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  • Django Template For Loop Removing <img> Self-Closing

    - by Zack
    Django's for loop seems to be removing all of my <img> tag's self-closing...ness (/>). In the Template, I have this code: {% for item in item_list %} <li> <a class="left" href="{{ item.url }}">{{ item.name }}</a> <a class="right" href="{{ item.url }}"> <img src="{{ item.icon.url }}" alt="{{ item.name }} Logo." /> </a> </li> {% endfor %} It outputs this: <li> <a class="left" href="/some-url/">This is an item</a> <a class="right" href="/some-url/"> <img src="/media/img/some-item.jpg" alt="This is an item Logo."> </a> </li> As you can see, the <img> tag is no longer closed, and thus the page doesn't validate. This isn't a huge issue since it'll still render properly in all browsers, but I'd like to know how to solve it. I've tried wrapping the whole for loop in {% autoescape off %}...{% endautoescape %} but that didn't change anything. All other self-closed <img> tags in the document outside the for loop still properly close.

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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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  • ViewController has wrong orientation after Landscape-only has been popped

    - by noroom
    In a navigation-based app, LandscapeViewController only supports landscape mode (all others support both modes). I also have a "loading screen" that advises the user to rotate the phone before continuing. This way I can make sure that when my landscape view loads, that it's in landscape mode. The problem comes when i rotate the phone to portrait mode while still showing LandscapeVC. I press the Back nagivation button to navigate up one level (to a VC that supports both landscape and portrait modes), but the upper level shows in landscape mode even though the phone is in portrait mode. I guess this is because when I left this view I was in portrait mode, I then rotated the phone while in another view, so this view has not received the notification. If I then proceed to rotate the phone to the other landscape mode (say the LandscapeVC was loaded on its right side, so I'd rotate the upper VC from portrait to the left landscape mode), it will update. My question is: how can I notify this upper view that the phone was rotated, so when the user goes up after putting the phone in portrait mode, the upper view shows correctly?

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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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  • WPF Buttons Style

    - by Polaris
    I have WPF Form which has many buttons with the same code. Appearance of all buttons must be the same For example, code for one of these buttons <Button x:Name="btnAddRelative" Width="120" Click="btnAddRelative_Click" > <Button.Content> <StackPanel Orientation="Horizontal"> <Image Height="26" HorizontalAlignment="Left"> <Image.Source> <BitmapImage UriSource="images/add.png" /> </Image.Source> </Image> <TextBlock Text=" Add Relative" Height="20" VerticalAlignment="Center"/> </StackPanel> </Button.Content> </Button> How can I create one style and use it for all my buttons. All buttons has the same png image, only their text different. How can I do this. I tried to do this with Style object in Resource Section: <UserControl.Resources> <Style TargetType="Button" x:Key="AddStyle"> <Setter Property="Content"> <Setter.Value> <StackPanel Orientation="Horizontal"> <Image Height="26" HorizontalAlignment="Left"> <Image.Source> <BitmapImage UriSource="images/add.png" /> </Image.Source> </Image> <TextBlock Text=" " Height="20" VerticalAlignment="Center"/> </StackPanel> </Setter.Value> </Setter> </Style> </UserControl.Resources> But this code not work. Can any body know how can I do this?

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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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  • Is there a sensible way of 'teaming' two ADSL connections?

    - by Tim Long
    I work in an office complex that has two seperate ADSL connections, which they use to provide two seperate networks (actually both the ADSL routers go into a Cisco managed switch with two VLANs, one for each ADSL connection). Circumstances have changed so that 95% of the users are all on one ADSL connection. It would be great if there were a way to join together both connections to emulate a single connection at double the speed, but the ISP doesn't support bonding. So, is there a sensible way to take two completely seperate ADSL lines and use them to provide a single internet gateway?

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  • Joining Windows 7 Professional to a Windows Server 2003 R2 x64 domain fails.

