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  • Android camera being landscaped in some devices

    - by nala4ever
    Im new to Android and I tried a tutorial for camera API. The tutorial works fine. When I use HTC desire I can see the camera view in both portrait and landscape, but when I use Samsung Galaxy I get a the camera view only in a landscaped view. I tried the following code to rotate the camera view as well.. Camera.Parameters parameters = camera.getParameters(); parameters.setRotation(90); then the camera doesn't work as expected. (screen splits into 4 and not clear). Does anyone have an idea for this issue ? Thanks.

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  • View, Control, Instruct with iTalc

    <b>Linux.com:</b> "If you work in an educational or training environment where you instruct users on the ins and outs of using computers, or you need to be able to (for whatever reason) control the PC user's use of a machine, the tools available are often quite expensive or quite difficult to use. Neither is the case in the Linux environment, where tools like iTalc are available."

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  • SQL VIEW Basics

    SQL Views are essential for the database developer. However, it is common to see them misued, or neglected. Joe Celko tackles an introduction to the subject, but there is something about the topic that makes it likely that even the experienced developer will find out something new from reading it. Get smart with SQL Backup ProGet faster, smaller backups with integrated verification.Quickly and easily DBCC CHECKDB your backups. Learn more.

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  • For a 1view/scene to 2view/scene app, what application should I choose in Xcode?

    - by Tony Xu
    The question may be simple to some others, but I have been struggling with this for a while. The app I want would be like this: first scene/view with two big buttons (no toolbar item), click each one to get into two new scenes. So totally three scenes. In Xcode, what application should I choose? And in storyboard how/should I drag/draw? Thanks. Update: thanks for the link, the big-number-user. I actually read that tutorial before I asked. A little update on what I got so far: 1, I selected "single view", so there's view controller 1 (VC1) in the storyboard. 2, dragged a navigation controller (NC), and move the initial view arrow pointing to NC 3, control-drag to link NC and VC1, selected "relationship segue root view controller" when some small dialog popup. IS THIS CORRECT? 4, created two additional VC, VC3 and VC4, control-drag link each to NC. selected "push", IS THIS CORRECT? 5, in VC1, I added two buttons, showVC3 and showVC4. NOW I DON'T KNOW how to add IBAction to button showVC3 and showVC4. I tried to control-drag it to ViewController.m file @interface and @end section, but failed. What should I do next?

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  • Simple Oracle File repository with folder hierarchy

    - by Ope
    I have an application that stores large amount of files (XML and binary) in folder hierarchies. Currently the main method is storing them in file system or using a legacy CMS, which we want to get rid of. The CMS supports Oracle and a customer wants to keep the files in Oracle because of enterprise policies (backup etc.) The question is: Is there a simple implementation of file repository with folder hierarchy for Oracle? What I am looking for is a small .Net component or example code (PL/SQL and/or .Net) that would have the following methods: Create, Delete, Exists Folder CRUD file Move and potentially Copy file or directory Access to files and folders with paths like "/root/folder1/folder2/file.xml" Ability to get all the files and folders in a folder and potentially also the entire directory tree Tree traversal, getting the parent, all children etc. needs to be fast. I need the implementation in .Net, but if it was just the stored procedures, I could create the .Net calling code. I have pointers to generic articles for creating hierarchies in DB, so if I need to do it from scratch, I know where to start. What I am asking here, is there already an implementation that I could take without doing this from scratch? It seems like such a generic requirement... If the answer is a CMS, Document management system or such it should be Open Source or at least quite cheap (some hundreds / server) and it should be possible to deploy it XCopy - hopefully only couple of DLL:s. I do not need - or want - a full featured big CMS with dozens of dlls and especially not an msi-installation. I have tried to google this, but the words "repository", "CMS", "file hierarchy" etc. give so many answers, the searches are pretty much useless. Thanks, OPe

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  • Android setContentView operation

    - by stormin986
    I've read that it's important to call setContentView() early in an activity since it builds the view objects that may be manipulated by subsequent code in onCreate(). In terms of lifecycle, does the view get drawn to screen as soon as setContentView() is called, or does it allow the onCreate() function to build/populate the information in the view objects, and wait to actually draw it after onCreate() completes? Thanks!

