React-Native:将React-Native View解除/退出回Native

Question

我有一个现有的应用程序,我正在努力为其中的一部分集成React-Native.我无法理解如何"退出"react-native并返回原生视图.

这是一些代码:

// Main objective-c code that bootstraps the react-native view. This view is loaded as a modal view.
MainViewController.m:

- (void)viewDidLoad {
    [super viewDidLoad];

    RCTRootView *rootView = [[RCTRootView alloc] initWithBundleURL:jsCodeLocation moduleName:@"IndexApp" initialProperties:props launchOptions:nil];

    rootView.frame = CGRectMake(0, 0, self.view.bounds.size.width, self.view.bounds.size.height-49);
    [self.view addSubview:rootView];

}

我最初的反应意见如下:

render() {
  return (
      <Navigator
          style={styles.container}
...

我在导航器上有一个右导航按钮,我想"关闭"反应视图和底层MainViewController本机视图.

我已经尝试从反应视图回调MainViewController,但没有用:

RCT_EXPORT_METHOD(dismiss:(NSString *)name location:(NSString *)location)
{
    NSLog(@"dismiss...");
    // these don't do anything
    //[self dismissViewControllerAnimated:YES completion:nil];
    //[self.navigationController popViewControllerAnimated:YES];
    // anything with _rootView doesn't do anything, i.e. _rootView removeFromSuperview];

}

任何有关"退出"反应本机视图并返回本机视图的方法的帮助将不胜感激.

Answer 1

注意：这不起作用 - 我将其留在这里作为不起作用的示例。请查看我的其他答案以了解可行的方法。

如果你以这种方式呈现 MainViewController ，popViewController:应该可以工作

NSURL *jsCodeLocation = [NSURL URLWithString:@"http://localhost:8081/index.ios.bundle?platform=ios&dev=true"];
RCTRootView *reactView = [[RCTRootView alloc] initWithBundleURL:jsCodeLocation
                                                    moduleName:@"SimpleApp"
                                             initialProperties:nil
                                                 launchOptions:nil];


MainViewController *rootViewController = [MainViewController new];
rootViewController.view = reactView;

[[self navigationController] pushViewController:rootViewController animated:YES];