    - by Vinko Vrsalovic
    I have a windows 7 professional (spanish) laptop trying to join a Windows Server 2003 (english) domain. It detect correctly the SRV record, finding the proper domain controller, but then the join fails with the error message (snippet, because the error is in spanish) An Active Directory Domain Controller for This Domain Could Not be Contacted The DNS is correctly set, and client can ping by name and IP the server, the server can ping the client by IP. I've tested with the FW down to no avail. A host of other XP Pro clients are connected to the domain. I've restarted Net Logon and checked that Windows Time is up. Also the times are in sync between the server and the client. I'll put below diagnostics output. I'm wondering if there's anything special to be done on either the server or the client to have a Win 7 Pro join a 2k3 R2 domain. The following diagnostic information follows: netdiag /q for the DC dcdiag on the DC ipconfig /all on the Win 7 client netdiag /q on the DC: .................................. Computer Name: HI-X2 DNS Host Name: hi-x2.hi.local System info : Microsoft Windows Server 2003 R2 (Build 3790) Processor : EM64T Family 6 Model 23 Stepping 10, GenuineIntel List of installed hotfixes : KB923561 KB924667-v2 KB925398_WMP64 KB925902 KB926122 KB927891 KB929123 KB930178 KB932168 KB936357 KB938127 KB941569 KB942830 KB942831 KB943055 KB943460 KB944338-v2 KB944653 KB945553 KB946026 KB948496 KB950760 KB950762 KB950974 KB951066 KB951748 KB952004 KB952069 KB952954 KB954155 KB954550-v7 KB955069 KB955759 KB956572 KB956802 KB956803 KB956844 KB958469 KB958644 KB958869 KB959426 KB960225 KB960803 KB960859 KB961063 KB961118 KB961501 KB967715 KB967723 KB968389 KB968816 KB969059 KB969947 KB970238 KB970430 KB970483 KB971032 KB971468 KB971657 KB971737 KB971961 KB971961-IE8 KB972270 KB973037 KB973354 KB973507 KB973540 KB973687 KB973815 KB973825 KB973869 KB973904 KB973917-v2 KB974112 KB974318 KB974392 KB974571 KB975025 KB975467 KB975560 KB975713 KB976662-IE8 KB977290 KB977816 KB977914 KB978037 KB978262 KB978338 KB978542 KB978601 KB978706 KB979306 KB979309 KB979683 KB980182 KB980182-IE8 KB980232 KB980302-IE8 KB981332-IE8 KB981350 Q147222 Per interface results: Adapter : Local Area Connection Host Name. . . . . . . . . : hi-x2.hi.local IP Address . . . . . . . . : 10.0.1.199 Subnet Mask. . . . . . . . : 255.0.0.0 Default Gateway. . . . . . : 10.0.1.1 Dns Servers. . . . . . . . : 10.0.1.199 WINS service test. . . . . : Skipped Global results: [WARNING] You don't have a single interface with the 'WorkStation Service', 'Messenger Service', 'WINS' names defined. DNS test . . . . . . . . . . . . . : Passed PASS - All the DNS entries for DC are registered on DNS server '10.0.1.199'. IP Security test . . . . . . . . . : Skipped The command completed successfully dcdiag on the DC: Domain Controller Diagnosis Performing initial setup: Done gathering initial info. Doing initial required tests Testing server: Default-First-Site-Name\HI-X2 Starting test: Connectivity ......................... HI-X2 passed test Connectivity Doing primary tests Testing server: Default-First-Site-Name\HI-X2 Starting test: Replications ......................... HI-X2 passed test Replications Starting test: NCSecDesc ......................... HI-X2 passed test NCSecDesc Starting test: NetLogons ......................... HI-X2 passed test NetLogons Starting test: Advertising ......................... HI-X2 passed test Advertising Starting test: KnowsOfRoleHolders ......................... HI-X2 passed test KnowsOfRoleHolders Starting test: RidManager ......................... HI-X2 passed test RidManager Starting test: MachineAccount ......................... HI-X2 passed test MachineAccount Starting test: Services ......................... HI-X2 passed test Services Starting test: ObjectsReplicated ......................... HI-X2 passed test ObjectsReplicated Starting test: frssysvol ......................... HI-X2 passed