Question

这是我的代码：

NSMutableArray *ratings = [[NSMutableArray alloc] init];
    NSMutableDictionary *eachRating = [[NSMutableDictionary alloc] init];

    for (UIView *subview in self.rateServiceView.subviews) {

        if ([subview isKindOfClass:[RSTapRateView class]]) {

            RSTapRateView *rs = (RSTapRateView *)subview;
            [eachRating setObject:rs.rateId forKey:@"iRatingId"];
            [eachRating setObject:[NSNumber numberWithInt:rs.rating] forKey:@"iRate"];

            [ratings addObject:eachRating];

        }

    }

而不是获取这些值：

{
        iRate = 1;
        iRatingId = 1;
    },
        {
        iRate = 5;
        iRatingId = 2;
    },
        {
        iRate = 2;
        iRatingId = 3;
    }

我得到了这些价值观：

{
        iRate = 2;
        iRatingId = 3;
    },
        {
        iRate = 2;
        iRatingId = 3;
    },
        {
        iRate = 2;
        iRatingId = 3;
    }

当我记录每次迭代的结果时，最后一个对象替换所有现有对象并为自己添加一个新对象。

Answer 1

移动这一行：

    NSMutableDictionary *eachRating = [[NSMutableDictionary alloc] init];

下面的这一行：

    for (UIView *subview in self.rateServiceView.subviews) {

这样，您就可以创建一个新的“eachRating”字典，并将其添加到“ratings”数组中。

Answer 2

是的，这是因为您要为同一个键分配不同的值，因此新值将替换该键的旧值。

因此请将代码更改为：

  NSMutableArray *ratings = [[NSMutableArray alloc] init];
    for (UIView *subview in self.rateServiceView.subviews){

        if ([subview isKindOfClass:[RSTapRateView class]]) {

            NSMutableDictionary *eachRating = [[NSMutableDictionary alloc] init];


            RSTapRateView *rs = (RSTapRateView *)subview;
            [eachRating setObject:rs.rateId forKey:@"iRatingId"];
            [eachRating setObject:[NSNumber numberWithInt:rs.rating] forKey:@"iRate"];

            [ratings addObject:eachRating];

        }

    }

Answer 3

如果你不需要在这个循环之后进一步改变单个词典，你可以通过这样编写来获得更加现代和紧凑：

NSMutableArray *ratings = [[NSMutableArray alloc] init];

for (UIView *subview in self.rateServiceView.subviews) {

    if ([subview isKindOfClass:[RSTapRateView class]]) {

        RSTapRateView *rs = (RSTapRateView *)subview;
        [ratings addObject:@{@"iRatingId":rs.rateId, @"iRate":@(rs.rating)}];

    }

}

对象替换了数组中的现有对象而不是添加

3 个答案: