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  • Adding a Windows Server 2012 Essentials server to an existing domain, without migrating the AD

    - by TiernanO
    I have an existing Active Directory in house, a mix between a Win2K8R2 and Win2K3 domain, and i would like to test out Windows Server 2012 Essentials BETA on the network. When walking though the install, it gives me the option of a new domain, or migrating from an existing domain. when clicking existing, it tells me i can only have one SBS server running on a domain at a time... So, i dont have any existing SBS servers in house (both are full standard or enterprise editions) but i do plan on keeping at least one of these extra servers running... So, how do i get a 2012 Essentials server to join a domain, and not migrate the existing domain? or if i do migrate, can i still get one of the other boxes to act as secondary controllers?

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  • Strange layout behaviour

    - by andrii
    I am a little bit confused. Here is an small web page. There are two main div-s: top and mainBlock. First contain image. The problem is that firebug shows that div#top's height is equal to 0 and because of that the next div mainBlock moves up. If I would delete this peace of code: div#logo{ float: left; } everything will start working fine and div#mainBlock will be below the div#top. Could you, please, explain me how it works and how to avoid this in proper way? Here is my code: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"/> <title>Paralab - Website &amp; interface design, web apps development, usability</title> <style text="text/css"> html, body{ } div#logo{ float: left; } #top{ width: 100%; } #mainBlock{ width:100%; } </style> </head> <body> <div id="top"> <div id="logo"> <img alt="logo" src="img/logo.png" /> </div> </div> <div id="mainBlock"> Contact Us </div> </body> </html>

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  • geting information from Treeview with HierarchicalDataTemplate

    - by lina
    Good day! I have such a template: <common:HierarchicalDataTemplate x:Key="my2ndPlusHierarchicalTemplate" ItemsSource="{Binding Children}"> <StackPanel Margin="0,2,5,2" Orientation="Vertical" Grid.Column="2"> <CheckBox IsTabStop="False" IsChecked="False" Click="ItemCheckbox_Click" Grid.Column="1" /> <TextBlock Text="{Binding Name}" FontSize="16" Foreground="#FF100101" HorizontalAlignment="Left" FontFamily="Verdana" FontWeight="Bold" /> <TextBlock Text="{Binding Description}" FontFamily="Verdana" FontSize="10" HorizontalAlignment="Left" Foreground="#FFA09A9A" FontStyle="Italic" /> <TextBox Width="100" Grid.Column="4" Height="24" LostFocus="TextBox_LostFocus" Name="tbNumber"></TextBox> </StackPanel> </common:HierarchicalDataTemplate> for a Treeview <controls:TreeView x:Name="tvServices" ItemTemplate="{StaticResource myHierarchicalTemplate}" ItemContainerStyle="{StaticResource expandedTreeViewItemStyle}" Grid.Column="1" Grid.Row="2" Grid.ColumnSpan="3" BorderBrush="#FFC1BCBC" FontFamily="Verdana" FontSize="14"> </controls:TreeView> I want to know the Name property of each TextBox in Treeview to make validation of each textbox such as: private void TextBox_LostFocus(object sender, RoutedEventArgs e) { tbNumber.ClearValidationError(); if ((!tbNumber.Text.IsZakazNumberValid()) && (tbNumber.Text != "")) { tbNumber.SetValidation(MyStrings.NumberError); tbNumber.RaiseValidationError(); isValid = false; } else { isValid = true; } } and I wnat to see what check boxes were checked how can I do it?

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  • iPhone TabbarController Switch Transition

    - by user269737
    I've implemented gestures (touchBegan-moved-ended) in order to allow for swiping through my tabs. It works. I'd like to add a slide-from-left and slide-from-right transition. It would be better if it could be part of the gesture if statement which tells me if the swipe is towards the right of left. Since I determine which tab is displayed from that, I could show that specific transition along with the new tab. So my question is this: what's the simplest way to simplement a slide transition at a specific instance. I don't want it to be for the whole tabbarcontrol since this is specifically for the swiping. Thanks for the help, much appreciated. For clarification purposes, this is snippet shows how I'm switching tabs: if(abs(diffx / diffy) > 2.5 && abs(diffx) > HORIZ_SWIPE_DRAG_MIN) { // It appears to be a swipe. if(isProcessingListMove) { // ignore move, we're currently processing the swipe return; } if (mystartTouchPosition.x < currentTouchPosition.x) { isProcessingListMove = YES; self.tabBarController.selectedViewController = [self.tabBarController.viewControllers objectAtIndex:0]; return; } else { isProcessingListMove = YES; self.tabBarController.selectedViewController = [self.tabBarController.viewControllers objectAtIndex:1 ]; return; }

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  • Presentation Issue in an Unordered List

