Question

所以，这是交易。我的xib中有16个按钮，每个按钮都在其控制器中实例化，代码如下：

IBOutlet UIButton *left1;
IBOutlet UIButton *left2;
IBOutlet UIButton *left3;
IBOutlet UIButton *left4;
IBOutlet UIButton *left5;
IBOutlet UIButton *left6;
IBOutlet UIButton *left7;
IBOutlet UIButton *left8;
IBOutlet UIButton *right1;
IBOutlet UIButton *right2;
IBOutlet UIButton *right3;
IBOutlet UIButton *right4;
IBOutlet UIButton *right5;
IBOutlet UIButton *right6;
IBOutlet UIButton *right7;
IBOutlet UIButton *right8;

当按下其中一个按钮时，它应该根据按下的按钮按下具有对象的视图。我是这样做的：

if (upperLower.selectedSegmentIndex == 0) {
    if (sender == left1) {
        aTooth = [aPatient.teeth objectForKey:@"11"];
    } else if (sender == left2) {
        aTooth = [aPatient.teeth objectForKey:@"12"];
    } else if (sender == left3) {
        aTooth = [aPatient.teeth objectForKey:@"13"];
    } else if (sender == left4) {
        aTooth = [aPatient.teeth objectForKey:@"14"];
    } else if (sender == left5) {
        aTooth = [aPatient.teeth objectForKey:@"15"];
    } else if (sender == left6) {
        aTooth = [aPatient.teeth objectForKey:@"16"];
    } else if (sender == left7) {
        aTooth = [aPatient.teeth objectForKey:@"17"];
    } else if (sender == left8) {
        aTooth = [aPatient.teeth objectForKey:@"18"];
    } else if (sender == right1) {
        aTooth = [aPatient.teeth objectForKey:@"21"];
    } else if (sender == right2) {
        aTooth = [aPatient.teeth objectForKey:@"22"];
    } else if (sender == right3) {
        aTooth = [aPatient.teeth objectForKey:@"23"];
    } else if (sender == right4) {
        aTooth = [aPatient.teeth objectForKey:@"24"];
    } else if (sender == right5) {
        aTooth = [aPatient.teeth objectForKey:@"25"];
    } else if (sender == right6) {
        aTooth = [aPatient.teeth objectForKey:@"26"];
    } else if (sender == right7) {
        aTooth = [aPatient.teeth objectForKey:@"27"];
    } else if (sender == right8) {
        aTooth = [aPatient.teeth objectForKey:@"28"];
    }
} 
else {
    if (sender == left1) {
        aTooth = [aPatient.teeth objectForKey:@"41"];
    } else if (sender == left2) {
            aTooth = [aPatient.teeth objectForKey:@"42"];
    } else if (sender == left3) {
            aTooth = [aPatient.teeth objectForKey:@"43"];
    } else if (sender == left4) {
            aTooth = [aPatient.teeth objectForKey:@"44"];
    } else  if (sender == left5) {
            aTooth = [aPatient.teeth objectForKey:@"45"];
    } else  if (sender == left6) {
            aTooth = [aPatient.teeth objectForKey:@"46"];
    } else  if (sender == left7) {
            aTooth = [aPatient.teeth objectForKey:@"47"];
    } else  if (sender == left8) {
            aTooth = [aPatient.teeth objectForKey:@"48"];
    } else  if (sender == right1) {
            aTooth = [aPatient.teeth objectForKey:@"31"];
    } else  if (sender == right2) {
            aTooth = [aPatient.teeth objectForKey:@"32"];
    } else  if (sender == right3) {
            aTooth = [aPatient.teeth objectForKey:@"33"];
    } else  if (sender == right4) {
            aTooth = [aPatient.teeth objectForKey:@"34"];
    } else  if (sender == right5) {
            aTooth = [aPatient.teeth objectForKey:@"35"];
    } else  if (sender == right6) {
            aTooth = [aPatient.teeth objectForKey:@"36"];
    } else  if (sender == right7) {
            aTooth = [aPatient.teeth objectForKey:@"37"];
    } else  if (sender == right8) {
            aTooth = [aPatient.teeth objectForKey:@"38"];
    }
}
toothController.aTooth = aTooth;
toothController.aPatient = aPatient;

if (aTooth) {
    [self.navigationController pushViewController:toothController animated:YES];    
} else {
    UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Tooth not found" message:@"The patient doesn't have that tooth" delegate:self cancelButtonTitle:@"Ok" otherButtonTitles:nil];
    [alert show];
    [alert release];
}
[aTooth release];

现在问题是我无法弄清楚为什么如果我按下left1，left2或left3它完美无缺。现在，如果我按下left4它会崩溃，有时会有异常（NSCFArray号码无法识别的选择器发送blalbal），有时没有它。编号为11,12,13和14的所有齿都存在于阵列中。我只尝试过那些因为我太懒而无法创造36个。但是为什么它只能用于14号呢？

我一直在努力解决这个问题几个小时失败。任何有帮助的帮助。

更新:(最后一次更新错误）我从解析牙齿的xml中删除了牙齿编号14，并添加了牙齿编号15.问题仍然存在于15号牙齿并显示正确的信息（牙齿没有存在）牙齿数14。

更新2：在[aTooth发布]之后我添加了“aTooth = nil”，现在我可以进入15号牙齿，但是我无法接触任何牙齿两次。一旦我找到了一颗牙齿，我就不能再回头去了。

Answer 1

objectForKey:是一种NSDictionary方法。您收到的异常表明teeth是NSArray。我相信你想要使用

aTooth = [aPatient.teeth objectAtIndex:some-index];

Answer 2

[aTooth release] - ???所以你在从字典中检索它后发布了对象？

试图用对象pushView的奇怪情况

2 个答案: