Search Results

Search found 25747 results on 1030 pages for 'tree view'.

Page 153/1030 | < Previous Page | 149 150 151 152 153 154 155 156 157 158 159 160  | Next Page >

  • Where should I put code that is supposed to fire AFTER the view has loaded?

    - by Timbo
    I’m writing an objective-c program that does some calculations based on time and will eventually updates UILabels each second. To explain the concept here’s some simplified code, which I’ve placed into the viewDidLoad of the class that handles the view. (void)viewDidLoad { [super viewDidLoad]; // how do i make this stuff happen AFTER the view has loaded?? int a = 1; while (a < 10) { NSLog(@"doing something"); a = a + 1; sleep(1); } } My problem is that the code halts the loading of the view until the loop is all complete (in this case 10 seconds). Where should I put code that I want to fire AFTER the view has finished loading? newbie question I know =/

    Read the article

  • How can I present a modal view controller after selecting a contact?

    - by barfoon
    Hey everyone, I'm trying to present a modal view controller after selecting a contact and it doesnt seem to be working. In my - (BOOL)peoplePickerNavigationController:(ABPeoplePickerNavigationController *)peoplePicker shouldContinueAfterSelectingPerson:(ABRecordRef)person method, I dismiss peoplePicker, create an instance of my new view controller, then present it with [self.navigationController presentModalViewController:newController animated:YES]; and it doesnt work. However if i PUSH the view controller it works with this code: [self.navigationController pushViewController:newController animated:YES]; How can I accomplish this? Thank you,

    Read the article

  • How do I control the background color during the iPhone flip view animation transition?

    - by Rob S.
    I have some pretty standing flipping action going on: [UIView beginAnimations:@"swapScreens" context:nil]; [UIView setAnimationTransition:UIViewAnimationTransitionFlipFromLeft forView:self.view cache:YES]; [UIView setAnimationDuration:1.0]; [self.view exchangeSubviewAtIndex:0 withSubviewAtIndex:1]; [UIView commitAnimations]; To Apple's credit, this style of animation is amazingly easy to work with. Very cool, and I've been able to animate transitions, flips, fades etc. throughout the app very easily. Question: During the flip transition, the background visible 'behind' the two views during the flip is white and I'd like it to be black. I've: Set the background of the containing view (self.view above) - no dice. I really thought that would work. Set the background of each view to black - no dice. I didn't think this would work although you give different things a shot to understand better :) Google'd like crazy; keep landing on Safari-related listings. Thanks in advance!

    Read the article

  • How can I render an in-memory UIViewController's view Landscape?

    - by Aaron
    I'm trying to render an in-memory (but not in hierarchy, yet) UIViewController's view into an in-memory image buffer so I can do some interesting transition animations. However, when I render the UIViewController's view into that buffer, it is always rendering as though the controller is in Portrait orientation, no matter the orientation of the rest of the app. How do I clue this controller in? My code in RootViewController looks like this: MyUIViewController* controller = [[MyUIViewController alloc] init]; int width = self.view.frame.size.width; int height = self.view.frame.size.height; int bitmapBytesPerRow = width * 4; unsigned char *offscreenData = calloc(bitmapBytesPerRow * height, sizeof(unsigned char)); CGColorSpaceRef colorSpace = CGColorSpaceCreateDeviceRGB(); CGContextRef offscreenContext = CGBitmapContextCreate(offscreenData, width, height, 8, bitmapBytesPerRow, colorSpace, kCGImageAlphaPremultipliedLast); CGContextTranslateCTM(offscreenContext, 0.0f, height); CGContextScaleCTM(offscreenContext, 1.0f, -1.0f); [(CALayer*)[controller.view layer] renderInContext:offscreenContext]; At that point, the offscreen memory buffers contents are portrait-oriented, even when the window is in landscape orientation. Ideas?

    Read the article

  • Assigning a view controller to be the delegate of a subview which is not directly descendent?

    - by ambertch
    I am writing an iphone app where in numerous cases a subview needs to talk to its superview. For example: View A has a table view that contains photos A has a subview B which allows users to add photos, upon which I want to auto append them to A's table view So far I have been creating a @protocol in B, and registering A as the delegate. The problem in my case is that B has a subview C that allows users to add content, and I want actions in C to invoke changes in its grandparent, A. Currently I am working around this by passing around a self pointer to my base view controller (C.delegate = B.delegate), but it doesn't seem very proper to me. Any thoughts? (and/or general advice on code organization when all sort of subviews needs to talk to superviews would be greatly appreciated) Thanks!

    Read the article

  • WPF MVVM View with varying number of objects. How to?