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  • Visual Studio Folder Structure

    - by nick
    I am not sure how this works. I am using Visual Studio 2008 and I created a Class Library (say the name is Test). I also selected the option to create a folder for the solution. Following is the directory structure I get: Test - Test - bin - Debug - obj - Debug - Properties - AassemblyInfo.cs - Test.cs - Test.csproj - Test.sln - Test.suo This is default and I have no problems running my code this way. My querry is I see other solutions (class libraries) created in the Subversion by others before have a different structure. The structure for that is as follows: Test - .svn - lib - <<Reference 1>> - <<Reference 2>> - .... - <<Reference N>> - src - bin - Debug - obj - Debug - Properties - AassemblyInfo.cs - Test.cs - Test.csproj - Test.sln - Test.suo My query is how to create this structure? All the references to other projects are maintained in lib folder and source code is maintained in src folder. This is not the case happening with me. When I open the solution in Visual Studio, I cannot see any such folder like lib or src. It shows the same way as mine. Kindly help and forgive me for being so elaborative. Thanks

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  • SQL View: Beyond the Basics

    Joe Celko delves into the main uses of views, explains how the WITH CHECK OPTION works, and demonstrates how the INSTEAD OF trigger can be used in those cases where views cannot be updatable. What are your servers really trying to tell you? Find out with new SQL Monitor 3.0, an easy-to-use tool built for no-nonsense database professionals.For effortless insights into SQL Server, download a free trial today.

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  • How to setup linux permissions the WWW folder?

    - by Xeoncross
    Updated Summery The /var/www directory is owned by root:root which means that no one can use it and it's entirely useless. Since we all want a web server that actually works (and no-one should be logging in as "root"), then we need to fix this. Only two entities need access. PHP/Perl/Ruby/Python all need access to the folders and files since they create many of them (i.e. /uploads/). These scripting languages should be running under nginx or apache (or even some other thing like FastCGI for PHP). The developers How do they get access? I know that someone, somewhere has done this before. With however-many billions of websites out there you would think that there would be more information on this topic. I know that 777 is full read/write/execute permission for owner/group/other. So this doesn't seem to be needed as it leaves random users full permissions. What permissions are need to be used on /var/www so that... Source control like git or svn Users in a group like "websites" (or even added to "www-data") Servers like apache or lighthttpd And PHP/Perl/Ruby can all read, create, and run files (and directories) there? If I'm correct, Ruby and PHP scripts are not "executed" directly - but passed to an interpreter. So there is no need for execute permission on files in /var/www...? Therefore, it seems like the correct permission would be chmod -R 1660 which would make all files shareable by these four entities all files non-executable by mistake block everyone else from the directory entirely set the permission mode to "sticky" for all future files Is this correct? Update: I just realized that files and directories might need different permissions - I was talking about files above so i'm not sure what the directory permissions would need to be. Update 2: The folder structure of /var/www changes drastically as one of the four entities above are always adding (and sometimes removing) folders and sub folders many levels deep. They also create and remove files that the other 3 entities might need read/write access to. Therefore, the permissions need to do the four things above for both files and directories. Since non of them should need execute permission (see question about ruby/php above) I would assume that rw-rw-r-- permission would be all that is needed and completely safe since these four entities are run by trusted personal (see #2) and all other users on the system only have read access. Update 3: This is for personal development machines and private company servers. No random "web customers" like a shared host. Update 4: This article by slicehost seems to be the best at explaining what is needed to setup permissions for your www folder. However, I'm not sure what user or group apache/nginx with PHP OR svn/git run as and how to change them. Update 5: I have (I think) finally found a way to get this all to work (answer below). However, I don't know if this is the correct and SECURE way to do this. Therefore I have started a bounty. The person that has the best method of securing and managing the www directory wins.

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  • Simulating remote environment?

    - by ropstah
    I'm building a .NET MVC application which will be deployed on a Windows 2003 server. The server has a folder @ c:\Website\Files which needs to be written to from the application. How do I cope with this in my development environment so that the MSI setup file, which I will compile, will work correctly when deployed? p.s. the folder is NOT located in a subdirectory of the application project

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  • MVC validation error with strongly typed view

