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  • Apache mod_rewrite : How to REWRITE (or whatever) child directories to parent?

    - by ????
    Actually i am trying to make a PHP MVC like application. A basic one. The current milestone i am reaching already includes: Basic RESTful Routing Means, if i type: www.example.com/items/book/8888 .. it properly just stays there as it is and i can already slice out the URL by slashes / and loads the responsible Controllers .... etc from the top single index.php file. I mean, so it is OK for the backend PHP. But the only problem is, it still CAN NOT process the REWRITES properly. For example, the CSS & JS are BROKEN as if i VIEW PAGE SOURCE of the page www.example.com/items/book/8888, the asset files are being called as: www.example.com/items/book/8888/css/main.css www.example.com/items/book/8888/js/jquery.js .. which really are PROBLEMS because in the code is like: <link type="text/css" rel="stylesheet" media="all" href="css/main.css"> <script type="text/javascript" src="js/jquery.js"></script> So the question is: How can i use Apache REWRITE (or whatever approach) to make sure every ASSET FILES to be correctly being called from the DOCROOT. For example, if i am in the URL: www.example.com/items/book/8888 My ASSET FILES should still be called as: www.example.com/css/main.css www.example.com/js/jquery.js Or is there any other methods i need to follow? Please kindly help suggest. Thank you.

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  • create child nodes from sibling nodes until a different sibbling occurs.

    - by user364939
    Hi does anyone know what the xsl would look like to transform this XML. There can be N nte's after pid and N nte's after pv1. The structure is guaranteed in that all nte that follows pid belongs to pid and all nte following pv1 belongs to pv1. From: <pid> </pid> <nte> <nte-1>1</nte-1> <nte-3>Note 1</nte-1> </nte> <nte></nte> <pv1></pv1> <nte> </nte> into: <pid> <nte> <nte-1>1</nte-1> <nte-3>Note 1</nte-1> </nte> <nte> </nte> </pid> <pv1> <nte> </nte> </pv1> Thanks!

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  • .NET (C#): Getting child windows when you only have a process handle or PID?

    - by shea241
    Kind of a special case problem: I start a process with System.Diagnostics.Process.Start(..) The process opens a splash screen -- this splash screen becomes the main window. The splash screen closes and the 'real' UI is shown. The main window (splash screen) is now invalid. I still have the Process object, and I can query its handle, module, etc. But the main window handle is now invalid. I need to get the process's UI (or UI handle) at this point. Assume I cannot change the behavior of the process to make this any easier (or saner). I have looked around online but I'll admit I didn't look for more than an hour. Seemed like it should be somewhat trivial :-( Thanks

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  • Is there any event sent to window when new tab opens from parent tab on click a anchor tag with target _blank?

    - by Justin John
    I need to write callback function on clicking an anchor with target="_blank". <a rel="external" href="http://twitter.com/" target="_blank"> <img src="twitter.png"> </a> I think, following code block will do it. $("a[target='_blank']").click(function() { // action required }); Is there any other way where we can get an event with new tab opens? For example: The unload event is sent to the window element when the user navigates away from the page. If I think same perspective, Is there any event sent to window when a new tab opens.

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  • exporting bind and keyframe bone poses from blender to use in OpenGL

