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  • How To Disconnect Non-Mapped UNC Path “Drives” in Windows

    - by The Geek
    Have you ever browsed over to another PC on your network using “network neighborhood”, and then connected to one of the file shares? Without a drive letter, how do you disconnect yourself once you’ve done so? Really confused as to what I’m talking about? Let’s walk through the process. First, imagine that you browse through and connect to a share, entering your username and password to gain access. The problem is that you stay connected, and there’s no visible way to disconnect yourself. If you try and shut down the other PC, you’ll receive a message that users are still connected. So let’s disconnect! Open up a command prompt, and then type in the following: net use This will give you a list of the connected drives, including the ones that aren’t actually mapped to a drive letter. To disconnect one of the connections, you can use the following command: net use /delete \\server\sharename For example, in this instance we’d disconnect like so: net use /delete \\192.168.1.205\root$ Now when you run the “net use” command again, you’ll see that you’ve been properly disconnected. If you wanted to actually connect to a share without mapping a drive letter, you can do the following: net use /user:Username \\server\sharename Password You could then just pop \\server\sharename into a Windows Explorer window and browse the files that way. Note that this technique should work exactly the same in any version of windows. Similar Articles Productive Geek Tips Remove "Map Network Drive" Menu Item from Windows Vista or XPDisable the Annoying "This page has an unspecified potential security risk" When Using Files on a Network ShareCopy Path of a File to the Clipboard in Windows 7 or VistaMap a Network Drive from XP to Windows 7Defrag Multiple Hard Drives At Once In Windows TouchFreeze Alternative in AutoHotkey The Icy Undertow Desktop Windows Home Server – Backup to LAN The Clear & Clean Desktop Use This Bookmarklet to Easily Get Albums Use AutoHotkey to Assign a Hotkey to a Specific Window Latest Software Reviews Tinyhacker Random Tips DVDFab 6 Revo Uninstaller Pro Registry Mechanic 9 for Windows PC Tools Internet Security Suite 2010 Enable DreamScene in Windows 7 Microsoft’s “How Do I ?” Videos Home Networks – How do they look like & the problems they cause Check Your IMAP Mail Offline In Thunderbird Follow Finder Finds You Twitter Users To Follow Combine MP3 Files Easily

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  • Spotlight on a career path: Paul, Business Development Consultant

    - by Maria Sandu
    I came to work for Oracle in November 2012 as a Customer Intelligence Representative and since then I was promoted to a Business Development Consultant, for Commercial Industries in the UK, based in Dublin. My background was primarily in Logistics, working for such companies as Indaver Ireland, Wincanton and P&O. I spent 10 years working in this industry and gained experience in negotiating with customers and suppliers in order to meet the needs of both, monitoring the quality and quantity of goods as well as the efficiency and organisation of the movement and storage of products. I decided to move from my logistics career in 2009 to study Information Technology in D.I.T. This was a challenge for me to move my career path; however the lectures at the college helped me significantly with the ability to understand how IT can have an effect on how businesses operate. Following on from college I came to work for Oracle. This also presented challenges but the training I received and the encouragement from management helped me understand that the same business rules apply no matter what background you come from. I have also learnt that using my past experience in working with customers and suppliers in Logistics has helped me understand how to meet customer’s needs. Oracle has offered me excellent training such as Sandler Sales Techniques and John Costigan. I continue to get all the training that I need to develop my career. If you’re interested in joining the Business Development Group visit http://bit.ly/oracledirectcareers or follow our CareersatOracle Facebook Community! /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin-top:0in; mso-para-margin-right:0in; mso-para-margin-bottom:10.0pt; mso-para-margin-left:0in; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;}

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  • Inkscape: what are "line" objects?

    - by Peter Mortensen
    What is a "line" object in Inkscape? Drawing lines in Inkscape is by using the tool "Draw Bezier curves and straight lines (Shift+F6)". This creates objects of another type, "path". Using Inkscape: is there a way to convert an object of type "line" into an object of the more general type "path"? I have imported a drawing (mostly lines, rectangles and text) that has been through Adobe Illustrator: originally made in Inkscape, imported into Illustrator, edited, saved from Illustrator as SVG, imported into Inkscape. Sample from the imported SVG file: <path id="path5855" stroke="#000000" d=" M320.198,275.935" /> <line fill="none" stroke="#000000" x1="348.553" y1="45.097" x2="348.553" y2="185.346" id="line3368" /> Update 1: I have inspected the original XML (SVG) file from 2006 and it does not contain any "line" XML tags. Thus it must be a crime of Adobe Illustrator. When a line is selected in this imported SVG file the bottom panel displays: "Line in root. Click selection to toggle scale/rotation handles.". When a line is selected that was drawn in Inkscape the bottom panel displays: "Path (2 nodes) in Layer 1. Click selection to toggle scale/rotation handles." What is the difference between "line" and "path"? Is "line" some kind of read-only/non-editable object? A generic term like "line" is not easy to use in search, but I have now found the definitions for "line" and "path": SVG line: http://www.w3schools.com/svg/svg_line.asp SVG path: http://www.w3schools.com/svg/svg_path.asp Platform: Inkscape v0.46 (2008-03-10), Windows XP 64 bit, 8 GB RAM.

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  • Finding a person in the forest

    - by PointsToShare
    © 2011 By: Dov Trietsch. All rights reserved finding a person in the forest or Limiting the AD result in SharePoint People Picker There are times when we need to limit the SharePoint audience of certain farms or servers or site collections to a particular audience. One of my experiences involved limiting access to US citizens, another to a particular location. Now, most of us – your humble servant included – are not Active Directory experts – but we must be able to handle the “audience restrictions” as required. So here is how it’s done in a nutshell. Important note. Not all could be done in PowerShell (at least not yet)! There are no Windows PowerShell commands to configure People Picker. The stsadm command is: stsadm -o setproperty -pn peoplepicker-searchadcustomquery -pv ADQuery –url http://somethingOrOther Note the long-hyphenated property name. Now to filling the ADQuery.   LDAP Query in a nutshell Syntax LDAP is no older than SQL and an LDAP query is actually a query against the LDAP Database. LDAP attributes are the equivalent of Database columns, so why do we have to learn a new query language? Beats me! But we must, so here it is. The syntax of an LDAP query string is made of individual statements with relational operators including: = Equal <= Lower than or equal >= Greater than or equal… and memberOf – a group membership. ! Not * Wildcard Equal and memberOf are the most commonly used. Checking for absence uses the ! – not and the * - wildcard Example: (SN=Grant) All whose last name – SurName – is Grant Example: (!(SN=Grant)) All except Grant Example: (!(SN=*)) all where there is no SurName i.e SurName is absent (probably Rappers). Example: (CN=MyGroup) Common Name is MyGroup.  Example: (GN=J*) all the Given Names that start with J (JJ, Jane, Jon, John, etc.) The cryptic SN, CN, GN, etc. are attributes and more about them later All the queries are enclosed in parentheses (Query). Complex queries are comprised of sets that are in AND or OR conditions. AND is denoted by the ampersand (&) and the OR is denoted by the vertical pipe (|). The general syntax is that of the Prefix polish notation where the operand precedes the variables. E.g +ab is the sum of a and b. In an LDAP query (&(A)(B)) will garner the objects for which both A and B are true. In an LDAP query (&(A)(B)(C)) will garner the objects for which A, B and C are true. There’s no limit to the number of conditions. In an LDAP query (|(A)(B)) will garner the objects for which either A or B are true. In an LDAP query (|(A)(B)(C)) will garner the objects for which at least one of A, B and C is true. There’s no limit to the number of conditions. More complex queries have both types of conditions and the parentheses determine the order of operations. Attributes Now let’s get into the SN, CN, GN, and other attributes of the query SN – is the SurName (last name) GN – is the Given Name (first name) CN – is the Common Name, usually GN followed by SN OU – is an Organization Unit such as division, department etc. DC – is a Domain Content in the AD forest l – lower case ‘L’ stands for location. Jerusalem anybody? Or Katmandu. UPN – User Principal Name, is usually the first part of an email address. By nature it is unique in the forest. Most systems set the UPN to be the first initial followed by the SN of the person involved. Some limit the total to 8 characters. If we have many ‘jsmith’ we have to somehow distinguish them from each other. DN – is the distinguished name – a name unique to AD forest in which it lives. Usually it’s a CN with some domain or group distinguishers. DN is important in conjunction with the memberOf relation. Groups have stricter requirement. Each group has to have a unique name - its CN and it has to be unique regardless of its place. See more below. All of the attributes are case insensitive. CN, cn, Cn, and cN are identical. objectCategory is an element that requires special consideration. AD contains many different object like computers, printers, and of course people and groups. In the queries below, we’re limiting our search to people (person). Putting it altogether Let’s get a list of all the Johns in the SPAdmin group of the Jerusalem that local domain. (&(objectCategory=person)(memberOf=cn=SPAdmin,ou=Jerusalem,dc=local)) The memberOf=cn=SPAdmin uses the cn (Common Name) of the SPAdmin group. This is how the memberOf relation is used. ‘SPAdmin’ is actually the DN of the group. Also the memberOf relation does not allow wild cards (*) in the group name. Also, you are limited to at most one ‘OU’ entry. Let’s add Marvin Minsky to the search above. |(&(objectCategory=person)(memberOf=cn=SPAdmin,ou=Jerusalem,dc=local))(CN=Marvin Minsky) Here I added the or pipeline at the beginning of the query and put the CN requirement for Minsky at the end. Note that if Marvin was already in the prior result, he’s not going to be listed twice. One last note: You may see a dryer but more complete list of attributes rules and examples in: http://www.tek-tips.com/faqs.cfm?fid=5667 And finally (thus negating the claim that my previous note was last), to the best of my knowledge there are 3 more ways to limit the audience. One is to use the peoplepicker-searchadcustomfilter property using the same ADQuery. This works only in SP1 and above. The second is to limit the search to users within this particular site collection – the property name is peoplepicker-onlysearchwithinsitecollection and the value is yes (-pv yes) And the third is –pn peoplepicker-serviceaccountdirectorypaths –pv “OU=ou1,DC=dc1…..” Again you are limited to at most one ‘OU’ phrase – no OU=ou1,OU=ou2… And now the real end. The main property discussed in this sprawling and seemingly endless monogram – peoplepicker-searchadcustomquery - is the most general way of getting the job done. Here are a few examples of command lines that worked and some that didn’t. Can you see why? C:\Program Files\Common Files\Microsoft Shared\Web Server Extensions\12\BIN>stsa dm -o setproperty -url http://somethingOrOther -pn peoplepicker-searchadcustomfi lter -pv (Title=David) Operation completed successfully. C:\Program Files\Common Files\Microsoft Shared\Web Server Extensions\12\BIN>stsa dm -o setproperty -url http://somethingOrOther -pn peoplepicker-searchadcustomfi lter -pv (!Title=David) Operation completed successfully. C:\Program Files\Common Files\Microsoft Shared\Web Server Extensions\12\BIN>stsa dm -o setproperty -url http://somethingOrOther -pn peoplepicker-searchadcustomfi lter -pv (OU=OURealName,OU=OUMid,OU=OUTop,DC=TopDC,DC=MidDC,DC=BottomDC) Command line error. Too many OUs C:\Program Files\Common Files\Microsoft Shared\Web Server Extensions\12\BIN>stsa dm -o setproperty -url http://somethingOrOther -pn peoplepicker-searchadcustomfi lter -pv (OU=OURealName) Operation completed successfully. C:\Program Files\Common Files\Microsoft Shared\Web Server Extensions\12\BIN>stsa dm -o setproperty -url http://somethingOrOther -pn peoplepicker-searchadcustomfi lter -pv (DC=TopDC,DC=MidDC,DC=BottomDC) Operation completed successfully. C:\Program Files\Common Files\Microsoft Shared\Web Server Extensions\12\BIN>stsa dm -o setproperty -url http://somethingOrOther -pn peoplepicker-searchadcustomfi lter -pv (OU=OURealName,DC=TopDC,DC=MidDC,DC=BottomDC) Operation completed successfully.   That’s all folks!