    - by phreeskier
    I'm having an issue with correctly presenting items in an unordered list. The labels are floating left and the related spans that are long in length are wrapping and showing below the label. I need a solution that keeps the related spans in their respective columns. In other words, I don't want long spans to show under the labels. What property can I take advantage of so that I get the desired layout in all of the popular browsers, including IE6? Thanks in advance for the help. My code is as follows: <ul> <li> <label>Name</label> <span><%= Html.Encode(Model.Name) %></span> </li> <li> <label>Entity</label> <span><%= Html.Encode(Model.Entity) %></span> </li> <li> <label>Phone</label> <span><%= Html.Encode(Model.Phone) %></span> </li> </ul> My CSS styling is as follows: ul { display:block; list-style-type:none; margin:0; padding:0; } ul li label { float:left; width:100px; }

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  • ASP: Assigning CSS to dynamically created Label in C#

    - by Tucker
    I'm trying to figure out how to apply CSS to a Label created in C#. Everything compiles and runs, it just doesn't seem to be applying the CSS. The CSS is in the file linked to in the site master page. Everything else in the CSS file is being applied as it should be. Codebehind: ... Label label = new Label(); SqlCommand command = new SqlCommand("SELECT Q_Text FROM HRA.dbo.Questions WHERE QID = 1"); command.Connection = connection; reader = command.ExecuteReader(); reader.Read(); label.Text = reader["Q_Text"].ToString(); label.ID = "rblabel"; label.CssClass = "rblabel"; reader.Close(); ... ASP: <asp:Content runat="server" ID="BodyContent" ContentPlaceHolderID="MainContent"> <asp:PlaceHolder ID="holder" runat="server"> </asp:PlaceHolder> </asp:Content> CSS: .rblabel { text-align:left; padding-left: 2em; font-size: 4em; }

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  • DVD ROM not working only in windows?

    - by Behrooz A
    I have an Asus N53SV laptop, My DVD rom doesn't read any type of DVD in windows , but I just installed windows 7 from a bootable DVD , I think this problem occurred after I tried to change my partitioning with paragon partition magic , I was trying to shrink , join and so , after that my windows showed an error that windows cannot find [weird address] on hard disk every time on starts. windows itself says the DVD drive works correctly and driver is working properly, but no DVD or CD can be read , note that the DVD I'm trying to read is the same DVD which I used to install windows some minutes ago. I don't know what's the problem

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  • Position absolute for rounded corners and problems in IE6

    - by danit
    Im using position absolute to give the top left corner of a DIV a rounded corner. HTML: <div id="MyDiv"> Some content <div class="topLeft">&nbsp</div> </div> CSS: #MyDiv { position: relative; padding: 12px; background: #fff url('graident.png') repeat-x top left; } .topLeft { position: absolute; top: 0; right: 0; width: 10px; height: 10px; background: transparent url('corner.png') no-repeat top right; } This works fine in all browsers expcept IE6. In IE6 the corner.png image seems to be about 1px out at the top corner, essentially not top: 0; and right: 0; but more like top: 1px; right: 1px; Can anyone explain why this might be happening only in IE6?

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  • HTML + CSS: fixed background image and body width/min-width (including fiddle)

    - by insertusernamehere
    So, here is my problem. I'm kinda stuck at the moment. I have a huge background image and content in the middle with those attributes: content is centered with margin auto and has a fixed width the content is related to the image (like the image is continued within the content) this relation is only horizontally (vertically scrolling moves everything around) This works actually fine (I'm only talking desktop, not mobile here :) ) with a position fixed on the huge background image. The problem that occurs is the following: When I resize the window to "smaller than the content" the background image gets it width from the body instead of the viewport. So the relation between content and image gets lost. Now I have this little JavaScript which does the trick, but this is of course some overhead I want to avoid: $(window).resize(function(){ img.css('left', (body.width() - img.width()) / 2 ); }); This works with a fixed positioned image, but can get a litty jumpy while calculating. I also tried things like that: <div id="test" style=" position: absolute; z-index: 0; top: 0; left: 0; width: 100%: height: 100%; background: transparent url(content/dummy/brand_backgroud_1600_1.jpg) no-repeat fixed center top; "></div> But this gets me back to my problem described. Is there any "script-less", elegant solution for this problem? UPDATE: now with Fiddle The one I'm trying to solve: http://jsfiddle.net/insertusernamehere/wPmrm/ The one with Javascript that works: http://jsfiddle.net/insertusernamehere/j5E8z/ NOTE The image size is always fixed. The image never gets scaled by the browser. In the JavaScript example it get's blown. So don't care about the size.

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  • How to insert and reterive key from registry editor

    - by deepa
    Hi.. I am new to cryptography.. i have to develop project based on cryptography..In part of my project I have to insert a key to the registry and afterwards i have to reterive the same key for decryption.. i done until getting the path of the registry .. Here i given my code.. import java.io.IOException; import java.io.InputStream; import java.io.StringWriter; public static final String readRegistry(String location, String key) { try { // Run reg query, then read output with StreamReader (internal class) Process process = Runtime.getRuntime().exec("reg query " + '"' + location + "\" /v " + key); StreamReader reader = new StreamReader(process.getInputStream()); reader.start(); process.waitFor(); reader.join(); String output = reader.getResult(); // Output has the following format: // \n<Version information>\n\n<key>\t<registry type>\t<value> if (!output.contains("\t")) { return null; } // Parse out the value String[] parsed = output.split("\t"); return parsed[parsed.length - 1]; } catch (Exception e) { return null; } } static class StreamReader extends Thread { private InputStream is; private StringWriter sw = new StringWriter(); ; public StreamReader(InputStream is) { this.is = is; } public void run() { try { int c; while ((c = is.read()) != -1) { System.out.println("Reading" + c); sw.write(c); } } catch (IOException e) { System.out.println("Exception in run() " + e); } } public String getResult() { System.out.println("Content " + sw.toString()); return sw.toString(); } } public static boolean addValue(String key, String valName, String val) { try { // Run reg query, then read output with StreamReader (internal class) Process process = Runtime.getRuntime().exec("reg add \"" + key + "\" /v \"" + valName + "\" /d \"\\\"" + val + "\\\"\" /f"); StreamReader reader = new StreamReader(process.getInputStream()); reader.start(); process.waitFor(); reader.join(); String output = reader.getResult(); System.out.println("Processing........ggggggggggggggggggggg." + output); // Output has the following format: // \n&lt;Version information&gt;\n\n&lt;key&gt;\t&lt;registry type&gt;\t&lt;value&gt; return output.contains("The operation completed successfully"); } catch (Exception e) { System.out.println("Exception in addValue() " + e); } return false; } public static void main(String[] args) { // Sample usage JAXRDeleteConcept hc = new JAXRDeleteConcept(); System.out.println("Before Insertion"); if (JAXRDeleteConcept.addValue("HKEY_CURRENT_USER\\Software\\Microsoft\\Windows\\CurrentVersion\\Explorer\\ComDlg32\\OpenSaveMRU", "REG_SZ", "Muthus")) { System.out.println("Inserted Successfully"); } String value = JAXRDeleteConcept.readRegistry("HKEY_CURRENT_USER\\Software\\Microsoft\\Windows\\CurrentVersion\\Explorer\\ComDlg32\\OpenSaveMRU" , "Project_Key"); System.out.println(value); } } But i dont know how to insert a key in a registry and read the particular key which i inserted..Please help me.. Thanks in advance..

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  • Image re sizing not working after rotation in Html5 canvas

    - by Deepu the Don
    In my HTML 5 + Javascript application, we can drag, re size and rotate image in Html 5 canvas. But after doing rotation, re sizing is not working. (I think it i related to finding dx,dy,not sure). Please help me to fix the code given below. Thanks in advance. <!doctype html> <html> <head> <style> #canvas{ border:red dashed #ccc; } </style> <script type="text/javascript" src="http://code.jquery.com/jquery.min.js"></script> <script> $(function(){ var canvas=document.getElementById("canvas"),ctx=canvas.getContext("2d"),canvasOffset=$("#canvas").offset(); var offsetX=canvasOffset.left,offsetY=canvasOffset.top,startX,startY,isDown=false,pi2=Math.PI*2; var resizerRadius=8,rr=resizerRadius*resizerRadius,draggingResizer={x:0,y:0},imageX=50,imageY=50; var imageWidth,imageHeight,imageRight,imageBottom,draggingImage=false,startX,startY,doRotation=false; var r=0,rotImg = new Image(); rotImg.src="rotation.jpg"; var img=new Image(); img.onload=function(){ imageWidth=img.width; imageHeight=img.height; imageRight=imageX+imageWidth; imageBottom=imageY+imageHeight; w=img.width/2; h=img.height/2; draw(true,false); } img.src="https://dl.dropboxusercontent.com/u/139992952/stackoverflow/facesSmall.png"; function draw(withAnchors,withBorders){ ctx.fillStyle="black"; ctx.clearRect(0,0,canvas.width,canvas.height); ctx.save(); ctx.translate(imageX,imageY); ctx.translate(imageWidth/2,imageHeight/2); ctx.rotate(r); ctx.translate(-imageX,-imageY); ctx.translate(-imageWidth/2,-imageHeight/2); ctx.drawImage(img,0,0,img.width,img.height,imageX,imageY,imageWidth,imageHeight); ctx.restore(); if(withAnchors){ drawDragAnchor(imageX,imageY); drawDragAnchor(imageRight,imageY); drawDragAnchor(imageRight,imageBottom); drawDragAnchor(imageX,imageBottom); } if(withBorders){ ctx.save(); ctx.translate(imageX,imageY); ctx.translate(imageWidth/2,imageHeight/2); ctx.rotate(r); ctx.translate(-imageX,-imageY); ctx.translate(-imageWidth/2,-imageHeight/2); ctx.beginPath(); ctx.moveTo(imageX,imageY); ctx.lineTo(imageRight,imageY); ctx.lineTo(imageRight,imageBottom); ctx.lineTo(imageX,imageBottom); ctx.closePath(); ctx.stroke(); ctx.restore(); } ctx.fillStyle="blue"; ctx.save(); ctx.translate(imageX,imageY); ctx.translate(imageWidth/2,imageHeight/2); ctx.rotate(r); ctx.translate(-imageX,-imageY); ctx.translate(-imageWidth/2,-imageHeight/2); ctx.beginPath(); ctx.moveTo(imageRight+15,imageY-10); ctx.lineTo(imageRight+45,imageY-10); ctx.lineTo(imageRight+45,imageY+20); ctx.lineTo(imageRight+15,imageY+20); ctx.fill(); ctx.closePath(); ctx.restore(); } function drawDragAnchor(x,y){ ctx.save(); ctx.translate(imageX,imageY); ctx.translate(imageWidth/2,imageHeight/2); ctx.rotate(r); ctx.translate(-imageX,-imageY); ctx.translate(-imageWidth/2,-imageHeight/2); ctx.beginPath(); ctx.arc(x,y,resizerRadius,0,pi2,false); ctx.closePath(); ctx.fill(); ctx.restore(); } function anchorHitTest(x,y){ var dx,dy; dx=x-imageX; dy=y-imageY; if(dx*dx+dy*dy<=rr){ return(0); } // top-right dx=x-imageRight; dy=y-imageY; if(dx*dx+dy*dy<=rr){ return(1); } // bottom-right dx=x-imageRight; dy=y-imageBottom; if(dx*dx+dy*dy<=rr){ return(2); } // bottom-left dx=x-imageX; dy=y-imageBottom; if(dx*dx+dy*dy<=rr){ return(3); } return(-1); } function hitImage(x,y){ return(x>imageX && x<imageX+imageWidth && y>imageY && y<imageY+imageHeight); } function handleMouseDown(e){ startX=parseInt(e.clientX-offsetX); startY=parseInt(e.clientY-offsetY); draggingResizer= anchorHitTest(startX,startY); draggingImage= draggingResizer<0 && hitImage(startX,startY); doRotation = draggingResizer<0 && !draggingImage && ctx.isPointInPath(startX,startY); } function handleMouseUp(e){ draggingResizer=-1; draggingImage=false; doRotation=false; draw(true,false); } function handleMouseOut(e){ handleMouseUp(e); } function handleMouseMove(e){ mouseX=parseInt(e.clientX-offsetX); mouseY=parseInt(e.clientY-offsetY); if(draggingResizer>-1){ switch(draggingResizer){ case 0: //top-left imageX=mouseX; imageWidth=imageRight-mouseX; imageY=mouseY; imageHeight=imageBottom-mouseY; break; case 1: //top-right imageY=mouseY; imageWidth=mouseX-imageX; imageHeight=imageBottom-mouseY; break; case 2: //bottom-right imageWidth=mouseX-imageX; imageHeight=mouseY-imageY; break; case 3: //bottom-left imageX=mouseX; imageWidth=imageRight-mouseX; imageHeight=mouseY-imageY; break; } if(imageWidth<25) imageWidth=25; if(imageHeight<25) imageHeight=25; imageRight=imageX+imageWidth; imageBottom=imageY+imageHeight; draw(true,true); }else if(draggingImage){ imageClick=false; var dx=mouseX-startX; var dy=mouseY-startY; imageX+=dx; imageY+=dy; imageRight+=dx; imageBottom+=dy; startX=mouseX; startY=mouseY; draw(false,true); }else if(doRotation){ var dx=mouseX-imageX; var dy=mouseY-imageY; r=Math.atan2(dy,dx); draw(false,true); } } $("#canvas").mousedown(function(e){handleMouseDown(e);}); $("#canvas").mousemove(function(e){handleMouseMove(e);}); $("#canvas").mouseup(function(e){handleMouseUp(e);}); $("#canvas").mouseout(function(e){handleMouseOut(e);}); }); </script> </head> <body> <canvas id="canvas" width=800 height=500></canvas> </body> </html>

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  • Navigating through code with keyboard shortcuts

