我已经添加到PeyloW的答案，以防你想实现一个previous/next按钮功能:

- (IBAction)moveThroughTextFields:(UIBarButtonItem *)sender 
{
    NSInteger nextTag;
    UITextView *currentTextField = [self.view findFirstResponderAndReturn];

    if (currentTextField != nil) {
        // I assigned tags to the buttons.  0 represent prev & 1 represents next
        if (sender.tag == 0) {
            nextTag = currentTextField.tag - 1;

        } else if (sender.tag == 1) {
            nextTag = currentTextField.tag + 1;
        }
    }
    // Try to find next responder
    UIResponder* nextResponder = [self.view viewWithTag:nextTag];
    if (nextResponder) {
        // Found next responder, so set it.
        // I added the resign here in case there's different keyboards in place.
        [currentTextField resignFirstResponder];
        [nextResponder becomeFirstResponder];
    } else {
        // Not found, so remove keyboard.
        [currentTextField resignFirstResponder];

    }
}

你像这样子类化UIView:

@implementation UIView (FindAndReturnFirstResponder)
- (UITextView *)findFirstResponderAndReturn
{
    for (UITextView *subView in self.subviews) {
        if (subView.isFirstResponder){
            return subView;
        }
    }
    return nil;
}
@end

在textFieldShouldReturn你应该检查你目前的文本域不是最后一个，当他们点击下一步，如果不是不要解散键盘..

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;    
  }
}

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
        }

    }

这是一个老帖子，但有一个很高的页面排名，所以我将附和我的解决方案。

我有一个类似的问题，最终创建了UIToolbar的子类来管理动态tableView中的下一个/前一个/完成功能:https://github.com/jday001/DataEntryToolbar

您将工具栏设置为文本字段的inputAccessoryView，并将它们添加到它的字典中。这允许您向前和向后循环浏览它们，即使是动态内容。有委托方法，如果你想触发你自己的功能时textField导航发生，但你不必处理管理任何标签或第一个响应器状态。

在GitHub链接中有代码片段和示例应用程序，以帮助实现细节。您需要自己的数据模型来跟踪字段中的值。

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 ;
}