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  • concurrency::index<N> from amp.h

    - by Daniel Moth
    Overview C++ AMP introduces a new template class index<N>, where N can be any value greater than zero, that represents a unique point in N-dimensional space, e.g. if N=2 then an index<2> object represents a point in 2-dimensional space. This class is essentially a coordinate vector of N integers representing a position in space relative to the origin of that space. It is ordered from most-significant to least-significant (so, if the 2-dimensional space is rows and columns, the first component represents the rows). The underlying type is a signed 32-bit integer, and component values can be negative. The rank field returns N. Creating an index The default parameterless constructor returns an index with each dimension set to zero, e.g. index<3> idx; //represents point (0,0,0) An index can also be created from another index through the copy constructor or assignment, e.g. index<3> idx2(idx); //or index<3> idx2 = idx; To create an index representing something other than 0, you call its constructor as per the following 4-dimensional example: int temp[4] = {2,4,-2,0}; index<4> idx(temp); Note that there are convenience constructors (that don’t require an array argument) for creating index objects of rank 1, 2, and 3, since those are the most common dimensions used, e.g. index<1> idx(3); index<2> idx(3, 6); index<3> idx(3, 6, 12); Accessing the component values You can access each component using the familiar subscript operator, e.g. One-dimensional example: index<1> idx(4); int i = idx[0]; // i=4 Two-dimensional example: index<2> idx(4,5); int i = idx[0]; // i=4 int j = idx[1]; // j=5 Three-dimensional example: index<3> idx(4,5,6); int i = idx[0]; // i=4 int j = idx[1]; // j=5 int k = idx[2]; // k=6 Basic operations Once you have your multi-dimensional point represented in the index, you can now treat it as a single entity, including performing common operations between it and an integer (through operator overloading): -- (pre- and post- decrement), ++ (pre- and post- increment), %=, *=, /=, +=, -=,%, *, /, +, -. There are also operator overloads for operations between index objects, i.e. ==, !=, +=, -=, +, –. Here is an example (where no assertions are broken): index<2> idx_a; index<2> idx_b(0, 0); index<2> idx_c(6, 9); _ASSERT(idx_a.rank == 2); _ASSERT(idx_a == idx_b); _ASSERT(idx_a != idx_c); idx_a += 5; idx_a[1] += 3; idx_a++; _ASSERT(idx_a != idx_b); _ASSERT(idx_a == idx_c); idx_b = idx_b + 10; idx_b -= index<2>(4, 1); _ASSERT(idx_a == idx_b); Usage You'll most commonly use index<N> objects to index into data types that we'll cover in future posts (namely array and array_view). Also when we look at the new parallel_for_each function we'll see that an index<N> object is the single parameter to the lambda, representing the (multi-dimensional) thread index… In the next post we'll go beyond being able to represent an N-dimensional point in space, and we'll see how to define the N-dimensional space itself through the extent<N> class. Comments about this post by Daniel Moth welcome at the original blog.

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  • Google Webmaster Tools Index dropped to Zero [closed]

    - by Brian Anderson
    Earlier this year I rebuilt my website using ZenCart. Immediately I saw a drop in index status from 59 to 0. I then signed up for Google Webmaster Tools and noticed the Index status took a dramatic drop and has never recovered. I have worked to add content and I know I am not done, but have not seen any recovery of this index since. What confuses me is when I look at the sitemap status under Optimization it shows me there are 1239 submitted and 1127 pages indexed. Most of my pages have fallen off page one for relevant search terms and some are as far back as page 7 or 8 where they used to be on the first page. I have made some changes in the past week to robots.txt and sitemap.xml, but have not seen any improvements. Can anyone tell me what might be going on here? My website is andersonpens.net. Thanks! Brian

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  • How to functionally generate a tree breadth-first. (With Haskell)