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  • Finding the normal of OBB face with an OBB penetrating

    - by Milo
    Below is an illustration: I have an OBB in an OBB (see below for OBB2D code if needed). What I need to determine is, what face it is in, and what direction do I point the normal? The goal is to get the OBB out of the OBB so the normal needs to face outward of the OBB. How could I go about: Finding what face the line is penetrating given the 4 corners of the OBB and the class below: if we define dx=x2-x1 and dy=y2-y1, then the normals are (-dy, dx) and (dy, -dx). Which normal points outward of the OBB? Thanks public class OBB2D { // Corners of the box, where 0 is the lower left. private Vector2D corner[] = new Vector2D[4]; private Vector2D center = new Vector2D(); private Vector2D extents = new Vector2D(); private RectF boundingRect = new RectF(); private float angle; //Two edges of the box extended away from corner[0]. private Vector2D axis[] = new Vector2D[2]; private double origin[] = new double[2]; public OBB2D(Vector2D center, float w, float h, float angle) { set(center,w,h,angle); } public OBB2D(float left, float top, float width, float height) { set(new Vector2D(left + (width / 2), top + (height / 2)),width,height,0.0f); } public void set(Vector2D center,float w, float h,float angle) { Vector2D X = new Vector2D( (float)Math.cos(angle), (float)Math.sin(angle)); Vector2D Y = new Vector2D((float)-Math.sin(angle), (float)Math.cos(angle)); X = X.multiply( w / 2); Y = Y.multiply( h / 2); corner[0] = center.subtract(X).subtract(Y); corner[1] = center.add(X).subtract(Y); corner[2] = center.add(X).add(Y); corner[3] = center.subtract(X).add(Y); computeAxes(); extents.x = w / 2; extents.y = h / 2; computeDimensions(center,angle); } private void computeDimensions(Vector2D center,float angle) { this.center.x = center.x; this.center.y = center.y; this.angle = angle; boundingRect.left = Math.min(Math.min(corner[0].x, corner[3].x), Math.min(corner[1].x, corner[2].x)); boundingRect.top = Math.min(Math.min(corner[0].y, corner[1].y),Math.min(corner[2].y, corner[3].y)); boundingRect.right = Math.max(Math.max(corner[1].x, corner[2].x), Math.max(corner[0].x, corner[3].x)); boundingRect.bottom = Math.max(Math.max(corner[2].y, corner[3].y),Math.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(new Vector2D(rect.centerX(),rect.centerY()),rect.width(),rect.height(),0.0f); } // Returns true if other overlaps one dimension of this. private boolean overlaps1Way(OBB2D other) { for (int a = 0; a < axis.length; ++a) { double t = other.corner[0].dot(axis[a]); // Find the extent of box 2 on axis a double tMin = t; double tMax = t; for (int c = 1; c < corner.length; ++c) { t = other.corner[c].dot(axis[a]); if (t < tMin) { tMin = t; } else if (t > tMax) { tMax = t; } } // We have to subtract off the origin // See if [tMin, tMax] intersects [0, 1] if ((tMin > 1 + origin[a]) || (tMax < origin[a])) { // There was no intersection along this dimension; // the boxes cannot possibly overlap. return false; } } // There was no dimension along which there is no intersection. // Therefore the boxes overlap. return true; } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0] = corner[1].subtract(corner[0]); axis[1] = corner[3].subtract(corner[0]); // Make the length of each axis 1/edge length so we know any // dot product must be less than 1 to fall within the edge. for (int a = 0; a < axis.length; ++a) { axis[a] = axis[a].divide((axis[a].length() * axis[a].length())); origin[a] = corner[0].dot(axis[a]); } } public void moveTo(Vector2D center) { Vector2D centroid = (corner[0].add(corner[1]).add(corner[2]).add(corner[3])).divide(4.0f); Vector2D translation = center.subtract(centroid); for (int c = 0; c < 4; ++c) { corner[c] = corner[c].add(translation); } computeAxes(); computeDimensions(center,angle); } // Returns true if the intersection of the boxes is non-empty. public boolean overlaps(OBB2D other) { if(right() < other.left()) { return false; } if(bottom() < other.top()) { return false; } if(left() > other.right()) { return false; } if(top() > other.bottom()) { return false; } if(other.getAngle() == 0.0f && getAngle() == 0.0f) { return true; } return overlaps1Way(other) && other.overlaps1Way(this); } public Vector2D getCenter() { return center; } public float getWidth() { return extents.x * 2; } public float getHeight() { return extents.y * 2; } public void setAngle(float angle) { set(center,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center,w,h,angle); } public float left() { return boundingRect.left; } public float right() { return boundingRect.right; } public float bottom() { return boundingRect.bottom; } public float top() { return boundingRect.top; } public RectF getBoundingRect() { return boundingRect; } public boolean overlaps(float left, float top, float right, float bottom) { if(right() < left) { return false; } if(bottom() < top) { return false; } if(left() > right) { return false; } if(top() > bottom) { return false; } return true; } };

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  • Finding the shortest path between two points on a grid, using Haskell.

    - by esperantist
    This is a problem that I can easily enough solve in a non-functional manner. But solving it in Haskell is giving me big problems. Me being inexperienced when it comes to functional programming is surely a reason. The problem: I have a 2D field divided into rectangles of equal size. A simple grid. Some rectangles are empty space (and can be passed through) while others are impassable. Given a starting rectangle A and a destination rectangle B, how would I calculate the shortest path between the two? Movement is possible only vertically and horizontally, in steps a single rectangle large. How would I go about accomplishing this in Haskell? Code snippets certainly welcome, but also certainly not neccessary. And links to further resources also very welcome! Thanks!

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  • how to merge multiple url/path into multidimensional array ?

    - by KelNoReem
    I have array like this: $path = array ( [0] => site\projects\terrace_and_balcony\mexico.jpg [1] => site\projects\terrace_and_balcony\new_york.jpg [2] => site\projects\terrace_and_balcony\berlin.jpg [3] => site\projects\terrace_and_balcony\Kentucky.jpg [4] => site\projects\terrace_and_balcony\Utah.jpg [5] => site\projects\terrace_and_balcony\Hawaii.jpg [6] => site\projects\private_gardens\mexico.jpg [7] => site\projects\private_gardens\new_york.jpg [8] => site\projects\private_gardens\berlin.jpg [9] => site\projects\private_gardens\Kentucky.jpg [10] => site\projects\private_gardens\Utah.jpg [11] => site\projects\private_gardens\Hawaii.jpg ) How to convert it to that: $path11 = array ( "site"=>array ( "projects"=>array ( "terrace_and_balcony"=>array ( "mexico.jpg", "new_york.jpg", "berlin.jpg", "Kentucky.jpg", "Utah.jpg", "Hawaii.jpg" ), "private_gardens"=>array ( "mexico.jpg", "new_york.jpg", "berlin.jpg", "Kentucky.jpg", "Utah.jpg", "Hawaii.jpg" ) ) ) );

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  • Clipping different parts of an image with path

    - by huggie
    I've recently asked a question about clipping an image via path at view's drawRect method. http://stackoverflow.com/questions/2570653/iphone-clip-image-with-path Krasnyk's code is copied below. - (void)drawRect:(CGRect)rect { CGContextRef context = UIGraphicsGetCurrentContext(); CGMutablePathRef path = CGPathCreateMutable(); //or for e.g. CGPathAddRect(path, NULL, CGRectInset([self bounds], 10, 20)); CGPathAddEllipseInRect(path, NULL, [self bounds]); CGContextAddPath(context, path); CGContextClip(context); CGPathRelease(path); [[UIImage imageNamed:@"GC.png"] drawInRect:[self bounds]]; } It works very well. However, when my image is larger than the view itself, how do I show different parts of the image? I tried tweaking around with translation on the locations (show as bounds above) of ellipse and/or UIImage drawInRect but some complex effects (Unwanted clipping, weird elipse size) I can't explain happens.

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  • How do find the longest path in a cyclic Graph between two nodes?

    - by Nazgulled
    Hi, I already solved most the questions posted here, all but the longest path one. I've read the Wikipedia article about longest paths and it seems any easy problem if the graph was acyclic, which mine is not. How do I solve the problem then? Brute force, by checking all possible paths? How do I even begin to do that? I know it's going to take A LOT on a Graph with ~18000. But I just want to develop it anyway, cause it's required for the project and I'll just test it and show it to the instructor on a smaller scale graph where the execution time is just a second or two. At least I did all tasks required and I have a running proof of concept that it works but there's no better way on cyclic graphs. But I don't have clue where to start checking all these paths...

