我如何通过iPhone键盘上的“下一步”按钮浏览所有的文本字段?

最后一个文本字段应该关闭键盘。

我已经设置了IB按钮(下一步/完成),但现在我卡住了。

我实现了textFieldShouldReturn动作,但现在下一步和完成按钮关闭键盘。


当前回答

退出一个文本字段后,调用[otherTextField becomeFirstResponder],下一个字段获得焦点。

这实际上是一个棘手的问题,因为通常你还想滚动屏幕或以其他方式调整文本字段的位置,以便在编辑时容易看到。只要确保做大量的测试,以不同的方式进入和退出文本字段,并提前离开(总是给用户一个选项,取消键盘,而不是进入下一个字段,通常在导航栏中有“完成”)

其他回答

如果有人想这样。我认为这是最接近问题所要求的要求

下面是我如何实现它

为需要设置的每个文本字段添加附件视图,使用

func setAccessoryViewFor(textField : UITextField)    {
    let toolBar = UIToolbar()
    toolBar.barStyle = .default
    toolBar.isTranslucent = true
    toolBar.sizeToFit()

    // Adds the buttons

    // Add previousButton
    let prevButton = UIBarButtonItem(title: "<", style: .plain, target: self, action: #selector(previousPressed(sender:)))
    prevButton.tag = textField.tag
    if getPreviousResponderFor(tag: textField.tag) == nil {
        prevButton.isEnabled = false
    }

    // Add nextButton
    let nextButton = UIBarButtonItem(title: ">", style: .plain, target: self, action: #selector(nextPressed(sender:)))
    nextButton.tag = textField.tag
    if getNextResponderFor(tag: textField.tag) == nil {
        nextButton.title = "Done"
    }

    let spaceButton = UIBarButtonItem(barButtonSystemItem: .flexibleSpace, target: nil, action: nil)
    toolBar.setItems([prevButton,spaceButton,nextButton], animated: false)
    toolBar.isUserInteractionEnabled = true
    textField.inputAccessoryView = toolBar
}

使用以下函数来处理轻拍

func nextPressed(sender : UIBarButtonItem) {
    if let nextResponder = getNextResponderFor(tag: sender.tag) {
        nextResponder.becomeFirstResponder()
    } else {
        self.view.endEditing(true)
    }

}

func previousPressed(sender : UIBarButtonItem) {
    if let previousResponder = getPreviousResponderFor(tag : sender.tag)  {
        previousResponder.becomeFirstResponder()
    }
}

func getNextResponderFor(tag : Int) -> UITextField? {
    return self.view.viewWithTag(tag + 1) as? UITextField
}

func getPreviousResponderFor(tag : Int) -> UITextField? {
    return self.view.viewWithTag(tag - 1) as? UITextField
}

您需要按您希望下一个/前一按钮响应的顺序给出textFields标记。

我用Michael G. Emmons的答案已经有一年了,效果很好。我最近注意到,立即调用resignFirstResponder,然后立即调用becomeFirstResponder会导致键盘“故障”,消失,然后立即出现。我稍微改变了他的版本,跳过resignFirstResponder如果nextField是可用的。

- (BOOL)textFieldShouldReturn:(UITextField *)textField
{ 

    if ([textField isKindOfClass:[NRTextField class]])
    {
        NRTextField *nText = (NRTextField*)textField;
        if ([nText nextField] != nil){
            dispatch_async(dispatch_get_main_queue(),
                           ^ { [[nText nextField] becomeFirstResponder]; });

        }
        else{
            [textField resignFirstResponder];
        }
    }
    else{
        [textField resignFirstResponder];
    }

    return true;

}

我喜欢Anth0和Answerbot已经提出的面向对象解决方案。然而,我正在开发一个快速而小型的POC,所以我不想让子类和类别使事情变得混乱。

另一个简单的解决方案是创建一个字段的NSArray,并在按下next时查找下一个字段。不是面向对象的解决方案,而是快速、简单且易于实现。此外,您可以一目了然地查看和修改排序。

下面是我的代码(基于这个线程中的其他答案):

@property (nonatomic) NSArray *fieldArray;

- (void)viewDidLoad {
    [super viewDidLoad];

    fieldArray = [NSArray arrayWithObjects: firstField, secondField, thirdField, nil];
}

- (BOOL) textFieldShouldReturn:(UITextField *) textField {
    BOOL didResign = [textField resignFirstResponder];
    if (!didResign) return NO;

    NSUInteger index = [self.fieldArray indexOfObject:textField];
    if (index == NSNotFound || index + 1 == fieldArray.count) return NO;

    id nextField = [fieldArray objectAtIndex:index + 1];
    activeField = nextField;
    [nextField becomeFirstResponder];

    return NO;
}

I always return NO because I don't want a line break inserted. Just thought I'd point that out since when I returned YES it would automatically exit the subsequent fields or insert a line break in my TextView. It took me a bit of time to figure that out. activeField keeps track of the active field in case scrolling is necessary to unobscure the field from the keyboard. If you have similar code, make sure you assign the activeField before changing the first responder. Changing first responder is immediate and will fire the KeyboardWasShown event immediately.

解决方案在Swift 3.1,连接你的文本字段IBOutlets设置你的文本字段委托在viewDidLoad,然后在textFieldShouldReturn导航你的动作

class YourViewController: UIViewController,UITextFieldDelegate {

        @IBOutlet weak var passwordTextField: UITextField!
        @IBOutlet weak var phoneTextField: UITextField!

        override func viewDidLoad() {
            super.viewDidLoad()
            self.passwordTextField.delegate = self
            self.phoneTextField.delegate = self
            // Set your return type
            self.phoneTextField.returnKeyType = .next
            self.passwordTextField.returnKeyType = .done
        }

        func textFieldShouldReturn(_ textField: UITextField) -> Bool{
            if textField == self.phoneTextField {
                self.passwordTextField.becomeFirstResponder()
            }else if textField == self.passwordTextField{
                // Call login api
                self.login()
            }
            return true
        }

    }

当按下“Done”按钮时,一个非常简单的方法是:

在头文件中创建一个新的IBAction

- (IBAction)textFieldDoneEditing:(id)sender;

在实现文件(。M文件)添加如下方法:

- (IBAction)textFieldDoneEditing:(id)sender 
{ 
  [sender resignFirstResponder];
}

然后,当你来链接IBAction到文本框-链接到'Did End On Exit'事件。