此处为:
- (IBAction)reset:(id) sender {
if ((boxHide1.hidden = YES) && (boxHide2.hidden = YES)) {
resetHide.hidden = NO;
iii
iii
我如何能够这样做? 我的发言共需要12个项目。 感谢!
此处为:
- (IBAction)reset:(id) sender {
if ((boxHide1.hidden = YES) && (boxHide2.hidden = YES)) {
resetHide.hidden = NO;
iii
iii
我如何能够这样做? 我的发言共需要12个项目。 感谢!
你可以使用所张贴的解决办法@Joe,但你可以把藏匿/编码的密码告诉大家,这种说法可能会变得非常大,难以读懂。
如果你想要使你的法典清白并易于理解/维持,我就把所有这些纽芬兰·阿雷拉放在一个安全的国家里,并通过它来决定你是否希望显示重新定点。
BOOL showResetButton = YES;
for (UIButton *button in buttonsArray)
{
if (button.hidden == NO) // If any of the buttons is not hidden do not show the reset button
showResetButton = NO;
}
resetButton.hidden = showResetButton;
确保你使用<条码>=条码>来比较数值,但由于这些数值已经过时,无需与<条码>YES代码”进行比较。 (&&
) 这样做是正确的,如果存在任何OR( >>>/code>)业务,则你可以放弃括号,否则,你将需要将适当的业务分类。
if (boxHide1.hidden &&
boxHide2.hidden &&
... &&
boxHide12.hidden)
{
resetHide.hidden = NO;
}
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