    - by Remnant
    I have a simple form that I would like to validate on form submission. Note I have stripped out the html for ease of viewing <%=Html.TextBox("LastName", "")%> //Lastname entry <%=Html.ValidationMessage("LastName")%> <%=Html.TextBox("FirstName", "")%>//Firstname entry <%=Html.ValidationMessage("FirstName")%> <%=Html.DropDownList("JobRole", Model.JobRoleList)%> //Dropdownlist of job roles <% foreach (var record in Model.Courses) // Checkboxes of different courses for user to select { %> <li><label><input type="checkbox" name="Courses" value="<%=record.CourseName%>" /><%= record.CourseName%></label></li> <% } %> On submission of this form I would like to check that both FirstName and LastName are populated (i.e. non-zero length). In my controller I have: public ActionResult Submit(string FirstName, string LastName) { if (FirstName.Trim().Length == 0) ModelState.AddModelError("FirstName", "You must enter a first name"); if (LastName.Trim().Length == 0) ModelState.AddModelError("LastName", "You must enter a first name"); if (ModelState.IsValid) { //Update database + redirect to action } return View(); //If ModelState not valid, return to View and show error messages } Unfortunately, this code logic produces an error that states that no objects are found for JobRole and Courses. If I remove the dropdownlist and checkboxes then all works fine. The issue appears to be that when I return the View the view is expecting objects for the dropwdownlist and checkboxes (which is sensible as that is what is in my View code) How can I overcome this problem? Things I have considered: In my controller I could create a JobRoleList object and Course object to pass to the View so that it has the objects to render. The issue with this is that it will overwrite any dropdownlist / checkbox selections that the user has already made. In the parameters of my controller method Submit I could aslo capture the JobRoleList object and Course object to pass back to the View. Again, not sure this would capture any items the user has already selected. I have done much googling and reading but I cannot find a good answer. When I look at examples in books or online (e.g. Nerddinner) all the validation examples involve simple forms with TextBox inputs and don't seems to show instances with multiple checkboxes and dropdownlists. Have I missed something obvious here? What would be best practice in this situation? Thanks

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  • MVC Partial View Not Refreshing when using JSON data

    - by 40-Love
    I have a dropdown that I'm using to refresh a div with checkboxes. I am trying to figure out why my view is not refreshing if I pass in data in JSON format. If I pass in just a regular string, the view refreshes. If I pass in JSON data, the view does not refresh. If I step through the code in the Partial view, I can see the correct number of records are being passed in, however the view doesn't get refreshed with the correct number of checkboxes. I tried to add some cache directives, it didn't work. This doesn't work: $(function () { $('#ddlMoveToListNames').change(function () { var item = $(this).val(); var selectedListID = $('#ddlListNames').val(); var checkValues = $('input[name=c]:checked').map(function () { return $(this).val(); }).toArray(); $.ajax({ url: '@Url.Action("Test1", "WordList")', type: 'POST', cache: false, data: JSON.stringify({ words: checkValues, moveToListID: item, selectedListID: selectedListID }), dataType: 'json', contentType: 'application/json; charset=utf-8', success: function (result) { } }) .done(function (partialViewResult) { $("#divCheckBoxes").replaceWith(partialViewResult); }); }); }); This works: $(function () { $('#ddlMoveToListNames').change(function () { var item = $(this).val(); var selectedListID = $('#ddlListNames').val(); var checkValues = $('input[name=c]:checked').map(function () { return $(this).val(); }).toArray(); $.ajax({ url: '@Url.Action("Test1", "WordList")', type: 'POST', cache: false, data: { selectedListID: item }, success: function (result) { } }) .done(function (partialViewResult) { $("#divCheckBoxes").replaceWith(partialViewResult); }); }); }); Partial View: @model WLWeb.Models.MyModel <div id="divCheckBoxes"> @foreach (var item in Model.vwWordList) { @Html.Raw("<label><input type='checkbox' value='" + @Html.DisplayFor(modelItem => item.Word) + "' name='c'> " + @Html.DisplayFor(modelItem => item.Word) + "</label>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;"); } </div> Controller: [AcceptVerbs(HttpVerbs.Post)] [OutputCache(NoStore = true, Duration = 0, VaryByParam = "*")] public PartialViewResult Test1(MyModel vm, string[] words, string moveToListID, string selectedListID) { int listNameID = Convert.ToInt32(moveToListID); List<vwWordList> lst = db.vwWordLists.Where(s => s.Word.StartsWith("wa") && s.ListID == listNameID).ToList(); vm.vwWordList = lst; return PartialView("Partial1", vm); } View: @Html.DropDownListFor(x => Model.FilterViewModel.MoveToListNameID, Model.FilterViewModel.MoveToListNameList, new { @id = "ddlMoveToListNames", style = "width:100px;" })