    - by Dennetik
    Say I have the following Haskell tree type, where "State" is a simple wrapper: data Tree a = Branch (State a) [Tree a] | Leaf (State a) deriving (Eq, Show) I also have a function "expand :: Tree a - Tree a" which takes a leaf node, and expands it into a branch, or takes a branch and returns it unaltered. This tree type represents an N-ary search-tree. Searching depth-first is a waste, as the search-space is obviously infinite, as I can easily keep on expanding the search-space with the use of expand on all the tree's leaf nodes, and the chances of accidentally missing the goal-state is huge... thus the only solution is a breadth-first search, implemented pretty decent over here, which will find the solution if it's there. What I want to generate, though, is the tree traversed up to finding the solution. This is a problem because I only know how to do this depth-first, which could be done by simply called the "expand" function again and again upon the first child node... until a goal-state is found. (This would really not generate anything other then a really uncomfortable list.) Could anyone give me any hints on how to do this (or an entire algorithm), or a verdict on whether or not it's possible with a decent complexity? (Or any sources on this, because I found rather few.)

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  • Mysql: create index on 1.4 billion records

    - by SiLent SoNG
    I have a table with 1.4 billion records. The table structure is as follows: CREATE TABLE text_page ( text VARCHAR(255), page_id INT UNSIGNED ) ENGINE=MYISAM DEFAULT CHARSET=ascii The requirement is to create an index over the column text. The table size is about 34G. I have tried to create the index by the following statement: ALTER TABLE text_page ADD KEY ix_text (text) After 10 hours' waiting I finally give up this approach. Is there any workable solution on this problem? UPDATE: the table is unlikely to be updated or inserted or deleted. The reason why to create index on the column text is because this kind of sql query would be frequently executed: SELECT page_id FROM text_page WHERE text = ?

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  • SQL SERVER – Identify Numbers of Non Clustered Index on Tables for Entire Database

    - by pinaldave
    Here is the script which will give you numbers of non clustered indexes on any table in entire database. SELECT COUNT(i.TYPE) NoOfIndex, [schema_name] = s.name, table_name = o.name FROM sys.indexes i INNER JOIN sys.objects o ON i.[object_id] = o.[object_id] INNER JOIN sys.schemas s ON o.[schema_id] = s.[schema_id] WHERE o.TYPE IN ('U') AND i.TYPE = 2 GROUP BY s.name, o.name ORDER BY schema_name, table_name Here is the small story behind why this script was needed. I recently went to meet my friend in his office and he introduced me to his colleague in office as someone who is an expert in SQL Server Indexing. I politely said I am yet learning about Indexing and have a long way to go. My friend’s colleague right away said – he had a suggestion for me with related to Index. According to him he was looking for a script which will count all the non clustered on all the tables in the database and he was not able to find that on SQLAuthority.com. I was a bit surprised as I really do not remember all the details about what I have written so far. I quickly pull up my phone and tried to look for the script on my custom search engine and he was correct. I never wrote a script which will count all the non clustered indexes on tables in the whole database. Excessive indexing is not recommended in general. If you have too many indexes it will definitely negatively affect your performance. The above query will quickly give you details of numbers of indexes on tables on your entire database. You can quickly glance and use the numbers as reference. Please note that the number of the index is not a indication of bad indexes. There is a lot of wisdom I can write here but that is not the scope of this blog post. There are many different rules with Indexes and many different scenarios. For example – a table which is heap (no clustered index) is often not recommended on OLTP workload (here is the blog post to identify them), drop unused indexes with careful observation (here is the script for it), identify missing indexes and after careful testing add them (here is the script for it). Even though I have given few links here it is just the tip of the iceberg. If you follow only above four advices your ship may still sink. Those who wants to learn the subject in depth can watch the videos here after logging in. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Index, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • Spanning-tree setup with incompatible switches

    - by wfaulk
    I have a set of eight HP ProCurve 2910al-48G Ethernet switches at my datacenter that are set up in a star topology with no physical loops. I want to partially mesh the switches for redundancy and manage the loops with a spanning-tree protocol. However, our connection to the datacenter is provided by two uplinks, each to a Cisco 3750. The datacenter's switches are handling the redundant connection using PVST spanning-tree, which is a Cisco-proprietary spanning-tree implementation that my HP switches do not support. It appears that my switches are not participating in the datacenter's spanning-tree domain, but are blindly passing the BPDUs between the two switchports on my side, which enables the datacenter's switches to recognize the loop and put one of the uplinks into the Blocking state. This is somewhat supposition, but I can confirm that, while my switches say that both of the uplink ports are forwarding, only one is passing any real quantity of data. (I am assuming that I cannot get the datacenter to move away from PVST. I don't know that I'd want them to make that significant of a change anyway.) The datacenter has also sent me this output from their switches (which I have expurgated of any identifiable info): 3750G-1#sh spanning-tree vlan nnn VLAN0nnn Spanning tree enabled protocol ieee Root ID Priority 10 Address 00d0.0114.xxxx Cost 4 Port 5 (GigabitEthernet1/0/5) Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Bridge ID Priority 32mmm (priority 32768 sys-id-ext nnn) Address 0018.73d3.yyyy Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Aging Time 300 sec Interface Role Sts Cost Prio.Nbr Type ------------------- ---- --- --------- -------- -------------------------------- Gi1/0/5 Root FWD 4 128.5 P2p Gi1/0/6 Altn BLK 4 128.6 P2p Gi1/0/8 Altn BLK 4 128.8 P2p and: 3750G-2#sh spanning-tree vlan nnn VLAN0nnn Spanning tree enabled protocol ieee Root ID Priority 10 Address 00d0.0114.xxxx Cost 4 Port 6 (GigabitEthernet1/0/6) Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Bridge ID Priority 32mmm (priority 32768 sys-id-ext nnn) Address 000f.f71e.zzzz Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Aging Time 300 sec Interface Role Sts Cost Prio.Nbr Type ------------------- ---- --- --------- -------- -------------------------------- Gi1/0/1 Desg FWD 4 128.1 P2p Gi1/0/5 Altn BLK 4 128.5 P2p Gi1/0/6 Root FWD 4 128.6 P2p Gi1/0/8 Desg FWD 4 128.8 P2p The uplinks to my switches are on Gi1/0/8 on both of their switches. The uplink ports are configured with a single tagged VLAN. I am also using a number of other tagged VLANs in my switch infrastructure. And, to be clear, I am passing the tagged VLAN I'm receiving from the datacenter to other ports on other switches in my infrastructure. My question is: how do I configure my switches so that I can use a spanning tree protocol inside my switch infrastructure without breaking the datacenter's spanning tree that I cannot participate in?

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  • Table index design

    - by Swoosh
    I would like to add index(s) to my table. I am looking for general ideas how to add more indexes to a table. Other than the PK clustered. I would like to know what to look for when I am doing this. So, my example: This table (let's call it TASK table) is going to be the biggest table of the whole application. Expecting millions records. IMPORTANT: massive bulk-insert is adding data in this table table has 27 columns: (so far, and counting :D ) int x 9 columns = id-s varchar x 10 columns bit x 2 columns datetime x 5 columns INT COLUMNS all of these are INT ID-s but from tables that are usually smaller than Task table (10-50 records max), example: Status table (with values like "open", "closed") or Priority table (with values like "important", "not so important", "normal") there is also a column like "parent-ID" (self - ID) join: all the "small" tables have PK, the usual way ... clustered STRING COLUMNS there is a (Company) column (string!) that is something like "5 characters long all the time" and every user will be restricted using this one. If in Task there are 15 different "Companies" the logged in user would only see one. So there's always a filter on this one. Might be a good idea to add an index to this column? DATE COLUMNS I think they don't index these ... right? Or can / should be?

