[英]iphone sdk conditional in switch function
我正在嘗試使隨機圖像出現在按鈕上。 因此,它生成一個隨機數,並且切換算法將所選圖像與imgview中的圖像交換。 但我想在設置應用中進行切換,以切換要使用的圖像集。 我非常知道該怎么做...只是不起作用。 我缺少一些語法的東西...
int Number = rand() %30;
NSString *toggleValue = [[NSUserDefaults standardUserDefaults] stringForKey:@"enabled_preference"];
switch (Number) {
if (*toggleValue == 0) {
case 0:
picture.image = [UIImage imageNamed:@"1.png"];
break;
case 1:
picture.image = [UIImage imageNamed:@"2.png"];
break;}
else {
case 0:
picture.image = [UIImage imageNamed:@"3.png"];
break;
case 1:
picture.image = [UIImage imageNamed:@"4.png"];
break;}
}
NSString *toggleValue = [[NSUserDefaults standardUserDefaults] stringForKey:@"enabled_preference"];
NSArray *imagesA = [NSArray arrayWithObjects:@"img1.png" , @"img2.png" , ... , nil];
NSArray *imagesB = [NSArray arrayWithObjects:@"img8.png" , @"img9.png" , ... , nil];
NSArray *images = [toggleValue integerValue] ? imagesA : imagesB;
NSString *name = [images objectAtIndex:rand() % [images count]];
picture.image = [UIImage imageNamed:name];
您不能將if放入這樣的開關中,而是嘗試使用以下語法:
if (*toggleValue == 0)
{
switch (Number)
{
case 0:picture.image = [UIImage imageNamed:@"1.png"]; break;
case 1:picture.image = [UIImage imageNamed:@"2.png"];break;
}
}
else
{
switch (Number)
{
case 0:picture.image = [UIImage imageNamed:@"3.png"];break;
case 1:picture.image = [UIImage imageNamed:@"4.png"];break;
}
}
int randomInt = rand() % 30;
if (toggleEnabled) {
switch (randomInt) {
case 0:
picture.image = [UIImage imageNamed:@"0.toggleEnabled.png"];
break;
case 1:
picture.image = [UIImage imageNamed:@"1.toggleEnabled.png"];
break;
// ...
}
else {
switch (randomInt) {
case 0:
picture.image = [UIImage imageNamed:@"0.toggleDisabled.png"];
break;
// ...
}
}
我在想您正在尋找一種僅使用數字打開圖像的方法。
在這種情況下,代碼如下:
NSString *theImage = [[NSBundle mainBundle] pathForResource:[NSString stringWithFormat:@"%d.toggleEnabled", sliceIndex] ofType:@"png"]; picture.image = [[UIImage alloc] initWithContentsOfFile: theImage];
如果您以不同的方式命名圖像,則也可以讓它處理切換。
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