Remove public iOS types from view.d.ts

This commit is contained in:
Hristo Hristov
2017-10-25 13:07:59 +03:00
parent 6cff5f4796
commit 2a10b7a220
3 changed files with 5 additions and 5 deletions

View File

@ -244,7 +244,7 @@ function getViewController(view: View): UIViewController {
if (viewController instanceof UIViewController) {
return viewController;
} else if (view.ios instanceof UIView) {
viewController = iosView.UILayoutViewController.initWithOwner(new WeakRef(view));
viewController = iosView.UILayoutViewController.initWithOwner(new WeakRef(view)) as UIViewController;
viewController.view.addSubview(view.ios);
return viewController;
} else {

View File

@ -645,9 +645,9 @@ export const isEnabledProperty: Property<View, boolean>;
export const isUserInteractionEnabledProperty: Property<View, boolean>;
export namespace ios {
export function updateConstraints(controller: UIViewController, owner: View): void;
export function layoutView(controller: UIViewController, owner: View): void;
export class UILayoutViewController extends UIViewController {
export function updateConstraints(controller: any /* UIViewController */, owner: View): void;
export function layoutView(controller: any /* UIViewController */, owner: View): void;
export class UILayoutViewController {
public static initWithOwner(owner: WeakRef<View>): UILayoutViewController;
}
}

View File

@ -296,7 +296,7 @@ export class TabView extends TabViewBase {
newController = item.view.ios.controller;
item.setViewController(newController, newController.view);
} else {
newController = iosView.UILayoutViewController.initWithOwner(new WeakRef(item.view));
newController = iosView.UILayoutViewController.initWithOwner(new WeakRef(item.view)) as UIViewController;
newController.view.addSubview(item.view.nativeViewProtected);
item.view.viewController = newController;
item.setViewController(newController, item.view.nativeViewProtected);