    - by MarceloRamires
    I'm starting to feel the need to run fastly through code with keyboard shortcuts, to arrive faster where I want to make any changes (avoiding use of mouse or long times holding [up], [left], [right] and [down]). I'm already using some: [home] - first position in current line [end] - last position in current line [ctrl] + [home] - first line of the entire code [ctrl] + [end] - last line of the entire code [pageup] - same vertical position, one screen above [pagedown] - same vertical position, one screen below [ctrl] + [pageup] - first line in current screen [ctrl] + [end] - last line in current screen [ctrl] + [left/right] - skipping word per word What have you got ? I use Visual Studio. (but I'm open to any answer, as I maybe can use others soon) obs: I've searched through stackoverflow and didn't find a nice question with this content, nor a list of keyboard code searching. If it's repeated, I'm sorry for not finding it, I'm here in my best intentions. This question is NOT about any shortcuts, and not only about visual studio, it's about running through code with shortcuts. Answers that suit the question so far: [Ctrl] + [-] - jumps to last cursor position [Ctrl] + [F3] - Jumps to next occurance of the word the curson is in [Shift] + [F3] - Same as the above, backwards. [F12] - Goes to definition of method/variable the cursor is in [Ctrl] + [ ] ] - Jumps to matching brace and select I'll ad more as there are answers.

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  • EC2 server in VPC stops responding after joining domain

    - by Geoff
    We have a EC2 Windows Server set up and running in our VPC, connected to our network via a Juniper 5GT. This is working well, with the tunnel up and stable. If I then join the server to our local domain, it appears to work - I can then log on using domain credentials, and use domain accounts when applying security to folders etc. After I log out, if I give it around an hour, the box becomes unresponsive. I can't ping it, although a tracert goes all the way barring the last hop - so the tunnel is ok. I can't RDP into it. If I reboot it, then it works for a while before doing the same thing. Un-joining it from the domain fixes the problem, and it stays up and stable. The event logs don't show anything obvious, at least to me. Any ideas?

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  • how to speed up the code??

    - by kaushik
    i have very huge code about 600 lines plus. cant post the whole thing here. but a particular code snippet is taking so much time,leading to problems. here i post that part of code please tell me what to do speed up the processing.. please suggest the part which may be the reason and measure to improve them if this small part of code is understandable. using_data={} def join_cost(a , b): global using_data #print a #print b save_a=[] save_b=[] print 1 #for i in range(len(m)): #if str(m[i][0])==str(a): save_a=database_index[a] #for i in range(len(m)): # if str(m[i][0])==str(b): #print 'save_a',save_a #print 'save_b',save_b print 2 save_b=database_index[b] using_data[save_a[0]]=save_a s=str(save_a[1]).replace('phone','text') s=str(s)+'.pm' p=os.path.join("c:/begpython/wavnk/",s) x=open(p , 'r') print 3 for i in range(6): x.readline() k2='a' j=0 o=[] while k2 is not '': k2=x.readline() k2=k2.rstrip('\n') oj=k2.split(' ') o=o+[oj] #print o[j] j=j+1 #print j #print o[2][0] temp=long(1232332) end_time=save_a[4] #print end_time k=(j-1) for i in range(k): diff=float(o[i][0])-float(end_time) if diff<0: diff=diff*(-1) if temp>diff: temp=diff pm_row=i #print pm_row #print temp #print o[pm_row] #pm_row=3 q=[] print 4 l=str(p).replace('.pm','.mcep') z=open(l ,'r') for i in range(pm_row): z.readline() k3=z.readline() k3=k3.rstrip('\n') q=k3.split(' ') #print q print 5 s=str(save_b[1]).replace('phone','text') s=str(s)+'.pm' p=os.path.join("c:/begpython/wavnk/",s) x=open(p , 'r') for i in range(6): x.readline() k2='a' j=0 o=[] while k2 is not '': k2=x.readline() k2=k2.rstrip('\n') oj=k2.split(' ') o=o+[oj] #print o[j] j=j+1 #print j #print o[2][0] temp=long(1232332) strt_time=save_b[3] #print strt_time k=(j-1) for i in range(k): diff=float(o[i][0])-float(strt_time) if diff<0: diff=diff*(-1) if temp>diff: temp=diff pm_row=i #print pm_row #print temp #print o[pm_row] #pm_row=3 w=[] l=str(p).replace('.pm','.mcep') z=open(l ,'r') for i in range(pm_row): z.readline() k3=z.readline() k3=k3.rstrip('\n') w=k3.split(' ') #print w cost=0 for i in range(12): #print q[i] #print w[i] h=float(q[i])-float(w[i]) cost=cost+math.pow(h,2) j_cost=math.sqrt(cost) #print cost return j_cost def target_cost(a , b): a=(b+1)*3 b=(a+1)*2 t_cost=(a+b)*5/2 return t_cost r1='shht:ra_77' r2='grx_18' g=[] nodes=[] nodes=nodes+[[r1]] for i in range(len(y_in_db_format)): g=y_in_db_format[i] #print g #print g[0] g.remove(str(g[0])) nodes=nodes+[g] nodes=nodes+[[r2]] print nodes print "lenght of nodes",len(nodes) lists=[] #lists=lists+[r1] for i in range(len(nodes)): for j in range(len(nodes[i])): lists=lists+[nodes[i][j]] #lists=lists+[r2] print lists distance={} for i in range(len(lists)): if i==0: distance[str(lists[i])]=0 else: distance[str(lists[i])]=long(123231223) #print distance group_dist=[] infinity=long(123232323) for i in range(len(nodes)): distances=[] for j in range(len(nodes[i])): #distances=[] if i==0: distances=distances+[[nodes[i][j], 0]] else: distances=distances+[[nodes[i][j],infinity]] group_dist=group_dist+[distances] #print distances print "group_distances",group_dist #print "check",group_dist[0][0][1] #costs={} #for i in range(len(lists)): #if i==0: # costs[str(lists[i])]=1 #else: # costs[str(lists[i])]=get_selfcost(lists[i]) path=[] for i in range(len(nodes)): mini=[] if i!=(len(nodes)-1): #temp=long(123234324) #Now calculate the cost between the current node and each of its neighbour for k in range(len(nodes[(i+1)])): for j in range(len(nodes[i])): current=nodes[i][j] #print "current_node",current j_distance=join_cost( current , nodes[i+1][k]) #t_distance=target_cost( current , nodes[i+1][k]) t_distance=34 #print distance #print "distance between current and neighbours",distance total_distance=(.5*(float(group_dist[i][j][1])+float(j_distance))+.5*(float(t_distance))) #print "total distance between the intial_nodes and current neighbour",total_distance if int(group_dist[i+1][k][1]) > int(total_distance): group_dist[i+1][k][1]=total_distance #print "updated distance",group_dist[i+1][k][1] a=current #print "the neighbour",nodes[i+1][k],"updated the value",a mini=mini+[[str(nodes[i+1][k]),a]] print mini

