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.

我更喜欢:

@interface MyViewController : UIViewController
@property (nonatomic, retain) IBOutletCollection(UIView) NSArray *inputFields;
@end

在NIB文件中，我以所需的顺序将textFields钩子到这个inputFields数组中。之后，我做了一个简单的UITextField的索引测试，报告用户点击返回:

// for UITextField
-(BOOL)textFieldShouldReturn:(UITextField*)textField {
    NSUInteger index = [_inputFields indexOfObject:textField];
    index++;
    if (index < _inputFields.count) {
        UIView *v = [_inputFields objectAtIndex:index];
        [v becomeFirstResponder];
    }
    return NO;
}

// for UITextView
-(BOOL)textView:(UITextView*)textView shouldChangeTextInRange:(NSRange)range replacementText:(NSString*)text {
    if ([@"\n" isEqualToString:text]) {
        NSUInteger index = [_inputFields indexOfObject:textView];
        index++;
        if (index < _inputFields.count) {
            UIView *v = [_inputFields objectAtIndex:index];
            [v becomeFirstResponder];
        } else {
            [self.view endEditing:YES];
        }
        return NO;
    }
    return YES;
}

Swift 3解决方案，使用UITextField的有序数组

func nextTextField() {
    let textFields = // Your textfields array

    for i in 0 ..< textFields.count{
        if let textfield = textFields[i], textfield.isFirstResponder{
            textfield.resignFirstResponder()
            if i+1 < textFields.count, let nextextfield = textFields[i+1]{
                nextextfield.becomeFirstResponder()
                return
            }
        }
    }
}

你可以使用IQKeyboardManager库来做到这一点。它处理所有的事情，你不需要任何额外的设置。IQKeyboardManager可以通过CocoaPods获得，要安装它，只需在Podfile中添加以下行:

pod 'IQKeyboardManager'

或 只需拖放IQKeyBoardManager目录从演示项目到您的项目。就是这样。 IQKeyBoardManager目录可以从https://github.com/hackiftekhar/IQKeyboardManager找到

一个快速扩展，应用mxcl的答案，使这特别容易(适应swift 2.3由旅行者):

extension UITextField {
    class func connectFields(fields:[UITextField]) -> Void {
        guard let last = fields.last else {
            return
        }
        for i in 0 ..< fields.count - 1 {
            fields[i].returnKeyType = .Next
            fields[i].addTarget(fields[i+1], action: "becomeFirstResponder", forControlEvents: .EditingDidEndOnExit)
        }
        last.returnKeyType = .Done
        last.addTarget(last, action: #selector(UIResponder.resignFirstResponder), forControlEvents: .EditingDidEndOnExit)
    }
}

它很容易使用:

UITextField.connectFields([field1, field2, field3])

扩展将设置返回按钮为“下一步”为所有但最后一个字段和“完成”为最后一个字段，并转移焦点/解散键盘时，这些被轻敲。

Swift < 2.3

extension UITextField {
    class func connectFields(fields:[UITextField]) -> Void {
        guard let last = fields.last else {
            return
        }
        for var i = 0; i < fields.count - 1; i += 1 {
            fields[i].returnKeyType = .Next
            fields[i].addTarget(fields[i+1], action: "becomeFirstResponder", forControlEvents: .EditingDidEndOnExit)
        }
        last.returnKeyType = .Done
        last.addTarget(last, action: "resignFirstResponder", forControlEvents: .EditingDidEndOnExit)
    }
}

斯威夫特3: 像这样使用-

UITextField.connectFields(fields: [field1, field2])

Extension:
    extension UITextField {
        class func connectFields(fields:[UITextField]) -> Void {
            guard let last = fields.last else {
                return
            }
            for i in 0 ..< fields.count - 1 {
                fields[i].returnKeyType = .next
                fields[i].addTarget(fields[i+1], action: #selector(UIResponder.becomeFirstResponder), for: .editingDidEndOnExit)
            }
            last.returnKeyType = .go
            last.addTarget(last, action: #selector(UIResponder.resignFirstResponder), for: .editingDidEndOnExit)
        }
    }

在Mac OS X的Cocoa中，你有下一个响应器链，在那里你可以询问文本字段下一个控件应该有焦点。这就是在文本字段之间进行标签操作的原因。但由于iOS设备没有键盘，只有触摸，所以这一概念没有在Cocoa touch的过渡中幸存下来。

这很容易做到，只要有两个假设:

所有“tabbable”UITextFields都在同一个父视图上。 它们的“制表符顺序”由tag属性定义。

假设你可以重写textFieldShouldReturn:如下:

-(BOOL)textFieldShouldReturn:(UITextField*)textField
{
  NSInteger nextTag = textField.tag + 1;
  // Try to find next responder
  UIResponder* nextResponder = [textField.superview viewWithTag:nextTag];
  if (nextResponder) {
    // Found next responder, so set it.
    [nextResponder becomeFirstResponder];
  } else {
    // Not found, so remove keyboard.
    [textField resignFirstResponder];
  }
  return NO; // We do not want UITextField to insert line-breaks.
}

添加更多的代码，也可以忽略这些假设。

斯威夫特4.0

 func textFieldShouldReturn(_ textField: UITextField) -> Bool {
    let nextTag = textField.tag + 1
    // Try to find next responder
    let nextResponder = textField.superview?.viewWithTag(nextTag) as UIResponder!

    if nextResponder != nil {
        // Found next responder, so set it
        nextResponder?.becomeFirstResponder()
    } else {
        // Not found, so remove keyboard
        textField.resignFirstResponder()
    }

    return false
}

如果文本字段的superview是一个UITableViewCell那么下一个responder将是

let nextResponder = textField.superview?.superview?.superview?.viewWithTag(nextTag) as UIResponder!