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  • shell script fun! how to perform an action on each subdirectory from a given path??

    - by pocketfullofcheese
    I am writing a shell script (which I suck at) and I need some help. Its a script that is moving things from git to CVS (not important). The thing is, i a file path: controllers/listbuilder/setup/SubmissionRolesListbuilderHandler.inc.php and I need to be able to do: cvs add controllers; cvs add controllers/listbuilder; cvs add controllers/listbuilder/setup; cvs add controllers/listbuilder/setup/SubmissionRolesListbuilderHandler.inc.php Can someone help me out? The best I've come up with so far is to recursively add ALL files in my working tree, but that seems overly inefficient.

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  • shell script fun! performing action on each subdirectory from a given path

    - by pocketfullofcheese
    I am writing a shell script (which I suck at) and I need some help. Its a script that is moving things from git to CVS (not important). The thing is, i a file path: controllers/listbuilder/setup/SubmissionRolesListbuilderHandler.inc.php and I need to be able to do: cvs add controllers; cvs add controllers/listbuilder; cvs add controllers/listbuilder/setup; cvs add controllers/listbuilder/setup/SubmissionRolesListbuilderHandler.inc.php Can someone help me out? The best I've come up with so far is to recursively add ALL files in my working tree, but that seems overly inefficient.

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  • How to set the path to "context path" for uploaded files using Apache Common fileupload?

    - by BugKiller
    Hi, I'm using Apache common fileupload library with Netbeans 6.8 + Glassfish.I'm trying to change the current upload path to be in the current context path of the servlet , something like this: WEB-INF/upload so I wrote : File uploadedFile = new File("WEB-INF/upload/"+fileName); session.setAttribute("path",uploadedFile.getAbsolutePath()); item.write(uploadedFile); but I notice that the library save the uploaded files into glassfish folder , here what I get when I print the absolute path of the uploaded file : C:\Program Files\sges-v3\glassfish\domains\domain1\WEB-INF\upload\xx.rar My Question : How can I force the common fileupload to save the uploaded file in a path relative to the current servlet path , so I don't need to specify the whole path ? is this possible ?

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  • Can't Use Path in ASP MVC Action

    - by user1477388
    I am trying to use Path() but it has a blue line under it and says, "local variable (path) cannot be referred to until it is declared." How can I use Path()? Imports System.Globalization Imports System.IO Public Class MessageController Inherits System.Web.Mvc.Controller <EmployeeAuthorize()> <HttpPost()> Function SendReply(ByVal id As Integer, ByVal message As String, ByVal files As IEnumerable(Of HttpPostedFileBase)) As JsonResult ' upload files For Each i In files If (i.ContentLength > 0) Then Dim fileName = path.GetFileName(i.FileName) Dim path = path.Combine(Server.MapPath("~/App_Data/uploads"), fileName) i.SaveAs(path) End If Next End Function End Class

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  • Python os module path functions

    - by broiyan
    From the documentation: os.path.realpath(path) Return the canonical path of the specified filename, eliminating any symbolic links encountered in the path (if they are supported by the operating system). When I invoke this with an extant file's name, I get the path to it: /home/myhome/myproject. When I invoke this with a 'nonsense.xxx' string argument, I still get a path to /home/myhome/myproject/nonsense.xxx. This is a little inconsistent because it looks like nonsense.xxx is taken to be a directory not a file (though it is neither: it does not exist). When I invoke this with a null string file name, I still get a path to /home/myhome/myproject. How can I account for this behaviour when the documentation says so little about realpath()? (I am using Python 2.5.) Edit: Somebody suggested a way to test if files exist. My concern is not to test if files exist. My concern is to account for behaviour.

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  • comments in svg path

    - by dr jerry
    I'm experimenting with jquery and svg, I make up my path but now I want to change that (based on some user input) I do something like var dpath = $('path').attr("d"); $('path').attr("d",dpath.replace("150 150", "450 450")); which works fine, but this not useful when my path grows, so I wonder is there a possibility to put a label or a comment in the path to serve as a replace hook? searching for svg path comments gives me all. I'm close to write my own "replaceble" pseudo svg path code, but is there an alternative possible in svg? regards, Jeroen.

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  • How to add folder to assembly search path at runtime in .NET?

    - by Valery Tydykov
    My DLLs are loaded by a third-party application, which we can not customize. My assemblies have to be located in their own folder. I can not put them into GAC (my application has a requirement to be deployed using XCOPY). When the root DLL tries to load resource or type from another DLL (in the same folder), the loading fails (FileNotFound). Is it possible to add the folder where my DLLs are located to the assembly search path programmatically (from the root DLL)? I am not allowed to change the configuration files of the application.

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  • finding the total number of distinct shortest paths between 2 nodes in undirected weighted graph in linear time?

    - by logan
    I was wondering, that if there is a weighted graph G(V,E), and I need to find a single shortest path between any two vertices S and T in it then I could have used the Dijkstras algorithm. but I am not sure how this can be done when we need to find all the distinct shortest paths from S to T. Is it solvable on O(n) time? I had one more question like if we assume that the weights of the edges in the graph can assume values only in certain range lets say 1 <=w(e)<=2 will this effect the time complexity?

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  • Where is the rand php function located? localhost php running, path of default functions?

    - by Bona Chon
    I have a local server installed on my Mac (MAMP). My question is, where I can find the php functions in my computer? What is the path to the functions folder? For instance, rand() time() isset() explode(). I would like to se the code to learn of it. I have a while looking for it, but I can't find it. Or is it that is already compiled? Can someone help me here? I'm kind of lost. Thanks people. EDIT: can you be a little bit more positive? I;m trying to learn here, forgive me if I'm not to smart for you. Thanks again. Explanations would help better than giving links I guess?

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  • Best method for creating absolute path in PHP? (See 3 methods listed inside)

    - by mark
    I can create paths with no problem, but I want to know which of these 3 methods is the most rock solid and reliable and will work on the most servers. Right now I am using method 1 in my script and some users are having path issues. I just want the method that will work on any version of php and almost any server config. 1. <?php echo $_SERVER['DOCUMENT_ROOT']; ?> 2. <?php echo getcwd(); ?> 3. <?php echo dirname(__FILE__); ?> Thank you so much for any expertise you can provide about this!

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  • Off The Beaten Path—Three Things Growing Midsize Companies are Thankful For

    - by Christine Randle
    By: Jim Lein, Senior Director, Oracle Accelerate Last Sunday I went on a walkabout.  That’s when I just step out the door of my Colorado home and hike through the mountains for hours with no predetermined destination. I favor “social trails”, the unmapped routes pioneered by both animal and human explorers.  These tracks  are usually more challenging than established, marked routes and you can’t be 100% sure of where you’re going to end up. But I’ve found the rewards to be much greater. For awhile, I pondered on how—depending upon your perspective—the current economic situation worldwide could be viewed as either a classic “the glass is half empty” or a “the glass is half full” scenario. Midsize companies buy Oracle to grow and so I’m continually amazed and fascinated by the success stories our customers relate to me.  Oracle’s successful midsize companies are growing via innovation, agility, and opportunity. For them, the glass isn’t half full—it’s overflowing. Growing Midsize Companies are Thankful for: Innovation The sun angling through the pine trees reminded me of a conversation with a European customer a year ago May.  You might not recognize the name but, chances are, your local evening weather report relies on this company’s weather observation, monitoring and measurement products.  For decades, the company was recognized in its industry for product innovation, but its recent rapid growth comes from tailoring end to end product and service solutions based on the needs of distinctly different customer groups across industrial, public sector, and defense sectors.  Hours after that phone call I was walking my dog in a local park and came upon a small white plastic box sprouting short antennas and dangling by a nylon cord from a tree branch.  I cut it down. The name of that customer’s company was stamped on the housing. “It’s a radiosonde from a high altitude weather balloon,” he told me the next day. “Keep it as a souvenir.”  It sits on my fireplace mantle and elicits many questions from guests. Growing Midsize Companies are Thankful for: Agility In July, I had another interesting discussion with the CFO of an Asia-Pacific company which owns and operates a large portfolio of leisure assets. They are best known for their epic outdoor theme parks. However, their primary growth today is coming from a chain of indoor amusement centers in the USA where billiards, bowling, and laser tag take the place of roller coasters, kiddy rides, and wave pools. With mountains and rivers right out my front door, I’m not much for theme parks, but I’ll take a spirited game of laser tag any day.  This company has grown dramatically since first implementing Oracle ERP more than a decade ago. Their profitable expansion into a completely foreign market is derived from the ability to replicate proven and efficient best business practices across diverse operating environments.  They recently went live on Oracle’s Fusion HCM and Taleo. Their CFO explained to me how, with thousands of employees in three countries, Fusion HCM and Taleo would enable them to remain incredibly agile by acting on trends linking individual employee performance to their management, establishing and maintaining those best practices. Growing Midsize Companies are Thankful for: Opportunity I have three GPS apps on my iPhone. I use them mainly to keep track of my stats—distance, time, and vertical gain. However, every once in awhile I need to find the most efficient route back home before dark from my current location (notice I didn’t use the word “lost”). In August I listened in on an interview with the CFO of another European company that designs and delivers telematics solutions—the integrated use of telecommunications and informatics—for managing the mobile workforce. These solutions enable customers to achieve evolutionary step-changes in their performance and service delivery. Forgive the overused metaphor, but this is route optimization on steroids.  The company’s executive team saw an opportunity in this emerging market and went “all in”. Consequently, they are being rewarded with tremendous growth results and market domination by providing the ability for their clients to collect and analyze performance information related to fuel consumption, service workforce safety, and asset productivity. This Thanksgiving, I’m thankful for health, family, friends, and a career with an innovative company that helps companies leverage top tier software to drive and manage growth. And I’m thankful to have learned the lesson that good things happen when you get off the beaten path—both when hiking and when forging new routes through a complex world economy. Halfway through my walkabout on Sunday, after scrambling up a long stretch of scree-covered hill, I crested a ridge with an obstructed view of 14,265 ft Mt Evans just a few miles to the west.  There, nowhere near a house or a trail, someone had placed a wooden lounge chair. Its wood was worn and faded but it was sturdy. I had lunch and a cold drink in my pack. Opportunity knocked and I seized it. Happy Thanksgiving.  