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  • Pecking order of pigeons?

    - by sc_ray
    I was going though problems on graph theory posted by Prof. Ericksson from my alma-mater and came across this rather unique question about pigeons and their innate tendency to form pecking orders. The question goes as follows: Whenever groups of pigeons gather, they instinctively establish a pecking order. For any pair of pigeons, one pigeon always pecks the other, driving it away from food or potential mates. The same pair of pigeons always chooses the same pecking order, even after years of separation, no matter what other pigeons are around. Surprisingly, the overall pecking order can contain cycles—for example, pigeon A pecks pigeon B, which pecks pigeon C, which pecks pigeon A. Prove that any finite set of pigeons can be arranged in a row from left to right so that every pigeon pecks the pigeon immediately to its left. Since this is a question on Graph theory, the first things that crossed my mind that is this just asking for a topological sort of a graphs of relationships(relationships being the pecking order). What made this a little more complex was the fact that there can be cyclic relationships between the pigeons. If we have a cyclic dependency as follows: A-B-C-A where A pecks on B,B pecks on C and C goes back and pecks on A If we represent it in the way suggested by the problem, we have something as follows: C B A But the above given row ordering does not factor in the pecking order between C and A. I had another idea of solving it by mathematical induction where the base case is for two pigeons arranged according to their pecking order, assuming the pecking order arrangement is valid for n pigeons and then proving it to be true for n+1 pigeons. I am not sure if I am going down the wrong track here. Some insights into how I should be analyzing this problem will be helpful. Thanks

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  • How to append \line into RTF using RichTextBox control

    - by Steve Sheldon
    When using the Microsoft RichTextBox control it is possible to add new lines like this... richtextbox.AppendText(System.Environment.NewLine); // appends \r\n However, if you now view the generated rtf the \r\n characters are converted to \par not \line How do I insert a \line control code into the generated RTF? What does't work: Token Replacement Hacks like inserting a token at the end of the string and then replacing it after the fact, so something like this: string text = "my text"; text = text.Replace("||" "|"); // replace any '|' chars with a double '||' so they aren't confused in the output. text = text.Replace("\r\n", "_|0|_"); // replace \r\n with a placeholder of |0| richtextbox.AppendText(text); string rtf = richtextbox.Rtf; rtf.Replace("_|0|_", "\\line"); // replace placeholder with \line rtf.Replace("||", "|"); // set back any || chars to | This almost worked, it breaks down if you have to support right to left text as the right to left control sequence always ends up in the middle of the placeholder. Sending Key Messages public void AppendNewLine() { Keys[] keys = new Keys[] {Keys.Shift, Keys.Return}; SendKeys(keys); } private void SendKeys(Keys[] keys) { foreach(Keys key in keys) { SendKeyDown(key); } } private void SendKeyDown(Keys key) { user32.SendMessage(this.Handle, Messages.WM_KEYDOWN, (int)key, 0); } private void SendKeyUp(Keys key) { user32.SendMessage(this.Handle, Messages.WM_KEYUP, (int)key, 0); } This also ends up being converted to a \par Is there a way to post a messaged directly to the msftedit control to insert a control character? I am totally stumped, any ideas guys? Thanks for your help!

