iphone

나는 또한이 문제가 있었고 해결책을 찾았습니다.

아래는 iOS 8.0 및 아래 버전에서 작동하는 코드입니다.

iOS 7 및 8.0 (Xcode 버전 6.0.1)에서 테스트했습니다.

- (void)addButtonToKeyboard
    {
    // create custom button
    self.doneButton = [UIButton buttonWithType:UIButtonTypeCustom];
            //This code will work on iOS 8.3 and 8.4. 
       if ([[[UIDevice currentDevice] systemVersion] floatValue] >= 8.3) {
            self.doneButton.frame = CGRectMake(0, [[UIScreen mainScreen]   bounds].size.height - 53, 106, 53);
      } else {
           self.doneButton.frame = CGRectMake(0, 163+44, 106, 53);
      }

    self.doneButton.adjustsImageWhenHighlighted = NO;
    [self.doneButton setTag:67123];
    [self.doneButton setImage:[UIImage imageNamed:@"doneup1.png"] forState:UIControlStateNormal];
    [self.doneButton setImage:[UIImage imageNamed:@"donedown1.png"] forState:UIControlStateHighlighted];

    [self.doneButton addTarget:self action:@selector(doneButton:) forControlEvents:UIControlEventTouchUpInside];

    // locate keyboard view
    int windowCount = [[[UIApplication sharedApplication] windows] count];
    if (windowCount < 2) {
        return;
    }

    UIWindow *tempWindow = [[[UIApplication sharedApplication] windows] objectAtIndex:1];
    UIView *keyboard;

    for (int i = 0; i < [tempWindow.subviews count]; i++) {
        keyboard = [tempWindow.subviews objectAtIndex:i];
             // keyboard found, add the button
          if ([[[UIDevice currentDevice] systemVersion] floatValue] >= 8.3) {

            UIButton *searchbtn = (UIButton *)[keyboard viewWithTag:67123];
            if (searchbtn == nil)
                [keyboard addSubview:self.doneButton];

            } else {
                if([[keyboard description] hasPrefix:@"<UIPeripheralHost"] == YES) {
              UIButton *searchbtn = (UIButton *)[keyboard viewWithTag:67123];
                   if (searchbtn == nil)//to avoid adding again and again as per my requirement (previous and next button on keyboard)
                [keyboard addSubview:self.doneButton];

        } //This code will work on iOS 8.0
        else if([[keyboard description] hasPrefix:@"<UIInputSetContainerView"] == YES) {

            for (int i = 0; i < [keyboard.subviews count]; i++)
            {
                UIView *hostkeyboard = [keyboard.subviews objectAtIndex:i];

                if([[hostkeyboard description] hasPrefix:@"<UIInputSetHost"] == YES) {
                    UIButton *donebtn = (UIButton *)[hostkeyboard viewWithTag:67123];
                    if (donebtn == nil)//to avoid adding again and again as per my requirement (previous and next button on keyboard)
                        [hostkeyboard addSubview:self.doneButton];
                }
            }
        }
      }
    }
 }

>

- (void)removedSearchButtonFromKeypad
    {

    int windowCount = [[[UIApplication sharedApplication] windows] count];
    if (windowCount < 2) {
        return;
    }

    UIWindow *tempWindow = [[[UIApplication sharedApplication] windows] objectAtIndex:1];

    for (int i = 0 ; i < [tempWindow.subviews count] ; i++)
    {
        UIView *keyboard = [tempWindow.subviews objectAtIndex:i];

           if ([[[UIDevice currentDevice] systemVersion] floatValue] >= 8.3){
                [self removeButton:keyboard];
            } else if([[keyboard description] hasPrefix:@"<UIPeripheralHost"] == YES) {
                  [self removeButton:keyboard];

             } else if([[keyboard description] hasPrefix:@"<UIInputSetContainerView"] == YES){

            for (int i = 0 ; i < [keyboard.subviews count] ; i++)
            {
                UIView *hostkeyboard = [keyboard.subviews objectAtIndex:i];

                if([[hostkeyboard description] hasPrefix:@"<UIInputSetHost"] == YES) {
                    [self removeButton:hostkeyboard];
                }
            }
        }
    }
}


- (void)removeButton:(UIView *)keypadView
    {
    UIButton *donebtn = (UIButton *)[keypadView viewWithTag:67123];
    if(donebtn) {
        [donebtn removeFromSuperview];
        donebtn = nil;
    }
}

도움이 되었기를 바랍니다.

하지만, 여전히이 경고가 표시됩니다.

키보드 iPhone-Portrait-NumberPad에 대해 유형 4를 지원하는 키 플레인을 찾을 수 없습니다. 3876877096_Portrait_iPhone-Simple-Pad_Default 사용

이 경고를 무시하고 작동합니다. 이 경고에서 구제를받을 수 있는지 알려주십시오.

답변