Question

做我正在做的事情的最佳方式是什么？ goodThingX，badThingX和nothin是UILabels。

NSString *todayNothing = [[todayArray objectAtIndex:0] objectForKey: @"nothing"];
if (todayNothing!=NULL) {
    goodThing1.hidden = YES;
    goodThing2.hidden = YES;
    goodThing3.hidden = YES;
    badThing1.hidden = YES;
    badThing2.hidden = YES;
    badThing3.hidden = YES;
    nothing.text = todayNothing;
    nothing.hidden = NO;
} else {
    goodThing1.hidden = NO;
    goodThing2.hidden = NO;
    goodThing3.hidden = NO;
    badThing1.hidden = NO;
    badThing2.hidden = NO;
    badThing3.hidden = NO;
    nothing.hidden = YES;
}

即。当todayNothing有一些文字时，我想隐藏6个标签，并显示nothing标签，否则相反。我没有打算优化这个，但可能会有比6更多的标签..

Answer 1

您可以将它们全部放在init或awakeFromNib或类似的数组中，以便以后进行迭代。

@class MyThing {

   UIView *theThings[NUM_THINGS]; // ...or use an NSArray if you like that

}
@end

- (id)init // maybe awakeFromNib would be a better place for UI elements
{
  self = [super init];
  if (self) {
    theThings[0] = goodThing1;
    theThings[1] = goodThing2;
    theThings[2] = goodThing3;
    :
  }
  return self;
}

...然后你会使用像这样的循环

for (int i=0; i<NUM_THINGS; i++)
  theThings[i].hidden = YES;

Answer 2

首先，你有一个叫做“布尔变量”的东西。

NSString *todayNothing = [[todayArray objectAtIndex:0] objectForKey: @"nothing"];
BOOL todayNothing_has_something = (todayNothing!=nil); // YES if todayNothing!=nil, NO otherwise
goodThing1.hidden = todayNothing_has_something;
goodThing2.hidden = todayNothing_has_something;
goodThing3.hidden = todayNothing_has_something;
badThing1.hidden = todayNothing_has_something;
badThing2.hidden = todayNothing_has_something;
badThing3.hidden = todayNothing_has_something;
if (todayNothing)
  nothing.text = todayNothing;
nothing.hidden = ! todayNothing_has_something;

其次，最好使用数组或NSArray来存储所有good - 和badThing s。（见epatel的答案。）

迭代对象操作或一次更改多个对象

2 个答案: