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我在弹出窗口中有一个 UIPickerView 。我只是想在显示弹出框之前设置选择器视图的默认/起始值。这是我用来创建和显示弹出框的代码（在处理触摸按钮操作的方法中）：

- (IBAction) handleClickStepTimeButton: (id)sender
{
    UIViewController *timePickerController = [UIViewController alloc];

    UIPickerView *timePicker = [[UIPickerView alloc] initWithFrame:CGRectMake(0, 0, 100, 180)];
    timePicker.delegate = self;
    timePicker.showsSelectionIndicator = YES;
    [timePickerController.view addSubview:timePicker];

    UIPopoverController *timePickerPopoverController = [[UIPopoverController alloc] initWithContentViewController:timePickerController];
    timePickerPopoverController.popoverContentSize = CGSizeMake(100, 200);

    UIButton *stepTimeButton = (UIButton *)sender;

    [timePickerPopoverController presentPopoverFromRect:CGRectMake(stepTimeButton.frame.size.width, (stepTimeButton.frame.size.height / 2), 1, 1) inView:stepTimeButton permittedArrowDirections:UIPopoverArrowDirectionAny animated:YES];

    [timePicker selectRow:(currentStep.length - 1) inComponent:1 animated:YES];
}

每次单击此按钮时，我都不担心创建弹出框和 UIPickerView，因为它不会经常发生，并且 UIPickerView 的值只是少数整数。这是我用来将值添加到视图的代码：

- (NSInteger)pickerView:(UIPickerView *)pickerView numberOfRowsInComponent:(NSInteger)component {
    return 8;
}
- (NSString *)pickerView:(UIPickerView *)pickerView titleForRow:(NSInteger)rowId forComponent:(NSInteger)component {
    return [NSString stringWithFormat:@"%d", (rowId + 1)];
}

我的问题是，每次我去显示弹出窗口时，我都会得到一个 NSRangeException 就[timePicker selectRow...行了。这种对我来说很有意义，因为 UIPickerView 还没有显示。但是使用该逻辑......如何为 UIPickerView 设置默认/起始值？

我确定这里有一个简单的解决方案，但我只是没有看到它......

谢谢

4

1 回答 1

2

检查组件的数量是否有效currentStep.length - 1且不大于8。

你的主要问题是你没有设置dataSource选择器的。添加

timePicker.dataSource = self;
于 2011-07-15T15:25:32.993 回答