, but is that making it harder than it is?

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  • How to implement dynamic binary search for search and insert operations of n element (C or C++)

    - by iecut
    The idea is to use multiple arrays, each of length 2^k, to store n elements, according to binary representation of n.Each array is sorted and different arrays are not ordered in any way. In the above mentioned data structure, SEARCH is carried out by a sequence of binary search on each array. INSERT is carried out by a sequence of merge of arrays of the same length until an empty array is reached. More Detail: Lets suppose we have a vertical array of length 2^k and to each node of that array there attached horizontal array of length 2^k. That is, to the first node of vertical array, a horizontal array of length 2^0=1 is connected,to the second node of vertical array, a horizontal array of length 2^1= 2 is connected and so on. So the insert is first carried out in the first horizontal array, for the second insert the first array becomes empty and second horizontal array is full with 2 elements, for the third insert 1st and 2nd array horiz. array are filled and so on. I implemented the normal binary search for search and insert as follows: int main() { int a[20]= {0}; int n, i, j, temp; int *beg, *end, *mid, target; printf(" enter the total integers you want to enter (make it less then 20):\n"); scanf("%d", &n); if (n = 20) return 0; printf(" enter the integer array elements:\n" ); for(i = 0; i < n; i++) { scanf("%d", &a[i]); } // sort the loaded array, binary search! for(i = 0; i < n-1; i++) { for(j = 0; j < n-i-1; j++) { if (a[j+1] < a[j]) { temp = a[j]; a[j] = a[j+1]; a[j+1] = temp; } } } printf(" the sorted numbers are:"); for(i = 0; i < n; i++) { printf("%d ", a[i]); } // point to beginning and end of the array beg = &a[0]; end = &a[n]; // use n = one element past the loaded array! // mid should point somewhere in the middle of these addresses mid = beg += n/2; printf("\n enter the number to be searched:"); scanf("%d",&target); // binary search, there is an AND in the middle of while()!!! while((beg <= end) && (*mid != target)) { // is the target in lower or upper half? if (target < *mid) { end = mid - 1; // new end n = n/2; mid = beg += n/2; // new middle } else { beg = mid + 1; // new beginning n = n/2; mid = beg += n/2; // new middle } } // find the target? if (*mid == target) { printf("\n %d found!", target); } else { printf("\n %d not found!", target); } getchar(); // trap enter getchar(); // wait return 0; } Could anyone please suggest how to modify this program or a new program to implement dynamic binary search that works as explained above!!

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  • Recursive Binary Search Tree Insert

    - by Nick Sinklier
    So this is my first java program, but I've done c++ for a few years. I wrote what I think should work, but in fact it does not. So I had a stipulation of having to write a method for this call: tree.insertNode(value); where value is an int. I wanted to write it recursively, for obvious reasons, so I had to do a work around: public void insertNode(int key) { Node temp = new Node(key); if(root == null) root = temp; else insertNode(temp); } public void insertNode(Node temp) { if(root == null) root = temp; else if(temp.getKey() <= root.getKey()) insertNode(root.getLeft()); else insertNode(root.getRight()); } Thanks for any advice.

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  • Reverse reading WORD from a binary file?

    - by Angel
    Hi, I have a structure: struct JFIF_HEADER { WORD marker[2]; // = 0xFFD8FFE0 WORD length; // = 0x0010 BYTE signature[5]; // = "JFIF\0" BYTE versionhi; // = 1 BYTE versionlo; // = 1 BYTE xyunits; // = 0 WORD xdensity; // = 1 WORD ydensity; // = 1 BYTE thumbnwidth; // = 0 BYTE thumbnheight; // = 0 }; This is how I read it from the file: HANDLE file = CreateFile(filename, GENERIC_READ, FILE_SHARE_READ, NULL, OPEN_EXISTING, FILE_ATTRIBUTE_NORMAL, 0); DWORD tmp = 0; DWORD size = GetFileSize(file, &tmp); BYTE *DATA = new BYTE[size]; ReadFile(file, DATA, size, &tmp, 0); JFIF_HEADER header; memcpy(&header, DATA, sizeof(JFIF_HEADER)); This is how the beginning of my file looks in hex editor: 0xFF 0xD8 0xFF 0xE0 0x00 0x10 0x4A 0x46 0x49 0x46 0x00 0x01 0x01 0x00 0x00 0x01 When I print header.marker, it shows exactly what it should (0xFFD8FFE0). But when I print header.length, it shows 0x1000 instead of 0x0010. The same thing is with xdensity and ydensity. Why do I get wrong data when reading a WORD? Thank you.

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  • Return parent of node in Binary Tree

    - by user188995
    I'm writing a code to return the parent of any node, but I'm getting stuck. I don't want to use any predefined ADTs. //Assume that nodes are represented by numbers from 1...n where 1=root and even //nos.=left child and odd nos=right child. public int parent(Node node){ if (node % 2 == 0){ if (root.left==node) return root; else return parent(root.left); } //same case for right } But this program is not working and giving wrong results. My basic algorithm is that the program starts from the root checks if it is on left or on the right. If it's the child or if the node that was queried else, recurses it with the child.

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  • Is there a search engine that indexes source code of a web-page?

    - by Dexter
    I need to search the web for sites that are in our industry that use the same Adwords management company, to ensure that the said company is not violating our contract, as they have been accused of doing. They use a tracking code in the template of every page which has a certain domain in the URL, and I'm wondering if it's possible "Google" the source code using some bot that crawls the code rather than the content? For example, I bought an unlimited license for an image gallery, and I was asked to type the license number in a comment just before the script. I thought it was just so a human could look at the source and find out if someone paid, but it turned out that it was actually that they had a crawler looking for their source code and that comment. If it ran across the code on your site, it would look for the comment, and if it found one, it would check to see if it was an existing one. If not, it would first notify you of your noncompliance, and then notify the owner of the script. Edit: I'm looking to index HTML and JavaScript only, not the server-side languages or Java.

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  • .NET binary serialization conditionally without ISerializable

    - by SillyWhy
    I got 2 classes, for example: public class A { private B b; ... } public class B { ... } I need to serialize an object A using BinaryFormatter. When remoting it shall include the field b, but not when serialize to file. Here is what I added: [Serializable] public class A : MarshalByRefObject { private B b; [OnSerializing] private void OnSerializing(StreamingContext context) { if (context.State == StreamingContextStates.File) { this.b = null; } } ... } [Serializable] public class B : MarshalByRefObject { ... } I think this is a bad design because if another class C also contains B, in class C we must add the duplicate OnSerializing() logic as in A. Class B should decide what to do, not class A or C. I don't want to use ISerializable interface because there are too many variables in class B have to be added to SerializationInfo. I can create a SerializationSurrogate for class B, which perform nothing in GetObjectData() & SetObjectData(), then use it when serializing to file. However the same maintenance issue because whoever modify class B can't notice what going to happen during serialization & the existence of SerializationSurrogate. Is there a better alternative?

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  • Write to a binary file?

    - by rick irby
    Here is data structure w/ variables: struct Part_record { char id_no[3]; int qoh; string desc; double price: }; --- (Using "cin" to input data) --- Part_record null_part = {" ", 0," ",0.0}; --- --- file.seekg( -(long)sizeof(Part_record), ios::cur); file.write( ( char *)&part, sizeof(Part_record) ); The three variables, qoh, Id_no & price, write out correctly, but the "desc" variable is not right. Do I need to initialize Part_record some other way? It should be 20 characters in length. If you have enough info here, pls share your advice,thanks.

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  • Error inserting data in binary tree

    - by chepe263
    I copied this code (in spanish) http://www.elrincondelc.com/nuevorincon/index.php?pag=codigos&id=4 and wrote a new one. This is my code: #include <cstdlib> #include <conio.h> #include <iostream> using namespace std; struct nodoarbol { int dato; struct nodoarbol *izq; struct nodoarbol *der; }; typedef nodoarbol Nodo; typedef Nodo *Arbol; void insertar(Arbol *, int); void inorden(Arbol); void postorden(Arbol); void preorden(Arbol); void insertar(Arbol *raiz, int nuevo){ if (*raiz==NULL){ *raiz = (Nodo *)malloc(sizeof(Nodo)); if (*raiz != NULL){ (*raiz)->dato=nuevo; (*raiz)->der=NULL; (*raiz)->izq=NULL; } else{ cout<<"No hay memoria suficiente u ocurrio un error"; } } else{ if (nuevo < (*raiz)->dato) insertar( &((*raiz)->izq), nuevo ); else if (nuevo > (*raiz)->dato) insertar(&((*raiz)->der), nuevo); } }//inseertar void inorden(Arbol raiz){ if (raiz != NULL){ inorden(raiz->izq); cout << raiz->dato << " "; inorden(raiz->der); } } void preorden(Arbol raiz){ if (raiz != NULL){ cout<< raiz->dato << " "; preorden(raiz->izq); preorden(raiz->der); } } void postorden(Arbol raiz){ if (raiz!=NULL){ postorden(raiz->izq); postorden(raiz->der); cout<<raiz->dato<<" "; } } int main() { int i; i=0; int val; Arbol raiz = NULL; for (i=0; i<10; i++){ cout<<"Inserte un numero"; cin>>val; insertar( (raiz), val); } cout<<"\nPreorden\n"; preorden(raiz); cout<<"\nIneorden\n"; inorden(raiz); cout<<"\nPostorden\n"; postorden(raiz); return 0; } I'm using netbeans 7.1.1, mingw32 compiler This is the output: make[2]: Leaving directory `/q/netbeans c++/NetBeansProjects/treek' make[1]: Leaving directory `/q/netbeans c++/NetBeansProjects/treek' main.cpp: In function 'int main()': main.cpp:110:30: error: cannot convert 'Arbol {aka nodoarbol*}' to 'Nodo** {aka nodoarbol**}' for argument '1' to 'void insertar(Nodo**, int)' make[2]: *** [build/Release/MinGW-Windows/main.o] Error 1 make[1]: *** [.build-conf] Error 2 make: *** [.build-impl] Error 2 BUILD FAILED (exit value 2, total time: 11s) I don't understand what's wrong since i just copied the code (and rewrite it to my own code). I'm really good in php, asp.net (vb) and other languages but c is a headche for me. I've been struggling with this problem for about an hour. Could somebody tell me what could it be?

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  • efficient video format/codec for sparse & binary blob tracking

    - by user391339
    I am working on a blob tracking project and have many high-definition videos that I would like to reduce in size for storage and downstream tracking/shape-analysis. I want to use a lossless method that takes advantage of the black and white nature of the video as well as the fact that not much is moving between individual frames. The videos are quite sparse, with 5 to 10 b&w blobs per frame occupying <30% of the space in total, with each blob moving <5-10% of the field of view between frames and not changing shape too much between 2-3 frames. I will work in Python, Matlab, or LabView for this project, and could use a batch utility if available. It may be worthwhile to export the files as compressed image stacks if a proper video format can't be found. What are the pros and cons of this? A video codec uses correlations between neighboring frames, so it should be more efficient, but not if the wrong one is chosen or if it is improperly configured.

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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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  • How many posibilities on a binary ?

    - by Val
    in hexadecimal "10 10 10 10" system you have 0-255 posibilities right? in total 256 different posibilities as there are 8 1s and 0s. how many different posibilities would i get? if i had 10 digits. instead of 8? or how would i calculate that in php ?

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  • Better way to download a binary file?

    - by geoff
    I have a site where a user can download a file. Some files are extremely large (the largest being 323 MB). When I test it to try and download this file I get an out of memory exception. The only way I know to download the file is below. The reason I'm using the code below is because the URL is encoded and I can't let the user link directly to the file. Is there another way to download this file without having to read the whole thing into a byte array? FileStream fs = new FileStream(context.Server.MapPath(url), FileMode.Open, FileAccess.Read); BinaryReader br = new BinaryReader(fs); long numBytes = new FileInfo(context.Server.MapPath(url)).Length; byte[] bytes = br.ReadBytes((int) numBytes); string filename = Path.GetFileName(url); context.Response.Buffer = true; context.Response.Charset = ""; context.Response.Cache.SetCacheability(HttpCacheability.NoCache); context.Response.ContentType = "application/x-rar-compressed"; context.Response.AddHeader("content-disposition", "attachment;filename=" + filename); context.Response.BinaryWrite(bytes); context.Response.Flush(); context.Response.End();

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  • Problems in Binary Search Tree

    - by user2782324
    This is my first ever trial at implementing the BST, and I am unable to get it done. Please help The problem is that When I delete the node if the node is in the right subtree from the root or if its a right child in the left subtree, then it works fine. But if the node is in the left subtree from root and its any left child, then it does not get deleted. Can someone show me what mistake am I doing?? the markedNode here gets allocated to the parent node of the node to be deleted. the minValueNode here gets allocated to a node whose left value child is the smallest value and it will be used to replace the value to be deleted. package DataStructures; class Node { int value; Node rightNode; Node leftNode; } class BST { Node rootOfTree = null; public void insertintoBST(int value) { Node markedNode = rootOfTree; if (rootOfTree == null) { Node newNode = new Node(); newNode.value = value; rootOfTree = newNode; newNode.rightNode = null; newNode.leftNode = null; } else { while (true) { if (value >= markedNode.value) { if (markedNode.rightNode != null) { markedNode = markedNode.rightNode; } else { Node newNode = new Node(); newNode.value = value; markedNode.rightNode = newNode; newNode.rightNode = null; newNode.leftNode = null; break; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { markedNode = markedNode.leftNode; } else { Node newNode = new Node(); newNode.value = value; markedNode.leftNode = newNode; newNode.rightNode = null; newNode.leftNode = null; break; } } } } } public void searchBST(int value) { Node markedNode = rootOfTree; if (rootOfTree == null) { System.out.println("Element Not Found"); } else { while (true) { if (value > markedNode.value) { if (markedNode.rightNode != null) { markedNode = markedNode.rightNode; } else { System.out.println("Element Not Found"); break; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { markedNode = markedNode.leftNode; } else { System.out.println("Element Not Found"); break; } } if (value == markedNode.value) { System.out.println("Element Found"); break; } } } } public void deleteFromBST(int value) { Node markedNode = rootOfTree; Node minValueNode = null; if (rootOfTree == null) { System.out.println("Element Not Found"); return; } if (rootOfTree.value == value) { if (rootOfTree.leftNode == null && rootOfTree.rightNode == null) { rootOfTree = null; return; } else if (rootOfTree.leftNode == null ^ rootOfTree.rightNode == null) { if (rootOfTree.rightNode != null) { rootOfTree = rootOfTree.rightNode; return; } else { rootOfTree = rootOfTree.leftNode; return; } } else { minValueNode = rootOfTree.rightNode; if (minValueNode.leftNode == null) { rootOfTree.rightNode.leftNode = rootOfTree.leftNode; rootOfTree = rootOfTree.rightNode; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node rootOfTree.value = minValueNode.leftNode.value; // The value has been swapped if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } } else { while (true) { if (value > markedNode.value) { if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { break; } else { markedNode = markedNode.rightNode; } } else { System.out.println("Element Not Found"); return; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { break; } else { markedNode = markedNode.leftNode; } } else { System.out.println("Element Not Found"); return; } } } // Parent of the required element found // //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { if (markedNode.rightNode.rightNode == null && markedNode.rightNode.leftNode == null) { markedNode.rightNode = null; return; } else if (markedNode.rightNode.rightNode == null ^ markedNode.rightNode.leftNode == null) { if (markedNode.rightNode.rightNode != null) { markedNode.rightNode = markedNode.rightNode.rightNode; return; } else { markedNode.rightNode = markedNode.rightNode.leftNode; return; } } else { if (markedNode.rightNode.value == value) { minValueNode = markedNode.rightNode.rightNode; } else { minValueNode = markedNode.leftNode.rightNode; } if (minValueNode.leftNode == null) { // MinNode has no left value markedNode.rightNode = minValueNode; return; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { markedNode.leftNode.value = minValueNode.leftNode.value; } } if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { markedNode.rightNode.value = minValueNode.leftNode.value; } } // MarkedNode exchanged if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } // //////////////////////////////////////////////////////////////////////////////////////////////////////////////// if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { if (markedNode.leftNode.rightNode == null && markedNode.leftNode.leftNode == null) { markedNode.leftNode = null; return; } else if (markedNode.leftNode.rightNode == null ^ markedNode.leftNode.leftNode == null) { if (markedNode.leftNode.rightNode != null) { markedNode.leftNode = markedNode.leftNode.rightNode; return; } else { markedNode.leftNode = markedNode.leftNode.leftNode; return; } } else { if (markedNode.rightNode.value == value) { minValueNode = markedNode.rightNode.rightNode; } else { minValueNode = markedNode.leftNode.rightNode; } if (minValueNode.leftNode == null) { // MinNode has no left value markedNode.leftNode = minValueNode; return; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { markedNode.leftNode.value = minValueNode.leftNode.value; } } if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { markedNode.rightNode.value = minValueNode.leftNode.value; } } // MarkedNode exchanged if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } } // //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// } } } } } } public class BSTImplementation { public static void main(String[] args) { BST newBst = new BST(); newBst.insertintoBST(19); newBst.insertintoBST(13); newBst.insertintoBST(10); newBst.insertintoBST(20); newBst.insertintoBST(5); newBst.insertintoBST(23); newBst.insertintoBST(28); newBst.insertintoBST(16); newBst.insertintoBST(27); newBst.insertintoBST(9); newBst.insertintoBST(4); newBst.insertintoBST(22); newBst.insertintoBST(17); newBst.insertintoBST(30); newBst.insertintoBST(40); newBst.deleteFromBST(5); newBst.deleteFromBST(4); newBst.deleteFromBST(9); newBst.deleteFromBST(10); newBst.deleteFromBST(13); newBst.deleteFromBST(16); newBst.deleteFromBST(17); newBst.searchBST(5); newBst.searchBST(4); newBst.searchBST(9); newBst.searchBST(10); newBst.searchBST(13); newBst.searchBST(16); newBst.searchBST(17); System.out.println(); newBst.deleteFromBST(20); newBst.deleteFromBST(23); newBst.deleteFromBST(27); newBst.deleteFromBST(28); newBst.deleteFromBST(30); newBst.deleteFromBST(40); newBst.searchBST(20); newBst.searchBST(23); newBst.searchBST(27); newBst.searchBST(28); newBst.searchBST(30); newBst.searchBST(40); } }

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  • Dlink search is hijacking my browser

    - by James
    For months now "DLink search" has been hijacking my search engines. I use google chrome, and I have organized my search engines in the handy dandy "manage search engines" tool about a TRILLION times. It never even says D-link is hacking my search engines. It does not show up! I have read many posts on this forum and others saying that to fix this problem from internet explorer: Setup, internet options, yadayada, magical fairies, and you are solved, but my browser is google chrome! How am I supposed to do this from there! I do not know how to re-setup my Dlink router, which is the cause of the problem! HOW? In those posts with the magical fairies fixing it, HUNDREDS responded saying, "yep, those fairies definitely fixed it right. :)" These people were so satisfied. IT WORKED FOR THEM, WHY NOT ME. I look at it and go ":(" because it does not help me. There are no options for anything to do with this in GOOGLE chrome. PLEASE EXPLAIN and HELP. I see no "SETUP" option, no "Internet Options" button, no anything. BTW the exact posts are these: "Uncheck Advanced DNS in the router internet setup. This will take care of it. I had this problem with my DLink router before." "I had this issue with my DIR-655 and unchecking the Advanced DNS setting in Setup - Internet - Manual Internet Connection Setup fixed it." "If this is just internet explorer, you can go to Tools Internet Options or Internet Options in Control Panel. From here, go to the advanced tab and click the Reset button." "I would set the router's DNS to a site like OpenDNS, and I would ensure the machines are set to get their DNS settings via DHCP or set the machine's DNS setting to OpenDNS. If the router's DNS looks like it was messed with, some bad software know the default passwords for routers and could have changed it. If you don't already I would make sure the password to the router is not default or easy to guess. I've had spyware change a machine's DNS, but the fact it is happening on all machines makes me wonder if it is the router." "Something got into your router and changed the dns server most likely, do a hard reset of the router and then change the password to something strong. Also check for a firmware update for the router and apply it as soon as possible."

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  • Rails, search item in different model?

    - by Danny McClelland
    Hi Everyone, I have a kase model which I am using a simple search form in. The problem I am having is some kases are linked to companies through a company model, and people through a people model. At the moment my search (in Kase model) looks like this: # SEARCH FACILITY def self.search(search) search_condition = "%" + search + "%" find(:all, :conditions => ['jobno LIKE ? OR casesubject LIKE ? OR transport LIKE ? OR goods LIKE ? OR comments LIKE ? OR invoicenumber LIKE ? OR netamount LIKE ? OR clientref LIKE ? OR kase_status LIKE ? OR lyingatlocationaddresscity LIKE ?', search_condition, search_condition, search_condition, search_condition, search_condition, search_condition, search_condition, search_condition, search_condition, search_condition]) end What I am trying to work out, is what condition can I add to allow a search by Company or Person to show the cases they are linked to. @kase.company.companyname and company.companyname don't work :( Is this possible? Thanks, Danny

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