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
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
其他回答
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!
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 ; }
这对我在Xamarin很有效。iOS / Monotouch。 将键盘按钮更改为Next,将控件传递给下一个UITextField,并将键盘隐藏在最后一个UITextField之后。
private void SetShouldReturnDelegates(IEnumerable<UIView> subViewsToScout )
{
foreach (var item in subViewsToScout.Where(item => item.GetType() == typeof (UITextField)))
{
(item as UITextField).ReturnKeyType = UIReturnKeyType.Next;
(item as UITextField).ShouldReturn += (textField) =>
{
nint nextTag = textField.Tag + 1;
var nextResponder = textField.Superview.ViewWithTag(nextTag);
if (null != nextResponder)
nextResponder.BecomeFirstResponder();
else
textField.Superview.EndEditing(true);
//You could also use textField.ResignFirstResponder();
return false; // We do not want UITextField to insert line-breaks.
};
}
}
在ViewDidLoad里面你会有:
如果你的TextFields没有标签设置它现在:
txtField1.Tag = 0;
txtField2.Tag = 1;
txtField3.Tag = 2;
//...
只是一个电话
SetShouldReturnDelegates(yourViewWithTxtFields.Subviews.ToList());
//If you are not sure of which view contains your fields you can also call it in a safer way:
SetShouldReturnDelegates(txtField1.Superview.Subviews.ToList());
//You can also reuse the same method with different containerViews in case your UITextField are under different views.
我更喜欢:
@interface MyViewController : UIViewController
@property (nonatomic, retain) IBOutletCollection(UIView) NSArray *inputFields;
@end
在NIB文件中,我以所需的顺序将textFields钩子到这个inputFields数组中。之后,我做了一个简单的UITextField的索引测试,报告用户点击返回:
// for UITextField
-(BOOL)textFieldShouldReturn:(UITextField*)textField {
NSUInteger index = [_inputFields indexOfObject:textField];
index++;
if (index < _inputFields.count) {
UIView *v = [_inputFields objectAtIndex:index];
[v becomeFirstResponder];
}
return NO;
}
// for UITextView
-(BOOL)textView:(UITextView*)textView shouldChangeTextInRange:(NSRange)range replacementText:(NSString*)text {
if ([@"\n" isEqualToString:text]) {
NSUInteger index = [_inputFields indexOfObject:textView];
index++;
if (index < _inputFields.count) {
UIView *v = [_inputFields objectAtIndex:index];
[v becomeFirstResponder];
} else {
[self.view endEditing:YES];
}
return NO;
}
return YES;
}