test frssysvol Starting test: frsevent ......................... HI-X2 passed test frsevent Starting test: kccevent ......................... HI-X2 passed test kccevent Starting test: systemlog ......................... HI-X2 passed test systemlog Starting test: VerifyReferences ......................... HI-X2 passed test VerifyReferences Running partition tests on : ForestDnsZones Starting test: CrossRefValidation ......................... ForestDnsZones passed test CrossRefValidation Starting test: CheckSDRefDom ......................... ForestDnsZones passed test CheckSDRefDom Running partition tests on : DomainDnsZones Starting test: CrossRefValidation ......................... DomainDnsZones passed test CrossRefValidation Starting test: CheckSDRefDom ......................... DomainDnsZones passed test CheckSDRefDom Running partition tests on : Schema Starting test: CrossRefValidation ......................... Schema passed test CrossRefValidation Starting test: CheckSDRefDom ......................... Schema passed test CheckSDRefDom Running partition tests on : Configuration Starting test: CrossRefValidation ......................... Configuration passed test CrossRefValidation Starting test: CheckSDRefDom ......................... Configuration passed test CheckSDRefDom Running partition tests on : hi Starting test: CrossRefValidation ......................... hi passed test CrossRefValidation Starting test: CheckSDRefDom ......................... hi passed test CheckSDRefDom Running enterprise tests on : hi.local Starting test: Intersite ......................... hi.local passed test Intersite Starting test: FsmoCheck ......................... hi.local passed test FsmoCheck ipconfig /all on the Windows 7 client: Configuraci¢n IP de Windows Nombre de host. . . . . . . . . : hi-p6 Sufijo DNS principal . . . . . : Tipo de nodo. . . . . . . . . . : h¡brido Enrutamiento IP habilitado. . . : no Proxy WINS habilitado . . . . . : no Adaptador de LAN inal mbrica Conexi¢n de red inal mbrica: Sufijo DNS espec¡fico para la conexi¢n. . : Descripci¢n . . . . . . . . . . . . . . . : Intel(R) WiFi Link 5100 AGN Direcci¢n f¡sica. . . . . . . . . . . . . : 00-22-FB-63-47-A0 DHCP habilitado . . . . . . . . . . . . . : no Configuraci¢n autom tica habilitada . . . : s¡ Direcci¢n IPv4. . . . . . . . . . . . . . : 10.0.1.42(Preferido) M scara de subred . . . . . . . . . . . . : 255.255.255.0 Puerta de enlace predeterminada . . . . . : 10.0.1.1 Servidores DNS. . . . . . . . . . . . . . : 10.0.1.199 NetBIOS sobre TCP/IP. . . . . . . . . . . : habilitado Adaptador de Ethernet Conexi¢n de  rea local: Estado de los medios. . . . . . . . . . . : medios desconectados Sufijo DNS espec¡fico para la conexi¢n. . : Descripci¢n . . . . . . . . . . . . . . . : Realtek PCIe GBE Family Controller Direcci¢n f¡sica. . . . . . . . . . . . . : 00-1E-33-1F-35-B1 DHCP habilitado . . . . . . . . . . . . . : s¡ Configuraci¢n autom tica habilitada . . . : s¡ Adaptador de t£nel isatap.{8926581E-09AC-4123-906B-DA6386AD2D60}: Estado de los medios. . . . . . . . . . . : medios desconectados Sufijo DNS espec¡fico para la conexi¢n. . : Descripci¢n . . . . . . . . . . . . . . . : Adaptador ISATAP de Microsoft Direcci¢n f¡sica. . . . . . . . . . . . . : 00-00-00-00-00-00-00-E0 DHCP habilitado . . . . . . . . . . . . . : no Configuraci¢n autom tica habilitada . . . : s¡ Adaptador de t£nel Teredo Tunneling Pseudo-Interface: Sufijo DNS espec¡fico para la conexi¢n. . : Descripci¢n . . . . . . . . . . . . . . . : Teredo Tunneling Pseudo-Interface Direcci¢n f¡sica. . . . . . . . . . . . . : 00-00-00-00-00-00-00-E0 DHCP habilitado . . . . . . . . . . . . . : no Configuraci¢n autom tica habilitada . . . : s¡ Direcci¢n IPv6 . . . . . . . . . . : 2001:0:5ef5:73ba:1cec:3883:f5ff:fed5(Preferido) V¡nculo: direcci¢n IPv6 local. . . : fe80::1cec:3883:f5ff:fed5%13(Preferido) Puerta de enlace predeterminada . . . . . : :: NetBIOS sobre TCP/IP. . . . . . . . . . . : deshabilitado