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  • Intermittent fillMode=kCAFillModeForwards bug using CAKeyframeAnimation with path

    - by Mark24x7
    I'm having an intermittent problem when I move a UIImageView around the screen using CAKeyframeAnimation. I want the position of the UIImageView to remain where the animation ends when it is done. This bug only happens for certain start and end points. When I use random points it works correctly most of the time, but about 5-15% of the time it fails and snaps back to the pre-animation position. The problem only appears when using CAKeyframeAnimation using the path property. If I use the values property the bug does not appear. I am setting removedOnCompletion = NO, and fillMode = kCAFillModeForwards. I have posted a link to a test Xcode below. Here is my code for setting up the animation. I have a property usePath. When this is YES, the bug appears. When I set usePath to NO, the snap back bug does not happen. In this case I am using a path that is a simple line, but once I resolve this bug with a simple path, I will use a more complex path with curves in it. // create the point CAKeyframeAnimation *moveAnimation = [CAKeyframeAnimation animationWithKeyPath:@"position"]; if (self.usePath) { CGMutablePathRef path = CGPathCreateMutable(); CGPathMoveToPoint(path, NULL, startPt.x, startPt.y); CGPathAddLineToPoint(path, NULL, endPt.x, endPt.y); moveAnimation.path = path; CGPathRelease(path); } else { moveAnimation.values = [NSArray arrayWithObjects: [NSValue valueWithCGPoint:startPt], [NSValue valueWithCGPoint:endPt], nil]; } moveAnimation.calculationMode = kCAAnimationPaced; moveAnimation.duration = 0.5f; moveAnimation.removedOnCompletion = NO; // leaves presentation layer in final state; preventing snap-back to original state moveAnimation.fillMode = kCAFillModeForwards; moveAnimation.timingFunction = [CAMediaTimingFunction functionWithName:kCAMediaTimingFunctionEaseOut]; // moveAnimation.delegate = self; // start the animation [ball.layer addAnimation:moveAnimation forKey:@"moveAnimation"]; To dl and view my test project goto test project (http://www.24x7digital.com/downloads/PathFillModeBug.zip) Tap the 'Move Ball' button to start the animation of the ball. I have hard coded a start and end point which causes the bug to happen every time. Use the switch to change usePath to YES or NO. When usePath is YES, you will see the snap back bug. When usePath is NO, you will not see the snap back bug. I'm using SDK 3.1.3, but I have seen this bug using SDK 3.0 as well, and I have seen the bug on the Sim and on my iPhone. Any idea on how to fix this or if I am doing something wrong are appreciated. Thanks, Mark.

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  • Build path issues learning Guice

    - by Preston
    I can't figure out why I'm getting this error below I have included all the appropriate jars as far as I can tell(I have included eclipses .classpath file below.) All of the classpath entries resolve just fine. What am I missing? The type javax.servlet.ServletContextListener cannot be resolved. It is indirectly referenced from required .class files on the "extends GuiceServletContextListener" line - import com.google.inject.Guice; import com.google.inject.Injector; import com.google.inject.servlet.GuiceServletContextListener; import com.google.inject.servlet.ServletModule; public class ServletConfig extends GuiceServletContextListener { @Override protected Injector getInjector() { return Guice.createInjector(new ServletModule(){ @Override protected void configureServlets() { // TODO: add necessary code to bind } }); } } .Classpath <?xml version="1.0" encoding="UTF-8"?> <classpath> <classpathentry kind="src" path="src"/> <classpathentry kind="con" path="org.eclipse.jdt.launching.JRE_CONTAINER/org.eclipse.jdt.internal.debug.ui.launcher.StandardVMType/jdk1.7.0_21"> <attributes> <attribute name="owner.project.facets" value="java"/> </attributes> </classpathentry> <classpathentry kind="con" path="oracle.eclipse.tools.glassfish.lib.system"> <attributes> <attribute name="owner.project.facets" value="jst.web"/> </attributes> </classpathentry> <classpathentry kind="con" path="org.eclipse.jst.j2ee.internal.web.container"/> <classpathentry kind="con" path="org.eclipse.jst.j2ee.internal.module.container"/> <classpathentry kind="lib" path="guice-3.0/aopalliance.jar"/> <classpathentry kind="lib" path="guice-3.0/guice-3.0.jar"/> <classpathentry kind="lib" path="guice-3.0/guice-servlet-3.0.jar"/> <classpathentry kind="lib" path="guice-3.0/javax.inject.jar"/> <classpathentry kind="output" path="build/classes"/> </classpath>

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  • Unauthorized Access Exception using Web Deploy to Site when the site root is a UNC path

