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
            }
        }
    }
}

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)
        }
    }

大家好，请看这个

- (void)nextPrevious:(id)sender
{

  UIView *responder = [self.view findFirstResponder];   

  if (nil == responder || ![responder isKindOfClass:[GroupTextField class]]) {
    return;
  }

  switch([(UISegmentedControl *)sender selectedSegmentIndex]) {
    case 0:
      // previous
      if (nil != ((GroupTextField *)responder).previousControl) {
        [((GroupTextField *)responder).previousControl becomeFirstResponder];
        DebugLog(@"currentControl: %i previousControl: %i",((GroupTextField *)responder).tag,((GroupTextField *)responder).previousControl.tag);
      }
      break;
    case 1:
      // next
      if (nil != ((GroupTextField *)responder).nextControl) {
        [((GroupTextField *)responder).nextControl becomeFirstResponder];
        DebugLog(@"currentControl: %i nextControl: %i",((GroupTextField *)responder).tag,((GroupTextField *)responder).nextControl.tag);
      }     
      break;    
  }
}

我尝试使用一种更复杂的方法来解决这个问题，该方法基于为UITableView中的每个单元格(或UITextField)分配一个稍后可以检索的唯一标签值: activate-next-uitextfield-in-uitableview-ios

我希望这能有所帮助!

if (cell == nil)
{
    cell = [[UITableViewCell alloc] initWithStyle:UITableViewCellStyleDefault reuseIdentifier:cellIdentifier];
    txt_Input = [[ UITextField alloc] initWithFrame:CGRectMake(0, 10, 150, 30)];
    txt_Input.tag = indexPath.row+1;
    [self.array_Textfields addObject:txt_Input]; // Initialize mutable array in ViewDidLoad
}

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

    int tag = ( int) textField.tag ;
    UITextField * txt = [  self.array_Textfields objectAtIndex:tag ] ;
    [ txt becomeFirstResponder] ;
    return YES ;
}

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!