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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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  • 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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  • 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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  • 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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  • access settings from whole jquery component

    - by Pacuraru Daniel
    I am trying to develop a jquery component for dialog modals and i dont know how to access the settings from all component functions. I need to access settings,zIndex from open function and it seems to not work. (function($) { var methods = { init: function(options) { var defaults = { bgClass: "fancy-dialog-bg", bgShow: null, zIndex: 100, show: null }; var settings = $.extend(defaults, options); return this.each(function() { var obj = $(this).hide().css("position", "fixed").css("z-index", settings.zIndex).css("left", "300px").css("top", "200px"); }); }, open: function() { // alert(settings.zIndex); not working var tes = $("<div></div>").css("backgroundColor", "#f00").css("position", "fixed").css("z-index", "99").css("width", "50%").css("height", "100%").css("left", "0").css("top", "0"); $('body').append(tes); var obj = $(this); obj.show(); }, close: function() { var obj = $(this); $("#fancy-dialog-bg-" + obj.attr('id')).remove(); obj.hide(); } }; $.fn.fancyDialog = function(method) { if (methods[method]) { return methods[method].apply(this, Array.prototype.slice.call(arguments, 1)); } else if (typeof method === 'object' || !method) { return methods.init.apply(this, arguments); } else { $.error('Method ' + method + ' does not exist.'); } }; })(jQuery);

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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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  • 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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  • 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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  • 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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  • Lights off effect and jquery placement on wordpress

    - by Alexander Santiago
    I'm trying to implement a lights on/off on single posts of my wordpress theme. I know that I have to put this code on my css, which I did already: #the_lights{ background-color:#000; height:1px; width:1px; position:absolute; top:0; left:0; display:none; } #standout{ padding:5px; background-color:white; position:relative; z-index:1000; } Now this is the code that I'm having trouble with: function getHeight() { if ($.browser.msie) { var $temp = $("").css("position", "absolute") .css("left", "-10000px") .append($("body").html()); $("body").append($temp); var h = $temp.height(); $temp.remove(); return h; } return $("body").height(); } $(document).ready(function () { $("#the_lights").fadeTo(1, 0); $("#turnoff").click(function () { $("#the_lights").css("width", "100%"); $("#the_lights").css("height", getHeight() + "px"); $("#the_lights").css({‘display’: ‘block’ }); $("#the_lights").fadeTo("slow", 1); }); $("#soft").click(function () { $("#the_lights").css("width", "100%"); $("#the_lights").css("height", getHeight() + "px"); $("#the_lights").css("display", "block"); $("#the_lights").fadeTo("slow", 0.8); }); $("#turnon").click(function () { $("#the_lights").css("width", "1px"); $("#the_lights").css("height", "1px"); $("#the_lights").css("display", "block"); $("#the_lights").fadeTo("slow", 0); }); }); I think it's a jquery. Where do I place it and how do I call it's function? Been stuck on this thing for 6 hours now and any help would be greatly appreciated...

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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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  • How do I share a home printer under Windows if the computer is part of a domain?

    - by Sorin Sbarnea
    I do have my work laptop at home and I want to share an USB printer in order to be able to print to it from my MacBook. Currently Windows 7 is refusing to share the printer because the computer is part of a domain and it tells me that I need to join a homegroup in order to be able to do that. Also it tells me that he cannot create this homegroup and this group has to be create by another Windows 7 computer from my network. As you can imagine I do have only one Windows computer on my home network. How can I solve this problem?

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  • little java help....

    - by jona
    Hi I am doing some practice problems and trying to print a diagonal line like the example below. I have writen the program you see below and I honestly dont understand what I am doing wrong. I m a java beginner and I cant see how to find the error. Example ( If you only see a straight line of stars...then imagine it diagonally....from top left to bottom right) * * * * * code: class Diagonal{ public static void main(String args[]) { int row, col; for(row = 1; row < 6; row++) { for(col = 1; col <= row; col++) { if(col==row){ System.out.print("*"); } else{ System.out.print(""); } System.out.println(); } } } } I am trying to learn for loops because they really confuse me. Another practice is to print a similar diagonal line but this time from right to left. I cant do that without getting this right however :( I believe they will be pretty similar? Above my reasining is this: As long as the column # is the same as the row number the print the line or otherwise leave a blank....what's wrong with how i did it? THANK YOU!

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  • client subscribed to a multicast group not receiving data

    - by Abruzzo Forte e Gentile
    I have a network that was setup for multicast traffic IN Machine A there is a server application generating multi-cast traffic. I have also different clients subscribing to that mulicast traffic -some client are in the same machine A -other clients are in machine B,C,D # Address I am using IP : 239.193.0.21 PORT: 20401 I don't know why but the client in machine A , even if they join the group, don't receive any data while (and this is the funny part) machine B,C and D receive everything. Checking with Wirshark I can see that the UDP traffic flow is there in Machine A. I am using LInux/Red-Hat Enterprise 6.2 What should I look at? Any setting in the os? Firewall or what?

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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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  • 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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  • 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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