    - by SaldaVonSchwartz
    I'm having a hard time trying to understand how exactly Blender's concept of bone transforms maps to the usual math of skinning (which I'm implementing in an OpenGL-based engine of sorts). Or I'm missing out something in the math.. It's gonna be long, but here's as much background as I can think of. First, a few notes and assumptions: I'm using column-major order and multiply from right to left. So for instance, vertex v transformed by matrix A and then further transformed by matrix B would be: v' = BAv. This also means whenever I export a matrix from blender through python, I export it (in text format) in 4 lines, each representing a column. This is so I can then I can read them back into my engine like this: if (fscanf(fileHandle, "%f %f %f %f", &skeleton.joints[currentJointIndex].inverseBindTransform.m[0], &skeleton.joints[currentJointIndex].inverseBindTransform.m[1], &skeleton.joints[currentJointIndex].inverseBindTransform.m[2], &skeleton.joints[currentJointIndex].inverseBindTransform.m[3])) { if (fscanf(fileHandle, "%f %f %f %f", &skeleton.joints[currentJointIndex].inverseBindTransform.m[4], &skeleton.joints[currentJointIndex].inverseBindTransform.m[5], &skeleton.joints[currentJointIndex].inverseBindTransform.m[6], &skeleton.joints[currentJointIndex].inverseBindTransform.m[7])) { if (fscanf(fileHandle, "%f %f %f %f", &skeleton.joints[currentJointIndex].inverseBindTransform.m[8], &skeleton.joints[currentJointIndex].inverseBindTransform.m[9], &skeleton.joints[currentJointIndex].inverseBindTransform.m[10], &skeleton.joints[currentJointIndex].inverseBindTransform.m[11])) { if (fscanf(fileHandle, "%f %f %f %f", &skeleton.joints[currentJointIndex].inverseBindTransform.m[12], &skeleton.joints[currentJointIndex].inverseBindTransform.m[13], &skeleton.joints[currentJointIndex].inverseBindTransform.m[14], &skeleton.joints[currentJointIndex].inverseBindTransform.m[15])) { I'm simplifying the code I show because otherwise it would make things unnecessarily harder (in the context of my question) to explain / follow. Please refrain from making remarks related to optimizations. This is not final code. Having said that, if I understand correctly, the basic idea of skinning/animation is: I have a a mesh made up of vertices I have the mesh model-world transform W I have my joints, which are really just transforms from each joint's space to its parent's space. I'll call these transforms Bj meaning matrix which takes from joint j's bind pose to joint j-1's bind pose. For each of these, I actually import their inverse to the engine, Bj^-1. I have keyframes each containing a set of current poses Cj for each joint J. These are initially imported to my engine in TQS format but after (S)LERPING them I compose them into Cj matrices which are equivalent to the Bjs (not the Bj^-1 ones) only that for the current spacial configurations of each joint at that frame. Given the above, the "skeletal animation algorithm is" On each frame: check how much time has elpased and compute the resulting current time in the animation, from 0 meaning frame 0 to 1, meaning the end of the animation. (Oh and I'm looping forever so the time is mod(total duration)) for each joint: 1 -calculate its world inverse bind pose, that is Bj_w^-1 = Bj^-1 Bj-1^-1 ... B0^-1 2 -use the current animation time to LERP the componets of the TQS and come up with an interpolated current pose matrix Cj which should transform from the joints current configuration space to world space. Similar to what I did to get the world version of the inverse bind poses, I come up with the joint's world current pose, Cj_w = C0 C1 ... Cj 3 -now that I have world versions of Bj and Cj, I store this joint's world- skinning matrix K_wj = Cj_w Bj_w^-1. The above is roughly implemented like so: - (void)update:(NSTimeInterval)elapsedTime { static double time = 0; time = fmod((time + elapsedTime),1.); uint16_t LERPKeyframeNumber = 60 * time; uint16_t lkeyframeNumber = 0; uint16_t lkeyframeIndex = 0; uint16_t rkeyframeNumber = 0; uint16_t rkeyframeIndex = 0; for (int i = 0; i < aClip.keyframesCount; i++) { uint16_t keyframeNumber = aClip.keyframes[i].number; if (keyframeNumber <= LERPKeyframeNumber) { lkeyframeIndex = i; lkeyframeNumber = keyframeNumber; } else { rkeyframeIndex = i; rkeyframeNumber = keyframeNumber; break; } } double lTime = lkeyframeNumber / 60.; double rTime = rkeyframeNumber / 60.; double blendFactor = (time - lTime) / (rTime - lTime); GLKMatrix4 bindPosePalette[aSkeleton.jointsCount]; GLKMatrix4 currentPosePalette[aSkeleton.jointsCount]; for (int i = 0; i < aSkeleton.jointsCount; i++) { F3DETQSType& lPose = aClip.keyframes[lkeyframeIndex].skeletonPose.jointPoses[i]; F3DETQSType& rPose = aClip.keyframes[rkeyframeIndex].skeletonPose.jointPoses[i]; GLKVector3 LERPTranslation = GLKVector3Lerp(lPose.t, rPose.t, blendFactor); GLKQuaternion SLERPRotation = GLKQuaternionSlerp(lPose.q, rPose.q, blendFactor); GLKVector3 LERPScaling = GLKVector3Lerp(lPose.s, rPose.s, blendFactor); GLKMatrix4 currentTransform = GLKMatrix4MakeWithQuaternion(SLERPRotation); currentTransform = GLKMatrix4Multiply(currentTransform, GLKMatrix4MakeTranslation(LERPTranslation.x, LERPTranslation.y, LERPTranslation.z)); currentTransform = GLKMatrix4Multiply(currentTransform, GLKMatrix4MakeScale(LERPScaling.x, LERPScaling.y, LERPScaling.z)); if (aSkeleton.joints[i].parentIndex == -1) { bindPosePalette[i] = aSkeleton.joints[i].inverseBindTransform; currentPosePalette[i] = currentTransform; } else { bindPosePalette[i] = GLKMatrix4Multiply(aSkeleton.joints[i].inverseBindTransform, bindPosePalette[aSkeleton.joints[i].parentIndex]); currentPosePalette[i] = GLKMatrix4Multiply(currentPosePalette[aSkeleton.joints[i].parentIndex], currentTransform); } aSkeleton.skinningPalette[i] = GLKMatrix4Multiply(currentPosePalette[i], bindPosePalette[i]); } } At this point, I should have my skinning palette. So on each frame in my vertex shader, I do: uniform mat4 modelMatrix; uniform mat4 projectionMatrix; uniform mat3 normalMatrix; uniform mat4 skinningPalette[6]; attribute vec4 position; attribute vec3 normal; attribute vec2 tCoordinates; attribute vec4 jointsWeights; attribute vec4 jointsIndices; varying highp vec2 tCoordinatesVarying; varying highp float lIntensity; void main() { vec3 eyeNormal = normalize(normalMatrix * normal); vec3 lightPosition = vec3(0., 0., 2.); lIntensity = max(0.0, dot(eyeNormal, normalize(lightPosition))); tCoordinatesVarying = tCoordinates; vec4 skinnedVertexPosition = vec4(0.); for (int i = 0; i < 4; i++) { skinnedVertexPosition += jointsWeights[i] * skinningPalette[int(jointsIndices[i])] * position; } gl_Position = projectionMatrix * modelMatrix * skinnedVertexPosition; } The result: The mesh parts that are supposed to animate do animate and follow the expected motion, however, the rotations are messed up in terms of orientations. That is, the mesh is not translated somewhere else or scaled in any way, but the orientations of rotations seem to be off. So a few observations: In the above shader notice I actually did not multiply the vertices by the mesh modelMatrix (the one which would take them to model or world or global space, whichever you prefer, since there is no parent to the mesh itself other than "the world") until after skinning. This is contrary to what I implied in the theory: if my skinning matrix takes vertices from model to joint and back to model space, I'd think the vertices should already be premultiplied by the mesh transform. But if I do so, I just get a black screen. As far as exporting the joints from Blender, my python script exports for each armature bone in bind pose, it's matrix in this way: def DFSJointTraversal(file, skeleton, jointList): for joint in jointList: poseJoint = skeleton.pose.bones[joint.name] jointTransform = poseJoint.matrix.inverted() file.write('Joint ' + joint.name + ' Transform {\n') for col in jointTransform.col: file.write('{:9f} {:9f} {:9f} {:9f}\n'.format(col[0], col[1], col[2], col[3])) DFSJointTraversal(file, skeleton, joint.children) file.write('}\n') And for current / keyframe poses (assuming I'm in the right keyframe): def exportAnimations(filepath): # Only one skeleton per scene objList = [object for object in bpy.context.scene.objects if object.type == 'ARMATURE'] if len(objList) == 0: return elif len(objList) > 1: return #raise exception? dialog box? skeleton = objList[0] jointNames = [bone.name for bone in skeleton.data.bones] for action in bpy.data.actions: # One animation clip per action in Blender, named as the action animationClipFilePath = filepath[0 : filepath.rindex('/') + 1] + action.name + ".aClip" file = open(animationClipFilePath, 'w') file.write('target skeleton: ' + skeleton.name + '\n') file.write('joints count: {:d}'.format(len(jointNames)) + '\n') skeleton.animation_data.action = action keyframeNum = max([len(fcurve.keyframe_points) for fcurve in action.fcurves]) keyframes = [] for fcurve in action.fcurves: for keyframe in fcurve.keyframe_points: keyframes.append(keyframe.co[0]) keyframes = set(keyframes) keyframes = [kf for kf in keyframes] keyframes.sort() file.write('keyframes count: {:d}'.format(len(keyframes)) + '\n') for kfIndex in keyframes: bpy.context.scene.frame_set(kfIndex) file.write('keyframe: {:d}\n'.format(int(kfIndex))) for i in range(0, len(skeleton.data.bones)): file.write('joint: {:d}\n'.format(i)) joint = skeleton.pose.bones[i] jointCurrentPoseTransform = joint.matrix translationV = jointCurrentPoseTransform.to_translation() rotationQ = jointCurrentPoseTransform.to_3x3().to_quaternion() scaleV = jointCurrentPoseTransform.to_scale() file.write('T {:9f} {:9f} {:9f}\n'.format(translationV[0], translationV[1], translationV[2])) file.write('Q {:9f} {:9f} {:9f} {:9f}\n'.format(rotationQ[1], rotationQ[2], rotationQ[3], rotationQ[0])) file.write('S {:9f} {:9f} {:9f}\n'.format(scaleV[0], scaleV[1], scaleV[2])) file.write('\n') file.close() Which I believe follow the theory explained at the beginning of my question. But then I checked out Blender's directX .x exporter for reference.. and what threw me off was that in the .x script they are exporting bind poses like so (transcribed using the same variable names I used so you can compare): if joint.parent: jointTransform = poseJoint.parent.matrix.inverted() else: jointTransform = Matrix() jointTransform *= poseJoint.matrix and exporting current keyframe poses like this: if joint.parent: jointCurrentPoseTransform = joint.parent.matrix.inverted() else: jointCurrentPoseTransform = Matrix() jointCurrentPoseTransform *= joint.matrix why are they using the parent's transform instead of the joint in question's? isn't the join transform assumed to exist in the context of a parent transform since after all it transforms from this joint's space to its parent's? Why are they concatenating in the same order for both bind poses and keyframe poses? If these two are then supposed to be concatenated with each other to cancel out the change of basis? Anyway, any ideas are appreciated.

