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
这是一个简单的解决方案在swift,没有标签使用,没有故事板技巧…
只需使用这个扩展:
extension UITextField{
func nextTextFieldField() -> UITextField?{
//field to return
var returnField : UITextField?
if self.superview != nil{
//for each view in superview
for (_, view) in self.superview!.subviews.enumerate(){
//if subview is a text's field
if view.isKindOfClass(UITextField){
//cast curent view as text field
let currentTextField = view as! UITextField
//if text field is after the current one
if currentTextField.frame.origin.y > self.frame.origin.y{
//if there is no text field to return already
if returnField == nil {
//set as default return
returnField = currentTextField
}
//else if this this less far than the other
else if currentTextField.frame.origin.y < returnField!.frame.origin.y{
//this is the field to return
returnField = currentTextField
}
}
}
}
}
//end of the mdethod
return returnField
}
}
并调用它像这样(例如)与你的textfield委托:
func textFieldShouldReturn(textField: UITextField) -> Bool {
textField.resignFirstResponder()
textField.nextTextFieldField()?.becomeFirstResponder()
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!
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;
}
}
这是一个简单的解决方案在swift,没有标签使用,没有故事板技巧…
只需使用这个扩展:
extension UITextField{
func nextTextFieldField() -> UITextField?{
//field to return
var returnField : UITextField?
if self.superview != nil{
//for each view in superview
for (_, view) in self.superview!.subviews.enumerate(){
//if subview is a text's field
if view.isKindOfClass(UITextField){
//cast curent view as text field
let currentTextField = view as! UITextField
//if text field is after the current one
if currentTextField.frame.origin.y > self.frame.origin.y{
//if there is no text field to return already
if returnField == nil {
//set as default return
returnField = currentTextField
}
//else if this this less far than the other
else if currentTextField.frame.origin.y < returnField!.frame.origin.y{
//this is the field to return
returnField = currentTextField
}
}
}
}
}
//end of the mdethod
return returnField
}
}
并调用它像这样(例如)与你的textfield委托:
func textFieldShouldReturn(textField: UITextField) -> Bool {
textField.resignFirstResponder()
textField.nextTextFieldField()?.becomeFirstResponder()
return true
}
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
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; }