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。
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
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
}
}
其他回答
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!
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。
下面是一个swift3版本的Anth0的答案。我把它贴在这里,以帮助任何想要利用他的伟大答案的敏捷开发人员!当您设置关联对象时,我擅自添加了一个返回键类型“Next”。
extension UITextField {
@nonobjc static var NextHashKey: UniChar = 0
var nextTextField: UITextField? {
get {
return objc_getAssociatedObject(self,
&UITextField.NextHashKey) as? UITextField
}
set(next) {
self.returnKeyType = UIReturnKeyType.next
objc_setAssociatedObject(self,
&UITextField.NextHashKey,next,.OBJC_ASSOCIATION_RETAIN_NONATOMIC)
}
}
}
下面是另一个扩展,它显示了使用上述代码循环遍历UITextFields列表的可能性。
extension UIViewController: UITextFieldDelegate {
public func textFieldShouldReturn(_ textField: UITextField) -> Bool {
guard let next = textField.nextTextField else {
textField.resignFirstResponder()
return true
}
next.becomeFirstResponder()
return false
}
}
然后在你的ViewController或者其他地方,你可以像这样设置你的文本框。
@IBOutlet fileprivate weak var textfield1: UITextField!
@IBOutlet fileprivate weak var textfield2: UITextField!
@IBOutlet fileprivate weak var textfield3: UITextField!
...
[textfield1, textfield2, textfield3].forEach{ $0?.delegate = self }
textfield1.nextTextField = textfield2
textfield2.nextTextField = textfield3
// We don't assign a nextTextField to textfield3 because we want
// textfield3 to be the last one and resignFirstResponder when
// the return button on the soft keyboard is tapped.
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.