    - by Peter LaComb Jr.
    I am trying to use Web Deploy to deploy a site where the Site is rooted on a UNC path instead of a local drive. This is because I want to have a shared configuration, and have all servers point to the same UNC for content. That would allow me to deploy to one server and have all servers updated at the same time. I've created a share with everyone and users read/write. The NTFS permissions have the ID of the appDomain account as full control, and that is the same account that is configured as the specific user in Management Service Delegation. I can log on to the destination server as that ID, access the share and create/delete files. However, I'm getting the following exception in my Microsoft Web Deploy log on the destination server: User: Client IP: 192.168.62.174 Content-Type: application/msdeploy Version: 9.0.0.0 MSDeploy.VersionMin: 7.1.600.0 MSDeploy.VersionMax: 9.0.1631.0 MSDeploy.Method: Sync MSDeploy.RequestId: c060c823-cdb4-4abe-8294-5ffbdc327d2e MSDeploy.RequestCulture: en-US MSDeploy.RequestUICulture: en-US ServerVersion: 9.0.1631.0 Skip: objectName="^configProtectedData$" Provider: auto, Path: A tracing deployment agent exception occurred that was propagated to the client. Request ID 'c060c823-cdb4-4abe-8294-5ffbdc327d2e'. Request Timestamp: '8/23/2012 11:01:56 AM'. Error Details: ERROR_INSUFFICIENT_ACCESS_TO_SITE_FOLDER Microsoft.Web.Deployment.DeploymentDetailedUnauthorizedAccessException: Unable to perform the operation ("Create Directory") for the specified directory ("\someserver.mydomain.local\sharename\sitename\applicationName"). This can occur if the server administrator has not authorized this operation for the user credentials you are using. Learn more at: http://go.microsoft.com/fwlink/?LinkId=221672#ERROR_INSUFFICIENT_ACCESS_TO_SITE_FOLDER. --- Microsoft.Web.Deployment.DeploymentException: The error code was 0x80070005. --- System.UnauthorizedAccessException: Access to the path '\someserver.mydomain.local\sharename\sitename\applicationName' is denied. at Microsoft.Web.Deployment.NativeMethods.RaiseIOExceptionFromErrorCode(Win32ErrorCode errorCode, String maybeFullPath) at Microsoft.Web.Deployment.DirectoryEx.CreateDirectory(String path) at Microsoft.Web.Deployment.DirPathProviderBase.CreateDirectory(String fullPath, DeploymentObject source) at Microsoft.Web.Deployment.DirPathProviderBase.Add(DeploymentObject source, Boolean whatIf) --- End of inner exception stack trace --- --- End of inner exception stack trace --- at Microsoft.Web.Deployment.FilePathProviderBase.HandleKnownRetryableExceptions(DeploymentBaseContext baseContext, Int32[] errorsToIgnore, Exception e, String path, String operation) at Microsoft.Web.Deployment.DirPathProviderBase.Add(DeploymentObject source, Boolean whatIf) at Microsoft.Web.Deployment.DeploymentObject.Add(DeploymentObject source, DeploymentSyncContext syncContext) at Microsoft.Web.Deployment.DeploymentSyncContext.HandleAdd(DeploymentObject destObject, DeploymentObject sourceObject) at Microsoft.Web.Deployment.DeploymentSyncContext.HandleUpdate(DeploymentObject destObject, DeploymentObject sourceObject) at Microsoft.Web.Deployment.DeploymentSyncContext.SyncChildrenNoOrder(DeploymentObject dest, DeploymentObject source) at Microsoft.Web.Deployment.DeploymentSyncContext.SyncChildrenNoOrder(DeploymentObject dest, DeploymentObject source) at Microsoft.Web.Deployment.DeploymentSyncContext.SyncChildrenOrder(DeploymentObject dest, DeploymentObject source) at Microsoft.Web.Deployment.DeploymentSyncContext.ProcessSync(DeploymentObject destinationObject, DeploymentObject sourceObject) at Microsoft.Web.Deployment.DeploymentObject.SyncToInternal(DeploymentObject destObject, DeploymentSyncOptions syncOptions, PayloadTable payloadTable, ContentRootTable contentRootTable, Nullable1 syncPassId) at Microsoft.Web.Deployment.DeploymentAgent.HandleSync(DeploymentAgentAsyncData asyncData, Nullable1 passId) at Microsoft.Web.Deployment.DeploymentAgent.HandleRequestWorker(DeploymentAgentAsyncData asyncData) at Microsoft.Web.Deployment.DeploymentAgent.HandleRequest(DeploymentAgentAsyncData asyncData) This is shown as the following on the console of the machine where I run the deployment: C:\Users\PLaComb"C:\Program Files (x86)\IIS\Microsoft Web Deploy V3\msdeploy.exe" -source:package='C:\Packages\Deployments\applicationName.zip' -dest:auto,computerName='https://SERVERNAME:8172/msdeploy.axd',includeAcls='True' -verb:sync -disableLink:AppPoolExtension -disableLink:ContentExtension -disableLink:CertificateExtension -setParamFile:"C:\Packages\Deployments\applicationName.SetParameters.xml" -allowUntrusted Info: Using ID 'c060c823-cdb4-4abe-8294-5ffbdc327d2e' for connections to the remote server. Info: Adding sitemanifest (sitemanifest). Info: Adding virtual path (JMS/admin) Info: Adding directory (JMS/admin). Error Code: ERROR_INSUFFICIENT_ACCESS_TO_SITE_FOLDER More Information: Unable to perform the operation ("Create Directory") for the specified directory ("\someserver.mydomain.local\sharename\sitename\applicationName"). This can occur if the server administrator has not authorized this operation for the user credentials you are using. Learn more at: http://go.microsoft.com/fwlink/?LinkId=221672#ERROR_INSUFFICIENT_ACCESS_TO_SITE_FOLDER. Error: The error code was 0x80070005. Error: Access to the path '\someserver.mydomain.local\sharename\sitename\applicationName' is denied. Error count: 1.

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