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  • Binary Search Tree Implementation

    - by Gabe
    I've searched the forum, and tried to implement the code in the threads I found. But I've been working on this real simple program since about 10am, and can't solve the seg. faults for the life of me. Any ideas on what I'm doing wrong would be greatly appreciated. BST.h (All the implementation problems should be in here.) #ifndef BST_H_ #define BST_H_ #include <stdexcept> #include <iostream> #include "btnode.h" using namespace std; /* A class to represent a templated binary search tree. */ template <typename T> class BST { private: //pointer to the root node in the tree BTNode<T>* root; public: //default constructor to make an empty tree BST(); /* You have to document these 4 functions */ void insert(T value); bool search(const T& value) const; bool search(BTNode<T>* node, const T& value) const; void printInOrder() const; void remove(const T& value); //function to print out a visual representation //of the tree (not just print the tree's values //on a single line) void print() const; private: //recursive helper function for "print()" void print(BTNode<T>* node,int depth) const; }; /* Default constructor to make an empty tree */ template <typename T> BST<T>::BST() { root = NULL; } template <typename T> void BST<T>::insert(T value) { BTNode<T>* newNode = new BTNode<T>(value); cout << newNode->data; if(root == NULL) { root = newNode; return; } BTNode<T>* current = new BTNode<T>(NULL); current = root; current->data = root->data; while(true) { if(current->left == NULL && current->right == NULL) break; if(current->right != NULL && current->left != NULL) { if(newNode->data > current->data) current = current->right; else if(newNode->data < current->data) current = current->left; } else if(current->right != NULL && current->left == NULL) { if(newNode->data < current->data) break; else if(newNode->data > current->data) current = current->right; } else if(current->right == NULL && current->left != NULL) { if(newNode->data > current->data) break; else if(newNode->data < current->data) current = current->left; } } if(current->data > newNode->data) current->left = newNode; else current->right = newNode; return; } //public helper function template <typename T> bool BST<T>::search(const T& value) const { return(search(root,value)); //start at the root } //recursive function template <typename T> bool BST<T>::search(BTNode<T>* node, const T& value) const { if(node == NULL || node->data == value) return(node != NULL); //found or couldn't find value else if(value < node->data) return search(node->left,value); //search left subtree else return search(node->right,value); //search right subtree } template <typename T> void BST<T>::printInOrder() const { //print out the value's in the tree in order // //You may need to use this function as a helper //and create a second recursive function //(see "print()" for an example) } template <typename T> void BST<T>::remove(const T& value) { if(root == NULL) { cout << "Tree is empty. No removal. "<<endl; return; } if(!search(value)) { cout << "Value is not in the tree. No removal." << endl; return; } BTNode<T>* current; BTNode<T>* parent; current = root; parent->left = NULL; parent->right = NULL; cout << root->left << "LEFT " << root->right << "RIGHT " << endl; cout << root->data << " ROOT" << endl; cout << current->data << "CURRENT BEFORE" << endl; while(current != NULL) { cout << "INTkhkjhbljkhblkjhlk " << endl; if(current->data == value) break; else if(value > current->data) { parent = current; current = current->right; } else { parent = current; current = current->left; } } cout << current->data << "CURRENT AFTER" << endl; // 3 cases : //We're looking at a leaf node if(current->left == NULL && current->right == NULL) // It's a leaf { if(parent->left == current) parent->left = NULL; else parent->right = NULL; delete current; cout << "The value " << value << " was removed." << endl; return; } // Node with single child if((current->left == NULL && current->right != NULL) || (current->left != NULL && current->right == NULL)) { if(current->left == NULL && current->right != NULL) { if(parent->left == current) { parent->left = current->right; cout << "The value " << value << " was removed." << endl; delete current; } else { parent->right = current->right; cout << "The value " << value << " was removed." << endl; delete current; } } else // left child present, no right child { if(parent->left == current) { parent->left = current->left; cout << "The value " << value << " was removed." << endl; delete current; } else { parent->right = current->left; cout << "The value " << value << " was removed." << endl; delete current; } } return; } //Node with 2 children - Replace node with smallest value in right subtree. if (current->left != NULL && current->right != NULL) { BTNode<T>* check; check = current->right; if((check->left == NULL) && (check->right == NULL)) { current = check; delete check; current->right = NULL; cout << "The value " << value << " was removed." << endl; } else // right child has children { //if the node's right child has a left child; Move all the way down left to locate smallest element if((current->right)->left != NULL) { BTNode<T>* leftCurrent; BTNode<T>* leftParent; leftParent = current->right; leftCurrent = (current->right)->left; while(leftCurrent->left != NULL) { leftParent = leftCurrent; leftCurrent = leftCurrent->left; } current->data = leftCurrent->data; delete leftCurrent; leftParent->left = NULL; cout << "The value " << value << " was removed." << endl; } else { BTNode<T>* temp; temp = current->right; current->data = temp->data; current->right = temp->right; delete temp; cout << "The value " << value << " was removed." << endl; } } return; } } /* Print out the values in the tree and their relationships visually. Sample output: 22 18 15 10 9 5 3 1 */ template <typename T> void BST<T>::print() const { print(root,0); } template <typename T> void BST<T>::print(BTNode<T>* node,int depth) const { if(node == NULL) { std::cout << std::endl; return; } print(node->right,depth+1); for(int i=0; i < depth; i++) { std::cout << "\t"; } std::cout << node->data << std::endl; print(node->left,depth+1); } #endif main.cpp #include "bst.h" #include <iostream> using namespace std; int main() { BST<int> tree; cout << endl << "LAB #13 - BINARY SEARCH TREE PROGRAM" << endl; cout << "----------------------------------------------------------" << endl; // Insert. cout << endl << "INSERT TESTS" << endl; // No duplicates allowed. tree.insert(0); tree.insert(5); tree.insert(15); tree.insert(25); tree.insert(20); // Search. cout << endl << "SEARCH TESTS" << endl; int x = 0; int y = 1; if(tree.search(x)) cout << "The value " << x << " is on the tree." << endl; else cout << "The value " << x << " is NOT on the tree." << endl; if(tree.search(y)) cout << "The value " << y << " is on the tree." << endl; else cout << "The value " << y << " is NOT on the tree." << endl; // Removal. cout << endl << "REMOVAL TESTS" << endl; tree.remove(0); tree.remove(1); tree.remove(20); // Print. cout << endl << "PRINTED DIAGRAM OF BINARY SEARCH TREE" << endl; cout << "----------------------------------------------------------" << endl; tree.print(); cout << endl << "Program terminated. Goodbye." << endl << endl; } BTNode.h #ifndef BTNODE_H_ #define BTNODE_H_ #include <iostream> /* A class to represent a node in a binary search tree. */ template <typename T> class BTNode { public: //constructor BTNode(T d); //the node's data value T data; //pointer to the node's left child BTNode<T>* left; //pointer to the node's right child BTNode<T>* right; }; /* Simple constructor. Sets the data value of the BTNode to "d" and defaults its left and right child pointers to NULL. */ template <typename T> BTNode<T>::BTNode(T d) : left(NULL), right(NULL) { data = d; } #endif Thanks.

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  • NHibernate Many-To-One on Joined Sublcass with Filter

    - by Nathan Roe
    I have a class setup that looks something like this: public abstract class Parent { public virtual bool IsDeleted { get; set; } } public class Child : Parent { } public class Other { public virtual ICollection<Child> Children { get; set; } } Child is mapped as a joined-subclass of Parent. Childen is mapped as a Many-To-One bag. The bag has a filter applied to it named SoftDeletableFilter. The filter mapping looks like: <filter-def name="SoftDeleteableFilter" condition="(IsDeleted = 0 or IsDeleted is null)" /> That problem is that when Other.Children is loaded the filter is being applied to the Child table and not the parent table. Is there any way to tell NHibernate to apply the filter to the parent class?

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  • T-SQL XML Query, how to seperate matching nodes into individual rows?

    - by FlySwat
    I have a table that has a column full of XML like: <parent> <child> <name>Sally</name> </child> <child> <name>Bobby</name> </child> </parent> I'm trying to extract all of the names of the children into seperate rows. My desired resultset would look like: Sally Bobby However, if I do something like: SELECT data.query('data(//parents/child/name)') FROM stuff I get back a single row that looks like Sally Bobby What am I doing wrong with the XPath?

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  • SWT Layout for absolute positioning with minimal-spanning composites

    - by pure.equal
    Hi, I'm writing a DND-editor where I can position elemtents (like buttons, images ...) freely via absolute positioning. Every element has a parent composite. These composites should span/grasp/embrace every element they contain. There can be two or more elements in the same composite and a composite can contain another composite. This image shows how it should look like. To achive this I wrote a custom layoutmanager: import org.eclipse.swt.SWT; import org.eclipse.swt.graphics.Point; import org.eclipse.swt.widgets.Composite; import org.eclipse.swt.widgets.Control; import org.eclipse.swt.widgets.Layout; public class SpanLayout extends Layout { Point[] sizes; int calcedHeight, calcedWidth, calcedX, calcedY; Point[] positions; /* * (non-Javadoc) * * @see * org.eclipse.swt.widgets.Layout#computeSize(org.eclipse.swt.widgets.Composite * , int, int, boolean) * * A composite calls computeSize() on its associated layout to determine the * minimum size it should occupy, while still holding all its child controls * at their minimum sizes. */ @Override protected Point computeSize(Composite composite, int wHint, int hHint, boolean flushCache) { int width = wHint, height = hHint; if (wHint == SWT.DEFAULT) width = composite.getBounds().width; if (hHint == SWT.DEFAULT) height = composite.getBounds().height; return new Point(width, height); } /* * (non-Javadoc) * * @see * org.eclipse.swt.widgets.Layout#layout(org.eclipse.swt.widgets.Composite, * boolean) * * Calculates the positions and sizes for the children of the passed * Composite, then places them accordingly by calling setBounds() on each * one. */ @Override protected void layout(Composite composite, boolean flushCache) { Control children[] = composite.getChildren(); for (int i = 0; i < children.length; i++) { calcedX = calcX(children[i]); calcedY = calcY(children[i]); calcedHeight = calcHeight(children[i]) - calcedY; calcedWidth = calcWidth(children[i]) - calcedX; if (composite instanceof Composite) { calcedX = calcedX - composite.getLocation().x; calcedY = calcedY - composite.getLocation().y; } children[i].setBounds(calcedX, calcedY, calcedWidth, calcedHeight); } } private int calcHeight(Control control) { int maximum = 0; if (control instanceof Composite) { if (((Composite) control).getChildren().length > 0) { for (Control child : ((Composite) control).getChildren()) { int calculatedHeight = calcHeight(child); if (calculatedHeight > maximum) { maximum = calculatedHeight; } } return maximum; } } return control.computeSize(SWT.DEFAULT, SWT.DEFAULT, true).y + control.getLocation().y; } private int calcWidth(Control control) { int maximum = 0; if (control instanceof Composite) { if (((Composite) control).getChildren().length > 0) { for (Control child : ((Composite) control).getChildren()) { int calculatedWidth = calcWidth(child); if (calculatedWidth > maximum) { maximum = calculatedWidth; } } return maximum; } } return control.computeSize(SWT.DEFAULT, SWT.DEFAULT, true).x + control.getLocation().x; } private int calcX(Control control) { int minimum = Integer.MAX_VALUE; if (control instanceof Composite) { if (((Composite) control).getChildren().length > 0) { for (Control child : ((Composite) control).getChildren()) { int calculatedX = calcX(child); if (calculatedX < minimum) { minimum = calculatedX; } } return minimum; } } return control.getLocation().x; } private int calcY(Control control) { int minimum = Integer.MAX_VALUE; if (control instanceof Composite) { if (((Composite) control).getChildren().length > 0) { for (Control child : ((Composite) control).getChildren()) { int calculatedY = calcY(child); if (calculatedY < minimum) { minimum = calculatedY; } } return minimum; } } return control.getLocation().y; } } The problem with it is that it always positions the composite at the position (0,0). This is because it tries to change the absolute positioning into a relative one. Lets say I position a image at position (100,100) and one at (200,200). Then it has to calculate the location of the composite to be at (100,100) and spanning the one at (200,200). But as all child positions are relative to their parents I have to change the positions of the children to remove the 100px offset of the parent. When the layout gets updated it moves everything to the top-left corner (as seen in the image) because the position of the image is not (100,100) but (0,0) since I tried to remove the 100px offset of the partent. Where is my error in reasoning? Is this maybe a totally wrong approach? Is there maybe an other way to achive the desired behavior? Thanks in advance! Best regards, Ed

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  • How can I use a TreeWalker to dump out the DOM

    - by michael
    Hi, In w3c dom interface, there is DOMTreeWalker to traverse the DOM. I find this example which uses DOMTreeWalker: http://www.java-tips.org/java-se-tips/org.w3c.dom/how-to-traverse-the-dom-tree-using-treewalker-2.html But my question is How can I use the DOMTreeWalk to dump the dom structure and value to a file? i.e. I need to know when a parent start and a parent end so that I can do when I dump the dom to a file (i need to insert after each parent dumps its children: <parent> <child a1="1" a2="2"/> <child a3="1" a4="3"/> <child a5="1" a6="5"/> </parent>

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  • Short hand for number of tags having lengthier [and almost same] Xpath

    - by infant programmer
    For example : this is an xslt <xsl:template match="/root/sub-root/parent/child/grand_child/dummy1/dummy2/dummy3/dummy4/node_1 |/root/sub-root/parent/child/grand_child/dummy1/dummy2/dummy3/dummy4/node_2 |/root/sub-root/parent/child/grand_child/dummy1/dummy2/dummy3/dummy4/node_3 . . |/root/sub-root/parent/child/grand_child/dummy1/dummy2/dummy3/dummy4/node_N"/> In the above code can I use the XPath /root/sub-root/parent/child/grand_child/dummy1/dummy2/dummy3/dummy4 only once [use the braces or whatever] and reduce the bulkiness of the code? As you can see, all the nodes are siblings of each-other, so except their name their Xpath is same. Is there any short-hand property ? Does XSLT 1.0 (or Xpath1.0) allows it?

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  • python: how to convert list of lists into a single nested list

    - by Bhuski
    I have a python list of lists as shown below: mylist=[ [['orphan1', ['some value1']]], [['parent1', ['child1', ['child', ['some value2']]]]], [['parent1', ['child2', ['child', ['some value3']]]]] ] I need to convert the above list to some thing like this: result=[ ['orphan1', ['some value1']], ['parent1', ['child1', ['child', ['some value2']]], ['child2', ['child', ['some value3']]]] ] Kindly help me approach this problem. I have given only simple list. In actual scenario here, in my list, even grand parents/grand childs are there. How much ever deep the input nested list is, I need to convert it to a single nested list, with common list elements (parents and grand parents) appearing only once. (but the next to innermost list element('child' in above example) should appear as many times it occurs in the input list. I have been trying to do this last two days, but did not end up with working solution :(. I need to use the output in django template filter: unordered_list so that the resultant nested list appears as a nested unordered list in my html page ..

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  • Jquery - Get all other menus to slide up

    - by Jova
    I have a drop down menu made with Jquery. Right now, if you hover over all the menu items, their submenus will all show at the same time. I would like that if I hover over one menu, then go to the next, that the previous menu's submenu will slide back up. As it is now, I have to hover out of the submenus for them to slide up. To put it simpler, when hovering over a parent menu, slide up all other parent menu's submenus if they're open. $("#DropDownMenu li.parent").hover(function() { $(this).find(".subMenu").slideDown('fast').show(); $(this).parent().find(".subMenu").hover(function() { }, function() { $(this).parent().find(".subMenu").slideUp('slow'); }); });

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  • Can Flash AS3 listbox data contain variable info?

    - by Anne
    I'm populating a listbox like this: dp.addItem( {label:"red dress", data:"OV4MP/23OL.swf"} ); Instead of data:"OV4MP/23OL.swf", I would like to make part of the data file name a variable from a dynamic textbox named centerPt that belongs to the parent movieclip, so I did this: dp.addItem( {label:"red dress", data:"OV4MP/23"+MovieClip(parent.parent).centerPt.text+".swf"} ); When I trace the selectedItem.data using: trace("you have selected: "+ overlays.selectedItem.data); trace(MovieClip(parent.parent).centerPt.text); I'm getting: you have selected: OV4MP/23.swf. What I should get is OV4MP/23OL.swf. It is not picking up what is in the dynamic centerPt.text field which are the letters OL eventhough that text field is tracing correctly. Is it possible that data can not hold a variable? Thank you in advance for any help. Anne

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  • JSF Render response programatically

    - by Shamik
    I have one parent page with a parentManagedBean (attached to Session Scope). On click of a button on this parent page, one popup comes which has a childManagedBean (attached to Request scope). Now ChildManagedBedan holds a reference to parentManaged bean thru JSFs managed Property feature. On this popup window, user selects some option which populates a large value object class. I use the managed property of childMnaagedBean to set the values from this large object to that of parentmanagedbean. Problem is - The parent page shows a link, on click of which a popup comes, on selection of the popup, the popup disappears and set the values to the parentManaged bean.So far so good, but the newly set values need to appear on the parent page. This is where I am stuck. How to programatically render the master page/render page when I am at the child managed bean... is there a way I can get handle of the parent page and refresh it ?

