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标题: ios - 带有滑出菜单的标签栏 [打印本页]

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作者: 菜鸟教程小白    时间: 2022-12-13 09:22
标题: ios - 带有滑出菜单的标签栏

RearViewController.m

-(IBAction)unwindFromViewControllerUIStoryboardSegue *)segue
{
    if ([segue.identifier isEqualToString"unwindToViewController"]) {
      ViewController *detail = [self.storyboard instantiateViewControllerWithIdentifier"ViewController"];
        [self presentViewController:detail animated:YES completion:nil];
}
}

如果我从我的 RearViewController 中单击返回，则不会显示 ViewController，而是会偷看并熄灭(再次显示 RearViewcontroller)。

图像2

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Best Answer-推荐答案

我给你一个解决方案。

      1.First remove triggered segue connection.
      2.After that give push segue connection to Back to your required view controller.
      3.Click Back Button Segue and give Identifier name as "goToMainViewController" or whatever you want just give there.Also segue should be "push".

谢谢

关于ios - 带有滑出菜单的标签栏，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/31695820/

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