objective-c - 我的选择器实际上并没有选择值?

标签 objective-c ios6 xcode4.5 picker

所以我有一个弹出的选择器而不是键盘,但是当您滚动到选择器上的不同点时它不会更改值。我通过显示警报显示选择的年龄来测试这一点,并且它与第一个选项相比永远不会改变。我认为警报已损坏,但选择器中当前所选选项上的蓝色选择器轮廓也没有显示。这是选择器和警报的代码:

- (void)viewDidLoad
{
    [super viewDidLoad];


    NSArray *array = [[NSArray alloc]        initWithObjects:@"18-29",@"30-39",@"40-49",@"50-59",@"60+",nil];
    self.pickerData = array;
}

-(IBAction)UITextFieldEditingDidBegin
{

    UIPickerView *picker = [[UIPickerView alloc] 
                            initWithFrame:CGRectMake(0, 244, 320, 270)];
    picker.delegate = self;
    picker.dataSource = self;
    [self.view addSubview:picker];
    [picker release];

}

-(IBAction)donePressed
{

NSInteger row = [agePicker selectedRowInComponent:0];
NSString *selected = [pickerData objectAtIndex:row];
NSString *title = [[NSString alloc] initWithFormat:
                   @"You selected %@!", selected];
UIAlertView *alert = [[UIAlertView alloc] initWithTitle:title
                                               message : @"Thank you for choosing."
                                               delegate:nil
                                     cancelButtonTitle :@"Okay!"             
                                     otherButtonTitles :nil];
[alert show];
[agePicker resignFirstResponder];
}

-(NSInteger)numberOfComponentsInPickerView:(UIPickerView *)pickerView
{
    return 1;
}

-(NSInteger)pickerView:(UIPickerView *)pickerView
numberOfRowsInComponent:(NSInteger)component
{
    return [pickerData count];
}

-(NSString *)pickerView:(UIPickerView *)pickerView
        titleForRow:(NSInteger)row
       forComponent:(NSInteger)component 
{
    return[pickerData objectAtIndex:row];
}

最佳答案

编辑, 帮助关闭您的选择器, 使其成为实例变量,在界面中声明您的选择器,

MyClass.h

@interface MyClass:UIViewController <UIPickerViewDelegate, UIPickerViewDataSource>{
UIPickerView *picker
}

MyClass.m

@interface MyClass()
property (nonatomic, retain) *selectedVal; //make it a NSString
@end



-(IBAction)UITextFieldEditingDidBegin
{

    picker = [[UIPickerView alloc] 
                            initWithFrame:CGRectMake(0, 244, 320, 270)];
    picker.delegate = self;
    picker.dataSource = self;
    [self.view addSubview:picker];
    //[picker release]; //Place your picker release in the dealloc method

}

您必须实现选择器的委托(delegate),

- (void)pickerView:(UIPickerView *)thePickerView didSelectRow:(NSInteger)row inComponent:(NSInteger)component {
 self.selectedVal = [pickerData objectAtIndex:row]
NSLog(@"Selected option: %@. ", self.selectedVal );
}

然后在完成按钮上,您的属性中就有选定的 val 可供使用

-(IBAction)donePressed
{


NSString *title = [[NSString alloc] initWithFormat:
                   @"You selected %@!", self.selectedVal];
UIAlertView *alert = [[UIAlertView alloc] initWithTitle:title
                                               message : @"Thank you for choosing."
                                               delegate:nil
                                     cancelButtonTitle :@"Okay!"             
                                     otherButtonTitles :nil];
[alert show];
[alert release]; //I noticed you release the picker, so not using ARC, so release             alert!!
[picker resignFirstResponder];

 //after done, remove the picker from the view
 [picker removeFromSuperview];

}

关于objective-c - 我的选择器实际上并没有选择值?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13520512/

相关文章:

ios - 使用 block 的 Sprite-kit Action

objective-c - 在 NSTableView 中添加行?

ios - 从数字中删除逗号并存储在 SQLite 中,然后从 SQLite 中检索数据并用逗号显示

ios - 如何在 iOS 中支持/评估所有文件类型 (UTI)?

ios - corebluetooth 读取 RSSI 错误 :The operation was cancelled

php - 将图像从 iOS 上传到 PHP

objective-c - 如何管理 iPhone 5 的背景图像?

ios - 如何使用 iOS 6 sdk 为 iPhone 4s 或 iPhone 4 制作 View Controller ?

objective-c - UICollectionView 单元格中的对象/项目而不是索引?

iphone - viewWillLayoutSubviews 被调用的方式太多了