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  • versioning fails for onetomany collection holder

    - by Alexander Vasiljev
    given parent entity @Entity public class Expenditure implements Serializable { ... @OneToMany(mappedBy = "expenditure", cascade = CascadeType.ALL, orphanRemoval = true) @OrderBy() private List<ExpenditurePeriod> periods = new ArrayList<ExpenditurePeriod>(); @Version private Integer version = 0; ... } and child one @Entity public class ExpenditurePeriod implements Serializable { ... @ManyToOne @JoinColumn(name="expenditure_id", nullable = false) private Expenditure expenditure; ... } While updating both parent and child in one transaction, org.hibernate.StaleObjectStateException is thrown: Row was updated or deleted by another transaction (or unsaved-value mapping was incorrect): Indeed, hibernate issues two sql updates: one changing parent properties and another changing child properties. Do you know a way to get rid of parent update changing child? The update results both in inefficiency and false positive for optimistic lock. Note, that both child and parent save their state in DB correctly. Hibernate version is 3.5.1-Final

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  • How to get predecessor and successors from an adjacency matrix

    - by NickTFried
    Hi I am am trying to complete an assignment, where it is ok to consult the online community. I have to create a graph class that ultimately can do Breadth First Search and Depth First Search. I have been able to implement those algorithms successfully however another requirement is to be able to get the successors and predecessors and detect if two vertices are either predecessors or successors for each other. I'm having trouble thinking of a way to do this. I will post my code below, if anyone has any suggestions it would be greatly appreciated. import java.util.ArrayList; import java.util.Iterator; import java.util.LinkedList; import java.util.Queue; import java.util.Stack; public class Graph<T> { public Vertex<T> root; public ArrayList<Vertex<T>> vertices=new ArrayList<Vertex<T>>(); public int[][] adjMatrix; int size; private ArrayList<Vertex<T>> dfsArrList; private ArrayList<Vertex<T>> bfsArrList; public void setRootVertex(Vertex<T> n) { this.root=n; } public Vertex<T> getRootVertex() { return this.root; } public void addVertex(Vertex<T> n) { vertices.add(n); } public void removeVertex(int loc){ vertices.remove(loc); } public void addEdge(Vertex<T> start,Vertex<T> end) { if(adjMatrix==null) { size=vertices.size(); adjMatrix=new int[size][size]; } int startIndex=vertices.indexOf(start); int endIndex=vertices.indexOf(end); adjMatrix[startIndex][endIndex]=1; adjMatrix[endIndex][startIndex]=1; } public void removeEdge(Vertex<T> v1, Vertex<T> v2){ int startIndex=vertices.indexOf(v1); int endIndex=vertices.indexOf(v2); adjMatrix[startIndex][endIndex]=1; adjMatrix[endIndex][startIndex]=1; } public int countVertices(){ int ver = vertices.size(); return ver; } /* public boolean isPredecessor( Vertex<T> a, Vertex<T> b){ for() return true; }*/ /* public boolean isSuccessor( Vertex<T> a, Vertex<T> b){ for() return true; }*/ public void getSuccessors(Vertex<T> v1){ } public void getPredessors(Vertex<T> v1){ } private Vertex<T> getUnvisitedChildNode(Vertex<T> n) { int index=vertices.indexOf(n); int j=0; while(j<size) { if(adjMatrix[index][j]==1 && vertices.get(j).visited==false) { return vertices.get(j); } j++; } return null; } public Iterator<Vertex<T>> bfs() { Queue<Vertex<T>> q=new LinkedList<Vertex<T>>(); q.add(this.root); printVertex(this.root); root.visited=true; while(!q.isEmpty()) { Vertex<T> n=q.remove(); Vertex<T> child=null; while((child=getUnvisitedChildNode(n))!=null) { child.visited=true; bfsArrList.add(child); q.add(child); } } clearVertices(); return bfsArrList.iterator(); } public Iterator<Vertex<T>> dfs() { Stack<Vertex<T>> s=new Stack<Vertex<T>>(); s.push(this.root); root.visited=true; printVertex(root); while(!s.isEmpty()) { Vertex<T> n=s.peek(); Vertex<T> child=getUnvisitedChildNode(n); if(child!=null) { child.visited=true; dfsArrList.add(child); s.push(child); } else { s.pop(); } } clearVertices(); return dfsArrList.iterator(); } private void clearVertices() { int i=0; while(i<size) { Vertex<T> n=vertices.get(i); n.visited=false; i++; } } private void printVertex(Vertex<T> n) { System.out.print(n.label+" "); } }

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  • Strange: Planner takes decision with lower cost, but (very) query long runtime

    - by S38
    Facts: PGSQL 8.4.2, Linux I make use of table inheritance Each Table contains 3 million rows Indexes on joining columns are set Table statistics (analyze, vacuum analyze) are up-to-date Only used table is "node" with varios partitioned sub-tables Recursive query (pg = 8.4) Now here is the explained query: WITH RECURSIVE rows AS ( SELECT * FROM ( SELECT r.id, r.set, r.parent, r.masterid FROM d_storage.node_dataset r WHERE masterid = 3533933 ) q UNION ALL SELECT * FROM ( SELECT c.id, c.set, c.parent, r.masterid FROM rows r JOIN a_storage.node c ON c.parent = r.id ) q ) SELECT r.masterid, r.id AS nodeid FROM rows r QUERY PLAN ----------------------------------------------------------------------------------------------------------------------------------------------------------------- CTE Scan on rows r (cost=2742105.92..2862119.94 rows=6000701 width=16) (actual time=0.033..172111.204 rows=4 loops=1) CTE rows -> Recursive Union (cost=0.00..2742105.92 rows=6000701 width=28) (actual time=0.029..172111.183 rows=4 loops=1) -> Index Scan using node_dataset_masterid on node_dataset r (cost=0.00..8.60 rows=1 width=28) (actual time=0.025..0.027 rows=1 loops=1) Index Cond: (masterid = 3533933) -> Hash Join (cost=0.33..262208.33 rows=600070 width=28) (actual time=40628.371..57370.361 rows=1 loops=3) Hash Cond: (c.parent = r.id) -> Append (cost=0.00..211202.04 rows=12001404 width=20) (actual time=0.011..46365.669 rows=12000004 loops=3) -> Seq Scan on node c (cost=0.00..24.00 rows=1400 width=20) (actual time=0.002..0.002 rows=0 loops=3) -> Seq Scan on node_dataset c (cost=0.00..55001.01 rows=3000001 width=20) (actual time=0.007..3426.593 rows=3000001 loops=3) -> Seq Scan on node_stammdaten c (cost=0.00..52059.01 rows=3000001 width=20) (actual time=0.008..9049.189 rows=3000001 loops=3) -> Seq Scan on node_stammdaten_adresse c (cost=0.00..52059.01 rows=3000001 width=20) (actual time=3.455..8381.725 rows=3000001 loops=3) -> Seq Scan on node_testdaten c (cost=0.00..52059.01 rows=3000001 width=20) (actual time=1.810..5259.178 rows=3000001 loops=3) -> Hash (cost=0.20..0.20 rows=10 width=16) (actual time=0.010..0.010 rows=1 loops=3) -> WorkTable Scan on rows r (cost=0.00..0.20 rows=10 width=16) (actual time=0.002..0.004 rows=1 loops=3) Total runtime: 172111.371 ms (16 rows) (END) So far so bad, the planner decides to choose hash joins (good) but no indexes (bad). Now after doing the following: SET enable_hashjoins TO false; The explained query looks like that: QUERY PLAN ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- CTE Scan on rows r (cost=15198247.00..15318261.02 rows=6000701 width=16) (actual time=0.038..49.221 rows=4 loops=1) CTE rows -> Recursive Union (cost=0.00..15198247.00 rows=6000701 width=28) (actual time=0.032..49.201 rows=4 loops=1) -> Index Scan using node_dataset_masterid on node_dataset r (cost=0.00..8.60 rows=1 width=28) (actual time=0.028..0.031 rows=1 loops=1) Index Cond: (masterid = 3533933) -> Nested Loop (cost=0.00..1507822.44 rows=600070 width=28) (actual time=10.384..16.382 rows=1 loops=3) Join Filter: (r.id = c.parent) -> WorkTable Scan on rows r (cost=0.00..0.20 rows=10 width=16) (actual time=0.001..0.003 rows=1 loops=3) -> Append (cost=0.00..113264.67 rows=3001404 width=20) (actual time=8.546..12.268 rows=1 loops=4) -> Seq Scan on node c (cost=0.00..24.00 rows=1400 width=20) (actual time=0.001..0.001 rows=0 loops=4) -> Bitmap Heap Scan on node_dataset c (cost=58213.87..113214.88 rows=3000001 width=20) (actual time=1.906..1.906 rows=0 loops=4) Recheck Cond: (c.parent = r.id) -> Bitmap Index Scan on node_dataset_parent (cost=0.00..57463.87 rows=3000001 width=0) (actual time=1.903..1.903 rows=0 loops=4) Index Cond: (c.parent = r.id) -> Index Scan using node_stammdaten_parent on node_stammdaten c (cost=0.00..8.60 rows=1 width=20) (actual time=3.272..3.273 rows=0 loops=4) Index Cond: (c.parent = r.id) -> Index Scan using node_stammdaten_adresse_parent on node_stammdaten_adresse c (cost=0.00..8.60 rows=1 width=20) (actual time=4.333..4.333 rows=0 loops=4) Index Cond: (c.parent = r.id) -> Index Scan using node_testdaten_parent on node_testdaten c (cost=0.00..8.60 rows=1 width=20) (actual time=2.745..2.746 rows=0 loops=4) Index Cond: (c.parent = r.id) Total runtime: 49.349 ms (21 rows) (END) - incredibly faster, because indexes were used. Notice: Cost of the second query ist somewhat higher than for the first query. So the main question is: Why does the planner make the first decision, instead of the second? Also interesing: Via SET enable_seqscan TO false; i temp. disabled seq scans. Than the planner used indexes and hash joins, and the query still was slow. So the problem seems to be the hash join. Maybe someone can help in this confusing situation? thx, R.

