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

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;    
  }
}

其他回答

没有usings标签,也没有为nextField/nextTextField添加属性,你可以尝试模拟TAB,其中"testInput"是你当前的活动字段:

if ([textInput isFirstResponder])
    [textInput.superview.subviews enumerateObjectsAtIndexes:
     [NSIndexSet indexSetWithIndexesInRange:
      NSMakeRange([textInput.superview.subviews indexOfObject:textInput]+1,
                  [textInput.superview.subviews count]-[textInput.superview.subviews indexOfObject:textInput]-1)]
                                                    options:0 usingBlock:^(UIView *obj, NSUInteger idx, BOOL *stop) {
                                                        *stop = !obj.hidden && [obj becomeFirstResponder];
                                                    }];
if ([textInput isFirstResponder])
    [textInput.superview.subviews enumerateObjectsAtIndexes:
     [NSIndexSet indexSetWithIndexesInRange:
      NSMakeRange(0,
                  [textInput.superview.subviews indexOfObject:textInput])]
                                                    options:0 usingBlock:^(UIView *obj, NSUInteger idx, BOOL *stop) {
                                                        *stop = !obj.hidden && [obj becomeFirstResponder];
                                                    }];

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.

我尝试了很多代码,最后,这在Swift 3.0最新版[2017年3月]对我有效

ViewController类应该继承UITextFieldDelegate以使这段代码正常工作。

class ViewController: UIViewController,UITextFieldDelegate  

使用适当的标记号添加文本字段,该标记号用于根据分配给它的增量标记号将控件转移到适当的文本字段。

override func viewDidLoad() {
    userNameTextField.delegate = self
    userNameTextField.tag = 0
    userNameTextField.returnKeyType = UIReturnKeyType.next
    passwordTextField.delegate = self
    passwordTextField.tag = 1
    passwordTextField.returnKeyType = UIReturnKeyType.go
}

在上面的代码中,returnKeyType = UIReturnKeyType。下一个将使键垫返回键显示为下一个你也有其他选项加入/走等,根据你的应用程序改变值。

这个textFieldShouldReturn是一个UITextFieldDelegate控制的方法,在这里我们有下一个字段选择基于标签值的增量

func textFieldShouldReturn(_ textField: UITextField) -> Bool {
    if let nextField = textField.superview?.viewWithTag(textField.tag + 1) as? UITextField {
        nextField.becomeFirstResponder()
    } else {
        textField.resignFirstResponder()
        return true;
    }
    return false
 }