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  • Obtain all keys of a Neo4j index

    - by MattiSG
    I have a Neo4j database whose content is generated dynamically from a big dataset. All “entry points” nodes are indexed on a named index (IndexManager.forNodes(…)). I can therefore look up a particular “entry point” node. However, I would now like to enumerate all those specific nodes, but I can't know on which key they were indexed. Is there any way to enumerate all keys of a Neo4j Index? If not, what would be the best way to store those keys, a data type that is eminently non-graph-oriented? UPDATE (thanks for asking details :) ): the list would be more than 2 million entries. The main use case would be to never update it after an initialization step, but other use cases might need it, so it has to be somewhat scalable. Also, I would really prefer avoiding killing my current resilience abilities, so storing all keys at once, as opposed to adding them incrementally, would be a last-resort solution.

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  • Google Webmaster Tools Index Status is 0 but sitemap URL shows indexed

    - by DD.
    I've added my site to Google Webmaster tools www.medexpress.co.uk. The site was submitted a few weeks ago. The index status shows 0 but it shows 6 URLs have been indexed in the sitemaps section. If I search in google I can see that the site is indexed and several pages appear: https://www.google.co.uk/search?q=site%3Awww.medexpress.co.uk&oq=site%3Awww.medexpress.co.uk&sourceid=chrome&ie=UTF-8 My question is why is the index status 0 when the sitemap section shows several indexed pages and also the pages appear in the search engine.

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  • HTML 5, Fluid Pages and Google Mobile Index

    - by Bob
    I am currently migrating my site to HTML5, at the same time designing pages so that they are "fluid" and are equally presentable for a mobile or a large screen. I took the fluid approach so as not to have to develop a separate application for mobile devices and I'm pleasantly surprised with the results that look equally as good on an iPhone as they do on a large screen. Then I went into the Google Webmaster Tools facility and became aware of the Google Mobile Index. I'm confused now as HTML5 doesn't seem to be supported by Google Mobile Indexing. Does this mean that when I go live with my new "pride and joy" HTML5 site on a mobile it won't appear on any Google searches as it's not in the Google Mobile Index?

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  • Did Google Delete my index?

    - by Sei
    I have a website that I haven't had much time to take care of. I updated it 4 times since last October, the contents are all original and informative. I can see those cache from Way back Machine and the site was indexed in Yahoo, but not in google. Did I get my index on Google deleted because I did not update it often? Normally Google crawl often and index fast, I just got really worried if I did something wrong. Is it possible that I set something wrong in the hosting or something? Please help me

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  • What is the difference between an Abstract Syntax Tree and a Concrete Syntax Tree?

    - by Jason Baker
    I've been reading a bit about how interpreters/compilers work, and one area where I'm getting confused is the difference between an AST and a CST. My understanding is that the parser makes a CST, hands it to the semantic analyzer which turns it into an AST. However, my understanding is that the semantic analyzer simply ensures that rules are followed. I don't really understand why it would actually make any changes to make it abstract rather than concrete. Is there something that I'm missing about the semantic analyzer, or is the difference between an AST and CST somewhat artificial?

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  • Deletion procedure for a Binary Search Tree

    - by Metz
    Consider the deletion procedure on a BST, when the node to delete has two children. Let's say i always replace it with the node holding the minimum key in its right subtree. The question is: is this procedure commutative? That is, deleting x and then y has the same result than deleting first y and then x? I think the answer is no, but i can't find a counterexample, nor figure out any valid reasoning. EDIT: Maybe i've got to be clearer. Consider the transplant(node x, node y) procedure: it replace x with y (and its subtree). So, if i want to delete a node (say x) which has two children i replace it with the node holding the minimum key in its right subtree: y = minimum(x.right) transplant(y, y.right) // extracts the minimum (it doesn't have left child) y.right = x.right y.left = x.left transplant(x,y) The question was how to prove the procedure above is not commutative.

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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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  • 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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  • De-index URL parameters by value

    - by Doug Firr
    Upon reading over this question is lengthy so allow me to provide a one sentence summary: I need to get Google to de-index URLs that have parameters with certain values appended I have a website example.com with language translations. There used to be many translations but I deleted them all so that only English (Default) and French options remain. When one selects a language option a parameter is aded to the URL. For example, the home page: https://example.com (default) https://example.com/main?l=fr_FR (French) I added a robots.txt to stop Google from crawling any of the language translations: # robots.txt generated at http://www.mcanerin.com User-agent: * Disallow: Disallow: /cgi-bin/ Disallow: /*?l= So any pages containing "?l=" should not be crawled. I checked in GWT using the robots testing tool. It works. But under html improvements the previously crawled language translation URLs remain indexed. The internet says to add a 404 to the header of the removed URLs so the Googles knows to de-index it. I checked to see what my CMS would throw up if I visited one of the URLs that should no longer exist. This URL was listed in GWT under duplicate title tags (One of the reasons I want to scrub up my URLS) https://example.com/reports/view/884?l=vi_VN&l=hy_AM This URL should not exist - I removed the language translations. The page loads when it should not! I played around. I typed example.com?whatever123 It seems that parameters always load as long as everything before the question mark is a real URL. So if Google has indexed all these URLS with parameters how do I remove them? I cannot check if a 404 is being generated because the page always loads because it's a parameter that needs to be de-indexed.

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  • De-index URL paremeters

    - by Doug Firr
    Upon reading over this question is lengthy so allow me to provide a one sentence summary: I need to get Google to de-index URLs that have certain parameters appended I have a website example.com with language translations. There used to be many translations but I deleted them all so that only English (Default) and French options remain. When one selects a language option a parameter is aded to the URL. For example, the home page: https://example.com (default) https://example.com/main?l=fr_FR (French) I added a robots.txt to stop Google from crawling any of the language translations: # robots.txt generated at http://www.mcanerin.com User-agent: * Disallow: Disallow: /cgi-bin/ Disallow: /*?l= So any pages containing "?l=" should not be crawled. I checked in GWT using the robots testing tool. It works. But under html improvements the previously crawled language translation URLs remain indexed. The internet says to add a 404 to the header of the removed URLs so the Googles knows to de-index it. I checked to see what my CMS would throw up if I visited one of the URLs that should no longer exist. This URL was listed in GWT under duplicate title tags (One of the reasons I want to scrub up my URLS) https://example.com/reports/view/884?l=vi_VN&l=hy_AM This URL should not exist - I removed the language translations. The page loads when it should not! I played around. I typed example.com?whatever123 It seems that parameters always load as long as everything before the question mark is a real URL. So if Google has indexed all these URLS with parameters how do I remove them? I cannot check if a 404 is being generated because the page always loads because it's a parameter that needs to be de-indexed.

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  • Excel Worksheet Index

    - by Ben
    I have the following code that works great but I am trying to modify it so that instead of replacing column 1 of the Index page with a new index I would rather start the range in Cell C11. Right now, the new Index starts in Cell A1 of the Index sheet. Here is the code: Private Sub Worksheet_Activate() Dim wSheet As Worksheet Dim l As Long l = 1 With Me .Columns(1).ClearContents .Cells(1, 1) = "INDEX" .Cells(1, 1).Name = "Index" End With For Each wSheet In Worksheets If wSheet.Name <> Me.Name Then l = l + 1 With wSheet .Range("A1").Name = "Start_" & wSheet.Index .Hyperlinks.Add Anchor:=.Range("A1"), Address:="", _ SubAddress:="Index", TextToDisplay:="Back to Index" End With Me.Hyperlinks.Add Anchor:=Me.Cells(l, 1), Address:="", _ SubAddress:="Start_" & wSheet.Index, TextToDisplay:=wSheet.Name End If Next wSheet End Sub I have successfully modified the code so that the link back to the index on each sheet is in cell A4 without trouble, but I can't figure out how to have the index be replaced starting at Cell C11

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  • Problem with building tree bottom up

    - by Esmond
    Hi, I have problems building a binary tree from the bottom up. THe input of the tree would be internal nodes of the trees with the children of this node being the leaves of the eventual tree. So initially if the tree is empty the root would be the first internal node. Afterwards, The next internal node to be added would be the new root(NR), with the old root(OR) being one of the child of NR. And so on. The problem i have is that whenever i add a NR, the children of the OR seems to be lost when i do a inOrder traversal. This is proven to be the case when i do a getSize() call which returns the same number of nodes before and after addNode(Tree,Node) Any help with resolving this problem is appreciated edited with the inclusion of node class code. both tree and node classes have the addChild methods because i'm not very sure where to put them for it to be appropriated. any comments on this would be appreciated too. The code is as follows: import java.util.*; public class Tree { Node root; int size; public Tree() { root = null; } public Tree(Node root) { this.root = root; } public static void setChild(Node parent, Node child, double weight) throws ItemNotFoundException { if (parent.child1 != null && parent.child2 != null) { throw new ItemNotFoundException("This Node already has 2 children"); } else if (parent.child1 != null) { parent.child2 = child; child.parent = parent; parent.c2Weight = weight; } else { parent.child1 = child; child.parent = parent; parent.c1Weight = weight; } } public static void setChild1(Node parent, Node child) { parent.child1 = child; child.parent = parent; } public static void setChild2(Node parent, Node child) { parent.child2 = child; child.parent = parent; } public static Tree addNode(Tree tree, Node node) throws ItemNotFoundException { Tree tree1; if (tree.root == null) { tree.root = node; } else if (tree.root.getSeq().equals(node.getChild1().getSeq()) || tree.root.getSeq().equals(node.getChild2().getSeq())) { Node oldRoot = tree.root; oldRoot.setParent(node); tree.root = node; } else { //form a disjoint tree and merge the 2 trees tree1 = new Tree(node); tree = mergeTree(tree, tree1); } System.out.print("addNode2 = "); if(tree.root != null ) { Tree.inOrder(tree.root); } System.out.println(); return tree; } public static Tree mergeTree(Tree tree, Tree tree1) { String root = "root"; Node node = new Node(root); tree.root.setParent(node); tree1.root.setParent(node); tree.root = node; return tree; } public static int getSize(Node root) { if (root != null) { return 1 + getSize(root.child1) + getSize(root.child2); } else { return 0; } } public static boolean isEmpty(Tree Tree) { return Tree.root == null; } public static void inOrder(Node root) { if (root != null) { inOrder(root.child1); System.out.print(root.sequence + " "); inOrder(root.child2); } } } public class Node { Node child1; Node child2; Node parent; double c1Weight; double c2Weight; String sequence; boolean isInternal; public Node(String seq) { sequence = seq; child1 = null; c1Weight = 0; child2 = null; c2Weight = 0; parent = null; isInternal = false; } public boolean hasChild() { if (this.child1 == null && this.child2 == null) { this.isInternal = false; return isInternal; } else { this.isInternal = true; return isInternal; } } public String getSeq() throws ItemNotFoundException { if (this.sequence == null) { throw new ItemNotFoundException("No such node"); } else { return this.sequence; } } public void setChild(Node child, double weight) throws ItemNotFoundException { if (this.child1 != null && this.child2 != null) { throw new ItemNotFoundException("This Node already has 2 children"); } else if (this.child1 != null) { this.child2 = child; this.c2Weight = weight; } else { this.child1 = child; this.c1Weight = weight; } } public static void setChild1(Node parent, Node child) { parent.child1 = child; child.parent = parent; } public static void setChild2(Node parent, Node child) { parent.child2 = child; child.parent = parent; } public void setParent(Node parent){ this.parent = parent; } public Node getParent() throws ItemNotFoundException { if (this.parent == null) { throw new ItemNotFoundException("This Node has no parent"); } else { return this.parent; } } public Node getChild1() throws ItemNotFoundException { if (this.child1 == null) { throw new ItemNotFoundException("There is no child1"); } else { return this.child1; } } public Node getChild2() throws ItemNotFoundException { if (this.child2 == null) { throw new ItemNotFoundException("There is no child2"); } else { return this.child2; } } }