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  • determine page UI culture throgh javascript

    - by mj-y
    I have two div tags in my html code as shown below. I want to change their float property depending on page UI culture ( Ui culture is en-US or fa-IR) ... I think I can use java script to do so. but I don't know how can I get the UI Culture through Javascript. I want a code in if condition to determine the Ui culture ... thx in advance for your help... <div id="zone1" style="float: left;"><img alt="" src="~/IconArrow.png" /> &nbsp;</div> <div id="zone2" style="float: left;"><img alt="" src="~/IconHome.png" /></div> <script type="text/javascript"> if(/* ui culture is fa-IR*/) { document.getElementById("zone1").style.float = "right"; document.getElementById("zone2").style.float = "right"; } </script>

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  • Got problem when uploading the html into the webview in iphone sdk.

    - by Monish Kumar
    Hi Guy's NSString* appendString=@""; appendString = [appendString stringByAppendingString:@"<body>"]; appendString =[appendString stringByAppendingString:@"<table background='footer.png' width='320' height='45' style='background-repeat:no-repeat'>"]; appendString =[appendString stringByAppendingString:@"<tr>"]; appendString =[appendString stringByAppendingString:@"<td align='left' width='57' height='31' style='padding: 6px 0 0 0' ><a href='/map/'><img src='details_Back.png'/></a></td>"]; appendString =[appendString stringByAppendingString:@"<td align='left' valign='middle' style='padding: 0 0 0 65px; font-family:Helvetica; font-size:21px ; font-weight:bold ; color:#FFF'>Details</td>"]; appendString =[appendString stringByAppendingString:@"</tr>"]; appendString =[appendString stringByAppendingString:@"</table>"]; appendString =[appendString stringByAppendingString:@"<br>"]; returnString = [returnString stringByReplacingOccurrencesOfString:@"<body>" withString:appendString]; printf("\n return string :%s",[returnString UTF8String]); [myWebView loadHTMLString:returnString baseURL:[NSURL URLWithString:@"http://abc.api.abcdefg.com/"]]; here in the above code the footer.png and details_back.png are the local images stored in my resource folder. Here the problem is I am gettin the background image from the server link I had passed to the webview as baseurl but the images footer.png and details_back.png which were stored in resource is not displayed. if I use the resource bundle as the baseurl then I am not displayed the background image from the server link. Can anyone please give me the suggestions to get rid of rid of this problem. thanks to all guy's, Monish.

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  • Using Mergesort to calculate number of inversions in C++

    - by Brown
    void MergeSort(int A[], int n, int B[], int C[]) { if(n > 1) { Copy(A,0,floor(n/2),B,0,floor(n/2)); Copy(A,floor(n/2),n-1,C,0,floor(n/2)-1); MergeSort(B,floor(n/2),B,C); MergeSort(C,floor(n/2),B,C); Merge(A,B,0,floor(n/2),C,0,floor(n/2)-1); } }; void Copy(int A[], int startIndexA, int endIndexA, int B[], int startIndexB, int endIndexB) { while(startIndexA < endIndexA && startIndexB < endIndexB) { B[startIndexB]=A[startIndexA]; startIndexA++; startIndexB++; } }; void Merge(int A[], int B[],int leftp, int rightp, int C[], int leftq, int rightq) //Here each sub array (B and C) have both left and right indices variables (B is an array with p elements and C is an element with q elements) { int i=0; int j=0; int k=0; while(i < rightp && j < rightq) { if(B[i] <=C[j]) { A[k]=B[i]; i++; } else { A[k]=C[j]; j++; inversions+=(rightp-leftp); //when placing an element from the right array, the number of inversions is the number of elements still in the left sub array. } k++; } if(i=rightp) Copy(A,k,rightp+rightq,C,j,rightq); else Copy(A,k,rightp+rightq,B,i,rightp); } I am specifically confused on the effect of the second 'B' and 'C' arguments in the MergeSort calls. I need them in there so I have access to them for Copy and and Merge, but

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  • How to set up my network/bridging using Apple Airport equipment?

    - by John
    I'd like to set up my network like this, and I want to make sure it's possible using the hardware I have. I think it should be... I've got my cable modem in one room. I want to plug it into an Apple Airport Express and create my wireless my wireless network here. The airport express will do the NAT and DHCP. By my TV there are a few things to be networked (Xbox and Tivo). I have an airport extreme here. I'd like to have the airport extreme join the wireless network and share the connection to the ethernet ports. Can anyone provide some assistance on the best way to configure to do this? Thanks!

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  • XNA Vector2 Rotation Question

    - by Tom Allen
    I'm messing about with some stuff in XNA and am trying to move an object around asteroids-style in that you press left and right to rotate and up/down to go forwards and backwards in the direction you are pointing. I've got the rotation of the sprite done, but i can't get the object to move in the direction you've pointed it, it always moves up and down on the x = 0 axis. I'm guessing this is straight forward but I just can't figure it out. My "ship" class has the following properties which are note worthy here: Vector2 Position float Rotation The "ship" class has an update method is where the input is handled and so far I've got the following: public void Update(GameTime gameTime) { KeyboardState keyboard = Keyboard.GetState(); GamePadState gamePad = GamePad.GetState(PlayerIndex.One); float x = Position.X; float y = Position.Y; if (keyboard.IsKeyDown(Keys.Left)) Rotation -= 0.1f; if (keyboard.IsKeyDown(Keys.Right)) Rotation += 0.1f; if (keyboard.IsKeyDown(Keys.Up)) y -= ??; if (keyboard.IsKeyDown(Keys.Down)) y += ??; this.Position = new Vector2(x, y); } Any help would be most appreciated!

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  • Paragraph with normal opacity within greyed-out div

    - by dmr
    I am greying out a web page when a user doesn't have permission to access it. In order to do that, I am placing a div with background-color white and a lowered opacity on top of the web page. I want to write some words in that div with the words having a normal opacity. As of now, the greyed out background is showing correctly. However, I can't seem to get the words to be a regular opacity. The derived styles on Firebug show the opacity on the words as normal, but it clearly isn't. What am I doing wrong? The HTML: <div class="noPermission"> <p>I'm sorry. You do not have permission to access this page.</p> </div> The CSS: div.noPermission { background-color: white; filter:alpha(opacity=50); /* IE */ opacity: 0.5; /* Safari, Opera */ -moz-opacity:0.50; /* FireFox */ z-index: 20; height: 100%; width: 100%; background-repeat:no-repeat; background-position:center; position:absolute; top: 0px; left: 0px; } div.noPermission p{ color: black; margin: 300px auto auto 50px; text-align: left; font-weight: bold; font-size: 18px; display: block; width: 250px; }

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  • What's the best way to format this simple HTML form using CSS?

    - by GregH
    I have have a simple HTML form with say four input widgets (see below)...two lines with two widgets on each line. However, when this renders it is pretty ugly. I want the whole form to be indented from the edge of the left page say 40px and I want the left edge of the widgets to line up with each other and the right edge of the labels to line up. I also want to be able to specify a minimum distance between the right edge of the first widget and the label of the widget next to it. How would I do this using CSS? Basically so it looks something like: Name: _____________ Common Names: _____________ Version: ____________ Status: ____________ See current un-formatted HTML below: <form name="detailData"> <div id="dataEntryForm"> <label> Name: <input type="text" class="input_text" name="ddName"/> Common Names: <input type="text" class="input_text" name="ddCommonNames"><P> Version: <input type="text" class="input_text" name="ddVer"/> Status: <select name="ddStatus"><option value="A" selected="selected">Active</option><option value="P">Planned</option><option value="D">Deprecated</option> </label> </div> </form>

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  • Cross-Forest Trust

    - by cdalley
    I am looking at testing a cross-domain trust we can have two domain controllers (with different forests and domain names) setup so we can move everyone onto the new domain. We do NOT run exchange on site and we do not have any links to O365 to AD currently. Onto the problem: I have setup two DCs in a Virtual Machine: They are on the same network 192.168.0.* The Windows 2003 server: Name: OLDSRVR "Clone" of our current Domain Controller IP: 192.168.0.1 Domain: internal.test.com The Windows 2012 server: Name: ADCTEST01 Brand new domain setup from scratch separate to internal.test.com Domain: internal.test2.com IP: 192.168.0.2 OLDSRVR can only see ADCTEST if it has dynamic IP set. If I set a static IP it cannot see it. If I try using the dynamic IP and try to join it gets to the end then complains "??The trust relationship between this workstation and the primary domain failed" Any ideas?

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  • wordpress -> showing custom data from child pages + pagination

    - by GaVrA
    Hello! You can see here what i am doing: http://www.arvag.net/otkrijte-svet/leto/ So its pulling 2 custom fields from child pages, one of them is just url for the image on the left, and the other field is that text showing on the right. Now, what i want to do is to add pagination there. The code i have now will just simply show all child pages, but i want to show only 5 child pages, so if user want to see 6th child page he would have to click on link for "Page 2" and so on. The code im using to display these child pages is this: <?php $pages = get_pages('child_of='.$post->ID.'&sort_column=post_title&sort_order=desc'); $count = 0; foreach($pages as $page) { $short_info = get_post_meta($page->ID,'info',true); $image = get_post_meta($page->ID,'slika',true); $count++; ?> <div class='preview_slika left'><img src="<?php echo $image ?>" alt="<?php echo $page->post_title ?>" /></div> <div class='preview_info right'> <h2><?php echo $page->post_title ?></h2> <p><?php echo $short_info ?></p> <a href="<?php echo get_page_link($page->ID) ?>">Više o >></a> </div> <div class='clear'></div> <?php } ?> So any idea what to do to get what i need?

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