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  • UIVIewController not released when view is dismissed

    - by Nelson Ko
    I have a main view, mainWindow, which presents a couple of buttons. Both buttons create a new UIViewController (mapViewController), but one will start a game and the other will resume it. Both buttons are linked via StoryBoard to the same View. They are segued to modal views as I'm not using the NavigationController. So in a typical game, if a person starts a game, but then goes back to the main menu, he triggers: [self dismissViewControllerAnimated:YES completion:nil ]; to return to the main menu. I would assume the view controller is released at this point. The user resumes the game with the second button by opening another instance of mapViewController. What is happening, tho, is some touch events will trigger methods on the original instance (and write status updates to them - therefore invisible to the current view). When I put a breakpoint in the mapViewController code, I can see the instance will be one or the other (one of which should be released). I have tried putting a delegate to the mainWindow clearing the view: [self.delegate clearMapView]; where in the mainWindow - (void) clearMapView{ gameWindow = nil; } I have also tried self.view=nil; in the mapViewController. The mapViewController code contains MVC code, where the model is static. I wonder if this may prevent ARC from releasing the view. The model.m contains: static CanShieldModel *sharedInstance; + (CanShieldModel *) sharedModel { @synchronized(self) { if (!sharedInstance) sharedInstance = [[CanShieldModel alloc] init]; return sharedInstance; } return sharedInstance; } Another post which may have a lead, but so far not successful, is UIViewController not being released when popped I have in ViewDidLoad: // checks to see if app goes inactive - saves. [[NSNotificationCenter defaultCenter] addObserver:self selector:@selector(resignActive) name:UIApplicationWillResignActiveNotification object:nil]; with the corresponding in ViewDidUnload: [[NSNotificationCenter defaultCenter] removeObserver:self name:UIApplicationWillResignActiveNotification object:nil]; Does anyone have any suggestions? EDIT: - (void) prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender{ NSString *identifier = segue.identifier; if ([identifier isEqualToString: @"Start Game"]){ gameWindow = (ViewController *)[segue destinationViewController]; gameWindow.newgame=-1; gameWindow.delegate = self; } else if ([identifier isEqualToString: @"Resume Game"]){ gameWindow = (ViewController *)[segue destinationViewController]; gameWindow.newgame=0; gameWindow.delegate = self;

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  • ImageView place at center on click in gallery view

    - by TGMCians
    i used gallery view in which i place multiple imageview dynamically but on click imageview place at center and second question how to start first imageview from left of screen. I do not want to change the place until user scroll horizontally by finger . Is there any way to achieve this. Please help for this.. private class ImageAdapter extends BaseAdapter{ public ImageAdapter() { //To set blank at bottom and make visible TextView textView = (TextView)findViewById(R.id.textView2); textView.setVisibility(View.VISIBLE); //To set the visibility visible of gallery myGallery.setVisibility(View.VISIBLE); } public int getCount() { return ProductItemArray.Image_URL.length; } public Object getItem(int position) { return null; } public long getItemId(int position) { return 0; } public View getView(int position, View arg1, ViewGroup arg2) { ImageView bottomImageView = new ImageView(context); if(Helper.isTablet(context)) bottomImageView.setLayoutParams(new Gallery.LayoutParams(VirtualMirrorActivity.convertDpToPixel(100, context), VirtualMirrorActivity.convertDpToPixel(100, context))); else bottomImageView.setLayoutParams(new Gallery.LayoutParams(VirtualMirrorActivity.convertDpToPixel(80, context), VirtualMirrorActivity.convertDpToPixel(80, context))); UrlImageViewHelper.setUrlDrawable(bottomImageView, ProductItemArray.Image_URL[position]); bottomImageView.setBackgroundResource(R.layout.border); return bottomImageView; } } myGallery.setAdapter(new ImageAdapter()); myGallery.setSelection(1); myGallery.setOnItemClickListener(new OnItemClickListener() { @Override public void onItemClick(AdapterView<?> parent, View view, final int position, long arg3) { linearLayout.removeView(frameImageView); Thread newThread = new Thread(new Runnable() { public void run() { URL url_1 = null; try { isAlreadyExistInWishlist = false; VMProductListPaging.productUrl = ProductItemArray.Image_small_URL[position]; VMProductListPaging.productId = ProductItemArray.productId[position]; VMProductListPaging.productName = ProductItemArray.product_Name[position]; url_1 = new URL(ProductItemArray.Image_small_URL[position]); bmp = BitmapFactory.decodeStream(url_1.openConnection().getInputStream()); isExecuted = true; bitmapHandler.sendMessage(bitmapHandler.obtainMessage()); } catch (Exception e) { //Toast.makeText(context,"Sorry!! This link appears to be broken",Toast.LENGTH_LONG).show(); } } }); newThread.start(); } }); Layout.xml <Gallery android:id="@+id/galleryView" android:layout_width="fill_parent" android:layout_height="wrap_content" android:spacing="5dp" android:layout_below="@+id/sendPhoto" android:layout_marginTop="10dp" android:visibility="gone"/>

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  • Finding folders

    - by gedO
    Hello. I want to find how much folders are in folder or I should say how much SubFolreds are in folder. So, how I should do that??? P.S. I'm programing with Delphi

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  • Mac OS X: Finder view options?

    - by trolle3000
    Hi there. In OS X 10.6, Finder usually looks something like this: The Finder window looks like that when you double-click on most folders or drives. However, whenever I mount a Truecrypt Volume and double-click that, it looks like this: Is there any way to default to the first view option for all types of folders? I tried view options in Finder, but it didn't seem to work.

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  • View/Find all compressed files on the server?

    - by Volodymyr
    I need to find all compressed files/folders regardless of file format on a Windows Server 2003 machine. Search options do not provide this capability. Is there a way to list/view all compressed files? Perhaps, this can be done by PowerShell using file/folder attributes and put into a txt file with file location. UPD: Under compressed files/folders - I mean files which appear in blue color in Explorer after changing file/folder attribute.

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  • BIND split-view DNS config problem

    - by organicveggie
    We have two DNS servers: one external server controlled by our ISP and one internal server controlled by us. I'd like internal requests for foo.example.com to map to 192.168.100.5 and external requests continue to map to 1.2.3.4, so I'm trying to configure a view in bind. Unfortunately, bind fails when I attempt to reload the configuration. I'm sure I'm missing something simple, but I can't figure out what it is. options { directory "/var/cache/bind"; forwarders { 8.8.8.8; 8.8.4.4; }; auth-nxdomain no; # conform to RFC1035 listen-on-v6 { any; }; }; zone "." { type hint; file "/etc/bind/db.root"; }; zone "localhost" { type master; file "/etc/bind/db.local"; }; zone "127.in-addr.arpa" { type master; file "/etc/bind/db.127"; }; zone "0.in-addr.arpa" { type master; file "/etc/bind/db.0"; }; zone "255.in-addr.arpa" { type master; file "/etc/bind/db.255"; }; view "internal" { zone "example.com" { type master; notify no; file "/etc/bind/db.example.com"; }; }; zone "example.corp" { type master; file "/etc/bind/db.example.corp"; }; zone "100.168.192.in-addr.arpa" { type master; notify no; file "/etc/bind/db.192"; }; I have excluded the entries in the view for allow-recursion and recursion in an attempt to simplify the configuration. If I remove the view and just load the example.com zone directly, it works fine. Any advice on what I might be missing?

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  • I can't see headers or footers on Word 2007 unless in full screen view

    - by kevyn
    I have a machine on a domain that does not show any headers or footers when viewing documents in word 2007, unless I switch to full screen mode. Other computers can see the headers and footers no problems. here is a video of what is happening: http://showmewhatswrong.com/play/c6fIjBVWT (expires in 6 days - but to summarize, it just shows me flicking between all the view options in word, and only when in full screen view can you see the headers and footers) any help greatly appreciated! Vista Business 32bit Office 2007

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  • Save a view in Windows Media Player

    - by Charles Roper
    I like to view my library in various ways in WMP. For example, I usually search for Podcast and order the result by date added. This gives me a list of my podcasts by date order, newest to oldest. Is there a way of saving this view so that I don't have recreate it each time I open WMP? If it's not possible to do this, can anyone suggest an app that does do it, and that handles syncing as well as WMP.

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  • VMware view for Remote Non Admin Users

    - by jcw248
    Has anyone came up with a way that I can connect to a view enviroment and not have to copy files or install software yet? I have heard of virtualizing the client. I would like to see it work similar to connecting to a view enviroment using a Linux machine. Where the desktop is launched through the browser.

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