    - by 0xDEAD BEEF
    HI! I want to design view which will contain multiple objects in different locations. For example - it would be great if viewmodel could have field like list of objects (rectangles) and when i change/add members to list, new rectangles appear on view in specified positions. How do i create such view/viewmodel?

    Read the article

  • How to import data from a Lotus private view into Access?

    - by PowerUser
    I'm experimenting with Lotus private views for the first time and I finally made a private view that (more or less) has the data I want. I just need to get it into MS Access. If this was a standard shared view, I'd just fire up the ODBC administrator, and make a DSN to the database using the NotesSQL driver. Been there, done that. But you can't get to private views that way. So, how do I import the data from my private view into Access? (Also, I'm not one of the IT gurus in our company, so I can't just make a view and share it, even if it's hidden)

    Read the article

  • How can I display the camera view in the main window in real time?

    - by Mongrel Warfare
    I'm trying to make an augmented reality application with a waypoint structure, like Yelp, and I'm wondering how to set up my main view so that it displays the camera view on the whole screen. I've heard of using the UIImagePickerController Class, but I'm unsure how to manipulate the code so that it doesn't actually take a picture, but just stays in the view mode. Any help would be appreciated, thanks!

    Read the article

  • How to launch a browser in view source mode?

    - by JorgeLarre
    I want to open a file in a web browser (anyone will do) and I want to see it in the view source mode instead of in the standard browser window. This can be done in two steps, by opening the file and then go to the view source window (different shortcuts in each browser), but I want to directly go to the view source window. I have not found any such command line argument for Firefox nor Chrome. Is this possible just with the base browser functionality?

    Read the article

  • Rails: how to represent available view actions in a stateful model?

    - by Greg
    I have a model that is stateful. In each state there are a selection of actions that the user might want to perform on an instance of the model. Currently I am translating the model state to actions that get represented in the view using a view helper. Something like this... in the model: Class Thing def state_is_A? state == 'A' end end In the helper: def display_available_actions(thing) if thing.state_is_A? link_to <action1> link_to <action2> end end And in the view: <%= display_available_actions(@thing) %> I don't like the fact that the model state is translated into view actions in the helper. I would like this to be incorporated into the model. On the other hand, it doesn't seem healthy for the model and view to get so coupled. Is there a Ruby or Rails idiom that suits this kind of situation better than my approach? Should each state be a separate model perhaps?

    Read the article

  • Is it bad to use a model directly from a view in codeigniter?

    - by jason
    I know normally the data is passed thru to the view with the controller. however, currently in my view I load my model ($this-load-model('Db_model');) so i can use it in a loop to retrieve a users profile picture path from a array of IDs that is passed from controller. Will loading the db model in the view to accomplish this make my site more vulnerable or bad form? To me it seems to be outside of MVC concept but its working atm. thanks

    Read the article

  • How can I hook up an IBAction method to a plain view for a touch up event?

    - by Thanks
    I have created a blank new view-based application project in Xcode. It generated a myProjectViewController and an nib for it. In that nib for that view controller, there is just one view. I wanted to test some event handling stuff and created an -(IBAction) method that will just log a "hello world" when I touch the view. But for some reason, IB doesn't give me a chance to hook up the action. What am I doing wrong there? I also tried to put a UIView as subview there. When I drag from that to File's Owner (whoose class is the myProjectViewController, where I have the IBAction in the header), doesn't even mention the IBAction. But it actually should, right?

    Read the article

  • How to access view items inside a ListView android?

    - by Yasir Khan
    I have ListView. i am successfully able to populate that ListView but what is want now is when user long press on ListItem it should make a button visible which i made invisible when i am populating ListView. here is snippet i have tried. mItemListView.setOnItemLongClickListener(new OnItemLongClickListener() { @Override public boolean onItemLongClick(AdapterView<?> adapterview, View arg1, int arg2, long arg3) { LinearLayout view=(LinearLayout) mItemListView.getChildAt(arg2); view.getChildAt(0).setVisibility(View.VISIBLE); return false; } }); My adapter is extending BaseAdapter

    Read the article

  • Retain a list of objects and pass it to the create/edit view when validation fails in ASP.NET MVC 2

