这是我对这个问题的解决方案。

为了解决这个问题(因为我讨厌依赖标签来做事情)，我决定向UITextField对象添加一个自定义属性。换句话说，我在UITextField上创建了一个这样的类别:

UITextField + Extended.h

@interface UITextField (Extended)

@property(retain, nonatomic)UITextField* nextTextField;

@end

UITextField +扩展。m

#import "UITextField+Extended.h"
#import <objc/runtime.h>

static char defaultHashKey;

@implementation UITextField (Extended)

- (UITextField*) nextTextField { 
    return objc_getAssociatedObject(self, &defaultHashKey); 
}

- (void) setNextTextField:(UITextField *)nextTextField{
    objc_setAssociatedObject(self, &defaultHashKey, nextTextField, OBJC_ASSOCIATION_RETAIN_NONATOMIC); 
}

@end

下面是我如何使用它:

UITextField *textField1 = ...init your textfield
UITextField *textField2 = ...init your textfield
UITextField *textField3 = ...init your textfield

textField1.nextTextField = textField2;
textField2.nextTextField = textField3;
textField3.nextTextField = nil;

实现textFieldShouldReturn方法:

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

    UITextField *next = theTextField.nextTextField;
    if (next) {
        [next becomeFirstResponder];
    } else {
        [theTextField resignFirstResponder];
    }

    return NO; 
}

我现在有一个UITextField的链表，每一个都知道下一个是谁。

希望能有所帮助。

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)
        }
    }

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

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。

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。

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;

}