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  • SQL Server – Learning SQL Server Performance: Indexing Basics – Video

    - by pinaldave
    Today I remember one of my older cartoon years ago created for Indexing and Performance. Every single time when Performance is discussed, Indexes are mentioned along with it. In recent times, data and application complexity is continuously growing.  The demand for faster query response, performance, and scalability by organizations is increasing and developers and DBAs need to now write efficient code to achieve this. DBA and Developers A DBA’s role is critical, because a production environment has to run 24×7, hence maintenance, trouble shooting, and quick resolutions are the need of the hour.  The first baby step into any performance tuning exercise in SQL Server involves creating, analysing, and maintaining indexes. Though we have learnt indexing concepts from our college days, indexing implementation inside SQL Server can vary.  Understanding this behaviour and designing our applications appropriately will make sure the application is performed to its highest potential. Video Learning Vinod Kumar and myself we often thought about this and realized that practical understanding of the indexes is very important. One can not master every single aspects of the index. However there are some minimum expertise one should gain if performance is one of the concern. We decided to build a course which just addresses the practical aspects of the performance. In this course, we explored some of these indexing fundamentals and we elaborated on how SQL Server goes about using indexes.  At the end of this course of you will know the basic structure of indexes, practical insights into implementation, and maintenance tips and tricks revolving around indexes.  Finally, we will introduce SQL Server 2012 column store indexes.  We have refrained from discussing internal storage structure of the indexes but have taken a more practical, demo-oriented approach to explain these core concepts. Course Outline Here are salient topics of the course. We have explained every single concept along with a practical demonstration. Additionally shared our personal scripts along with the same. Introduction Fundamentals of Indexing Index Fundamentals Index Fundamentals – Visual Representation Practical Indexing Implementation Techniques Primary Key Over Indexing Duplicate Index Clustered Index Unique Index Included Columns Filtered Index Disabled Index Index Maintenance and Defragmentation Introduction to Columnstore Index Indexing Practical Performance Tips and Tricks Index and Page Types Index and Non Deterministic Columns Index and SET Values Importance of Clustered Index Effect of Compression and Fillfactor Index and Functions Dynamic Management Views (DMV) – Fillfactor Table Scan, Index Scan and Index Seek Index and Order of Columns Final Checklist: Index and Performance Well, we believe we have done our part, now waiting for your comments and feedback. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Index, SQL Performance, SQL Query, SQL Server, SQL Tips and Tricks, SQLServer, T SQL, Technology, Video

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  • Capturing index operations using a DDL trigger

    - by AaronBertrand
    Today on twitter the following question came up on the #sqlhelp hash tag, from DaveH0ward : Is there a DMV that can tell me the last time an index was rebuilt? SQL 2008 My initial response: I don't believe so, you'd have to be monitoring for that ... perhaps a DDL trigger capturing ALTER_INDEX? Then I remembered that the default trace in SQL Server ( as long as it is enabled ) will capture these events. My follow-up response: You can get it from the default trace, blog post forthcoming So here is...(read more)

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  • Google webmaster Index Status. Total Indexed=0

    - by hammad
    I previously changed my domain from www.visualstudiolearn.blogspot.com to www.visualstudiolearn.com... i had around 300 posts with the previous domain name and most of them where showing up on Google. Now that i have changed my domain name the index status shows total indexed as 0 and when i go to the advanced tab it says 304(not selected) and 217 blocked my robots. Im really depressed because of this situation. could you please help out???

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  • Behaviour tree code example?

    - by jokoon
    http://altdevblogaday.org/2011/02/24/introduction-to-behavior-trees/ Obviously the most interesting article I found on this website. What do you think about it ? It lacks some code example, don't you know any ? I also read that state machines are not very flexible compared to behaviour trees... On top of that I'm not sure if there is a true link between state machines and the state pattern... is there ?

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