從另一種方法獲得價值

Question

我試圖使用我已經放在同一個類上的數據，但使用了不同的方法（我相信這是術語），但由於某些奇怪的原因，我似乎無法訪問這些數據。我可以這樣做，當我把它放在一起（在計時器中），但會消耗TONNE的內存，如果它不會反覆設置這些數據。無論如何，這是到目前爲止我的代碼：

/////////////////////////////////SET TIME/DAYS/////////////////// 
    currentTime = [NSDate date]; 
    NSDateFormatter *timeFormatter = [[NSDateFormatter alloc] init]; 
    [timeFormatter setTimeStyle:NSDateFormatterShortStyle]; 
    self.timeNow.text = [timeFormatter stringFromDate:currentTime]; 

    NSDate *lesson1Start = [timeFormatter dateFromString:@"8:40"]; 
    NSDate *lesson2Start = [timeFormatter dateFromString:@"9:35"]; 
    NSDate *recessStart = [timeFormatter dateFromString:@"10:30"]; 
    NSDate *caregroupStart = [timeFormatter dateFromString:@"10:50"]; 
    NSDate *lesson3Start = [timeFormatter dateFromString:@"11:00"]; 
    NSDate *lesson4Start = [timeFormatter dateFromString:@"11:55"]; 
    NSDate *lunchStart = [timeFormatter dateFromString:@"12:50"]; 
    NSDate *lesson5Start = [timeFormatter dateFromString:@"1:30"]; 
    NSDate *lesson6Start = [timeFormatter dateFromString:@"2:20"]; 
    NSDate *endDay = [timeFormatter dateFromString:@"3:10"]; 



    } 
    //////////////////////////END SET TIMES///////////////////////// 


- (void)updateTime { 

    NSString *mondayDay = (@"Monday"); 
    NSString *tuesdayDay = (@"Tuesday"); 
    NSString *wednesdayDay = (@"Wednesday"); 
    NSString *thursdayDay = (@"Thursday"); 
    NSString *fridayDay = (@"Friday"); 

    [updateTimer invalidate]; 
    updateTimer = nil; 

    currentTime = [NSDate date]; 
    NSDateFormatter *timeFormatter = [[NSDateFormatter alloc] init]; 
    [timeFormatter setTimeStyle:NSDateFormatterShortStyle]; 
    self.timeNow.text = [timeFormatter stringFromDate:currentTime]; 

    updateTimer = [NSTimer scheduledTimerWithTimeInterval:0.01 
          target:self 
          selector:@selector(updateTime) 
          userInfo:Nil repeats:YES]; 

    ///////Check day/time & update lesson///// 

    if (self.dayNow.text == mondayDay && self.timeNow.text == lesson1Start) { 
     NSLog(@""); 
    } 

正如你所看到的，我不能用「lesson1Start」，因爲它是不是在- (void)updateTime。

如果我將它捆綁在一起，它會一直反覆設置這些數據，因爲它被放置在計時器中，這樣會好嗎？

PS：我想檢查課程是否開始，如果時間是課程開始，那麼課程將出現。在這種情況下，我只有NSLog(@"")。

謝謝！

Answer 1

你可以這樣做... 讓我們考慮你的代碼是在類ClassRoom中;

@interface ClassRoom 
{ 
NSArray *scheduleArray; //instance variable 
} 

可以舉行//你必須分配和初始化它的實現文件

NSDate *lesson1Start = [timeFormatter dateFromString:@"8:40"]; 
NSDate *lesson2Start = [timeFormatter dateFromString:@"9:35"]; 
NSDate *recessStart = [timeFormatter dateFromString:@"10:30"]; 
NSDate *caregroupStart = [timeFormatter dateFromString:@"10:50"]; 
NSDate *lesson3Start = [timeFormatter dateFromString:@"11:00"]; 
NSDate *lesson4Start = [timeFormatter dateFromString:@"11:55"]; 
NSDate *lunchStart = [timeFormatter dateFromString:@"12:50"]; 
NSDate *lesson5Start = [timeFormatter dateFromString:@"1:30"]; 
NSDate *lesson6Start = [timeFormatter dateFromString:@"2:20"]; 
NSDate *endDay = [timeFormatter dateFromString:@"3:10"]; 

添加所有上述日期在scheduleArray創建NSArray *scheduleArray;。（希望你知道如何在NSArray中添加項目。）

現在你可以寫錄入爲

- (void)updateTime { 

     NSString *mondayDay = (@"Monday"); 
     NSString *tuesdayDay = (@"Tuesday"); 
     NSString *wednesdayDay = (@"Wednesday"); 
     NSString *thursdayDay = (@"Thursday"); 
     NSString *fridayDay = (@"Friday"); 

     [updateTimer invalidate]; 
     updateTimer = nil; 

     currentTime = [NSDate date]; 
     NSDateFormatter *timeFormatter = [[NSDateFormatter alloc] init]; 
     [timeFormatter setTimeStyle:NSDateFormatterShortStyle]; 
     self.timeNow.text = [timeFormatter stringFromDate:currentTime]; 

     updateTimer = [NSTimer scheduledTimerWithTimeInterval:0.01 
           target:self 
           selector:@selector(updateTime) 
           userInfo:Nil repeats:YES]; 

     ///////Check day/time & update lesson///// 

for(NSDate * date in scheduleArray){ 

    here you will get above saved dates use them to your need. 
} 

} 

希望能幫助到你 ！