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  • Refresh page isnt working in asp.net using treeview

    - by Greg
    Hi, I am trying to refresh an asp.net page using this command: <meta http-equiv="Refresh" content="10"/> On that page I have 2 treeviews. The refresh works ok when I just open the page, but when I click on one of the treeviews and expand it, the refresh stopps working and the page isnt being refreshed. Any ideas why this can happen? Is there any connection to the treeview being expanded? Here is the full code of the page: public partial class Results : System.Web.UI.Page { protected void Page_Load(object sender, EventArgs e) { } // Function that moves reviewed yellow card to reviewed tree protected void ycActiveTree_SelectedNodeChanged(object sender, EventArgs e) { ycActiveTree.SelectedNode.Text = "Move To Active"; ycReviewedTree.PopulateNodesFromClient = false; ycReviewedTree.Nodes[ycReviewedTree.Nodes.Count - 1].ChildNodes.Add(ycActiveTree.SelectedNode.Parent); Application["reviewedTree"] = new ArrayList(); int count = ((ArrayList)Application["activeTree"]).Count; // Move all the nodes from activeTree application to reviewedTree application for (int i = 0; Application["activeTree"] != null && i < count; i++) { ((ArrayList)Application["reviewedTree"]).Add(((ArrayList)Application["activeTree"])[i]); ((ArrayList)Application["activeTree"]).RemoveAt(0); } } protected void ycActiveTree_TreeNodePopulate(object sender, TreeNodeEventArgs e) { if (Application["idList"] != null && e.Node.Depth == 0) { string[] words = ((String)Application["idList"]).Split(' '); // Yellow Card details TreeNode child = new TreeNode(""); // Go over all the yellow card details and populate the treeview for (int i = 1; i < words.Length; i++) { child.SelectAction = TreeNodeSelectAction.None; // Same yellow card if (words[i] != "*") { // End of details and start of point ip's if (words[i] == "$") { // Add the yellow card node TreeNode yellowCardNode = new TreeNode(child.Text); yellowCardNode.SelectAction = TreeNodeSelectAction.Expand; e.Node.ChildNodes.Add(yellowCardNode); child.Text = ""; } // yellow card details else { child.Text = child.Text + words[i] + " "; } } // End of yellow card else { child.PopulateOnDemand = false; child.SelectAction = TreeNodeSelectAction.None; // Populate the yellow card node e.Node.ChildNodes[e.Node.ChildNodes.Count - 1].ChildNodes.Add(child); TreeNode moveChild = new TreeNode("Move To Reviewed"); moveChild.PopulateOnDemand = false; moveChild.SelectAction = TreeNodeSelectAction.Select; e.Node.ChildNodes[e.Node.ChildNodes.Count - 1].ChildNodes.Add(moveChild); child = new TreeNode(""); Application["activeTree"] = new ArrayList(); ((ArrayList)Application["activeTree"]).Add(e.Node.ChildNodes[e.Node.ChildNodes.Count - 1]); } } } // If there arent new yellow cards else if (Application["activeTree"] != null) { // Populate the active tree for (int i = 0; i < ((ArrayList)Application["activeTree"]).Count; i++) { e.Node.ChildNodes.Add((TreeNode)((ArrayList)Application["activeTree"])[i]); } } // If there were new yellow cards and nodes that moved from reviewed tree to active tree if (Application["idList"] != null && Application["activeTree"] != null && e.Node.ChildNodes.Count != ((ArrayList)Application["activeTree"]).Count) { for (int i = e.Node.ChildNodes.Count; i < ((ArrayList)Application["activeTree"]).Count; i++) { e.Node.ChildNodes.Add((TreeNode)((ArrayList)Application["activeTree"])[i]); } } // Nullify the yellow card id's Application["idList"] = null; } protected void ycReviewedTree_SelectedNodeChanged(object sender, EventArgs e) { ycActiveTree.PopulateNodesFromClient = false; ycReviewedTree.SelectedNode.Text = "Move To Reviewed"; ycActiveTree.Nodes[ycActiveTree.Nodes.Count - 1].ChildNodes.Add(ycReviewedTree.SelectedNode.Parent); int count = ((ArrayList)Application["reviewedTree"]).Count; // Move all the nodes from reviewedTree application to activeTree application for (int i = 0; Application["reviewedTree"] != null && i < count; i++) { ((ArrayList)Application["activeTree"]).Add(((ArrayList)Application["reviewedTree"])[i]); ((ArrayList)Application["reviewedTree"]).RemoveAt(0); } } protected void ycReviewedTree_TreeNodePopulate(object sender, TreeNodeEventArgs e) { if (Application["reviewedTree"] != null) { // Populate the reviewed tree for (int i = 0; i < ((ArrayList)Application["reviewedTree"]).Count; i++) { e.Node.ChildNodes.Add((TreeNode)((ArrayList)Application["reviewedTree"])[i]); } } } } Thanks, Greg

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  • how to register a selectionProvider for the Viewer from a superclass

    - by Tom
    I have a View which extends a View provided by another plugin like this; public class View2 extends some.other.package.View1 { public void createPartControl(Composite parent) { super.createPartControl(parent); //I want to do this, but I can't because its private //getSite().setSelectionProvider(treeViewer); } } But I can't edit the super class, so I need to access the viewer from the Control. public class some.other.package.View1 extends ViewPart { private TreeViewer treeViewer; public void createPartControl(Composite parent) { treeViewer = new TreeViewer(parent, SWT.V_SCROLL); } } I tried using parent.getChildren(); to get a Tree object, but I can't see that it can be cooerced into an ISElectionProvider as required

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  • WSS 3.0 to SharePoint 2010: Tips for delaying the Visual Upgrade

    - by Kelly Jones
    My most recent project has been to migrate a bunch of sites from WSS 3.0 (SharePoint 2007) to SharePoint Server 2010.  The users are currently working with WSS 3.0 and Office 2003, so the new ribbon based UI in 2010 will be completely new.  My client wants to avoid the new SharePoint 2010 look and feel until they’ve had time to train their users, so we’ve been testing the upgrades by keeping them with the 2007 user interface. Permission to perform the Visual Upgrade One of the first things we noticed was the default permissions for who was allowed to switch the UI from 2007 to 2010.  By default, site collection administrators and site owners can do this.  Since we wanted to more tightly control the timing of the new UI, I added a few lines to the PowerShell script that we are using to perform the migration.  This script creates the web application, sets the User Policy, and then does a Mount-SPDatabase to attach the old 2007 content database to the 2010 farm.  I added the following steps after the Mount-SPDatabase step: #Remove the visual upgrade option for site owners # it remains for Site Collection administrators foreach ($sc in $WebApp.Sites){ foreach ($web in $sc.AllWebs){ #Visual Upgrade permissions for the site/subsite (web) $web.UIversionConfigurationEnabled = $false; $web.Update(); } } These script steps loop through each Site Collection in a particular web application ($WebApp) and then it loops through each subsite ($web) in the Site Collection ($sc) and disables the Site Owner’s permission to perform the Visual Upgrade. This is equivalent to going to the Site Collection administrator settings page –> Visual Upgrade and selecting “Hide Visual Upgrade”. Since only IT people have Site Collection administrator privileges, this will allow IT to control the timing of the new 2010 UI rollout. Newly created subsites Our next issue was brought to our attention by SharePoint Joel’s blog post last week (http://www.sharepointjoel.com/Lists/Posts/Post.aspx?ID=524 ).  In it, he lists some updates about the 2010 upgrade, and his fourth point was one that I hadn’t seen yet: 4. If a 2007 upgraded site has not been visually upgraded, the sites created underneath it will look like 2010 sites – While this is something I’ve been aware of, I think many don’t realize how this impacts common look and feel for master pages, and how it impacts good navigation and UI. As well depending on your patch level you may see hanging behavior in the list picker. The site and list creation Silverlight control in Internet Explorer is looking for resources that don’t exist in the galleries in the 2007 site, and hence it continues to spin and spin and eventually time out. The work around is to upgrade to SP1, or use Chrome or Firefox which won’t attempt to render the Silverlight control. When the root site collection is a 2007 site and has it’s set of galleries and the children are 2010 sites there is some strange behavior linked to the way that the galleries work and pull from the parent. Our production SharePoint 2010 Farm has SP1 installed, as well as the December 2011 Cumulative Update, so I think the “hanging behavior” he mentions won’t affect us. However, since we want to control the roll out of the UI, we are concerned that new subsites will have the 2010 look and feel, no matter what the parent site has. Ok, time to dust off my developer skills. I first looked into using feature stapling, but I couldn’t get that to work (although I’m pretty sure I had everything wired up correctly).  Then I stumbled upon SharePoint 2010’s web events – a great way to handle this. Using Visual Studio 2010, I created a new SharePoint project and added a Web Event Receiver: In the Event Receiver class, I used the WebProvisioned method to check if the parent site is a 2007 site (UIVersion = 3), and if so, then set the newly created site to 2007:   /// <summary> /// A site was provisioned. /// </summary> public override void WebProvisioned(SPWebEventProperties properties) { base.WebProvisioned(properties);   try { SPWeb curweb = properties.Web;   if (curweb.ParentWeb != null) {   //check if the parent website has the 2007 look and feel if (curweb.ParentWeb.UIVersion == 3) { //since parent site has 2007 look and feel // we'll apply that look and feel to the current web curweb.UIVersion = 3; curweb.Update(); } } } catch (Exception) { //TODO: Add logging for errors } }   This event is part of a Feature that is scoped to the Site Level (Site Collection).  I added a couple of lines to my migration PowerShell script to activate the Feature for any site collections that we migrate. Plan Going Forward The plan going forward is to perform the visual upgrade after the users for a particular site collection have gone through 2010 training. If we need to do several site collections at once, we’ll use a PowerShell script to loop through each site collection to update the sites to 2010.  If it’s just one or two, we’ll be using the “Update All Sites” button on the Visual Upgrade page for Site Collection Administrators. The custom code for newly created sites won’t need to be changed, since it relies on the UI version of the parent site.  If the parent is 2010, then the new site will look 2010.

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  • Translatable behavior not working if used with Containable

    - by bakerjr
    An example: $this-Parent-Behaviors-attach('Containable'); $parent = $this->Parent->find('first', array( 'contain' => array('Child' => array( 'order' => 'Child.order ASC', )), 'conditions' => array('Parent.id' => $parentId) ) ); Child has translated data and parent isn't using translatable. When I fetch the child data this way it's not translated. Is Translatable (and also SmoothTranslate) working with Containable? What solutions would you guys recommend? Thanks in advance

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