    - by brainnovative
    I am binding a Foreign key property in my model. I am passing a list of possible values for that property in my model. The model looks something like this: public class UserModel { public bool Email { get; set; } public bool Name { get; set; } public RoleModel Role { get; set; } public IList<RoleModel> Roles { get; set; } } public class RoleModel { public string RoleName { get; set; } } This is what I have in the controller: public ActionResult Create() { IList<RoleModel> roles = RoleModel.FromArray(_userService.GetAllRoles()); UserModel model = new UserModel() { Roles = roles }; return View(model); } In the view I have: <div class="editor-label"> <%= Html.LabelFor(model => model.Role) %> </div> <div class="editor-field"> <%= Html.DropDownListFor(model => model.Role, new SelectList(Model.Roles, "RoleName", "RoleName", Model.Role))%> <%= Html.ValidationMessageFor(model => model.Role)%> </div> What do I need to do to get the list of roles back to my controller to pass it again to the view when validation fails. This is what I need: [HttpPost] public ActionResult Create(UserModel model) { if (ModelState.IsValid) { // insert logic here } //the validation fails so I pass the model again to the view for user to update data but model.Roles is null :( return View(model); } As written in the comments above I need to pass the model with the list of roles again to my view but model.Roles is null. Currently I ask the service again for the roles (model.Roles = RoleModel.FromArray(_userService.GetAllRoles());) but I don't want to add an extra overhead of getting the list from DB when I have already done that.. Anyone knows how to do it?

    Read the article

  • How can I prevent a view from covering my tab controller in my tab based application?

    - by helloJello
    I have an application with a Tab Bar Controller that has three tabs. In tab 1 there is a view (view1) with a button that when clicked transitions the user to a new view (view2) still within tab 1. However when this new view (view2) is loaded it covers my tab bar controller. What is the best approach for me to take to still display tab bar controller as well as keep tab 1 highlighted?

    Read the article

  • Oracle: Is there a way to get the column data types for a view?

    - by rally25rs
    For a table in oracle, I can query "all_tab_columns" and get table column information, like the data type, precision, whether or not the column is nullable. In SQL Developer or TOAD, you can click on a view in the GUI and it will spit out a list of the columns that the view returns and the same set of data (data type, precision, nullable, etc). So my question is, is there a way to query this column definition for a view, the way you can for a table? How do the GUI tools do it?

    Read the article

  • Send copy of class to view class so it can render him? ( iPhone )

    - by Johannes Jensen
    I'm making a game for the iPhone, and I have a class called Robot. Then I have a class called View, which renders everything. I want to send a copy of my Robot, which I defined in my ViewController, and I send it to gameView (which is View *gameView), like this: robot = [Robot new]; [gameView setRobot: [robot copy]]; I tried to make a copy but that didn't work, I could also do it with a pointer to Robot (&robot) but sometimes it just crashes ? I tried this in my View.h @interface definition: @property (copy) Robot* robot; but I get the error /RobotsAdventure/Classes/View.h:24: error: setter '-robot' argument type does not match property type :/ Help? I'm pretty new at this, heh.

    Read the article

  • How can I inherit an ASP.NET MVC controller and change only the view?

    - by AlexWalker
    I have a controller that's inheriting from a base controller, and I'm wondering how I can utilize all of the logic from the base controller, but return a different view than the base controller uses. The base controller populates a model object and passes that model object to its view, but I'm not sure how I can access that model object in the child controller so that I can pass it to the child controller's view.

    Read the article

  • Loading an external NIB, how do I set the view property?

    - by Sheehan Alam
    If I am loading a view from another NIB, how do I set the File's Owner view property? IB is not letting me hook it up to my View Controller which is loading the external NIB. My NIB looks like this: File's Owner - Identity is set to LBRootViewController First Responder LBTableViewController - Identity is set to LBTableViewController, NIB Name is LBTableViewController

    Read the article

  • Qt: QStackedWidget solution

    - by Martin
    I'm building a Qt application that have about 30 different views (QWidgets). My idea is to use a QStackedWidget to make it easy to switch between the different views in the application. I have two different solutions of how to implement this and use as little memory as possible when the user navigates through the application. Solution 1: Everytime I need to show a view I check if it is already in the stack. (The user might open the same view many times, maybe a view showing an item from a database). If the view is in the stack already it doesn't need to be created again and I can just show the view. The good thing with this solution is that I reuse the views (widgets) so they only need to be created once. This is good as the UI and other stuff should look the same everytime the user show a view, so why not reuse it? The problem with this solution is that every view has childrens. Maybe an object, a QList with objects or other things. A good thing with Qt is that you can use the parent-children mechanism so that the children will be deleted when the parent is deleted. As I never delete the parent (view) I need to handle this myself as the children might need to be deleted from different times when the view is shown. (Maybe the view show a list with objects and the list should be updated from a database each time the view is shown.) Solution 2: Everytime I need to show a QWidget I create a new one and show it. When it is not shown anymore, I delete it from memory. This is a quite easy solution. And as I delete the views when they are not shown both the view and it's children should be deleted from memory so it shouldn't increase memory, am I right? Which one of the solutions do you recommend?

    Read the article

< Previous Page | 149 150 151 152 153 154 155 156 157 158 159 160  | Next Page >