NSMutableArray 和 NSObject 问题

Posted 2023-02-26

【中文标题】NSMutableArray 和 NSObject 问题

【英文标题】：NSMutableArray and NSObject Issue

【发布时间】：2011-05-28 10:01:45

【问题描述】：

Checkin *checkinsA = [[Checkin alloc] init]; 

NSDictionary *decodedJson = 结果； NSArray *users = [decodedJson objectForKey:@"data"]; Checkin *test = [[Checkin alloc] init];

for(NSDictionary *user in users) 
    NSLog(@"Created item: %@ \n", [user objectForKey:@"created_time"]);
     checkinsA.time = [NSString stringWithFormat:@"%@",[user objectForKey:@"created_time"]];
    NSDictionary *fromData = [user objectForKey:@"from"];
     NSLog(@"user id is: %@ \n", [fromData objectForKey:@"id"]);
      checkinsA.profID = [fromData objectForKey:@"id"];

     NSLog(@"user name is: %@\n ", [fromData objectForKey:@"name"]);
      checkinsA.name =[fromData objectForKey:@"name"];

    NSDictionary *placeData = [user objectForKey:@"place"];
    NSDictionary *locationData = [placeData objectForKey:@"location"];
     NSLog(@"City: %@ \n", [locationData objectForKey:@"city"]);
      checkinsA.city = [locationData objectForKey:@"city"];

     NSLog(@"Country: %@ \n", [locationData objectForKey:@"country"]);
      checkinsA.country = [locationData objectForKey:@"country"];

     NSLog(@"Latitude: %@ \n", [locationData objectForKey:@"latitude"]);
     checkinsA.lat = [locationData objectForKey:@"latitude"];

     NSLog(@"Longitude: %@ \n", [locationData objectForKey:@"longitude"]);
     checkinsA.lon = [locationData objectForKey:@"longitude"];

     NSLog(@"Place name: %@ \n", [placeData objectForKey:@"name"]);
     checkinsA.place = [placeData objectForKey:@"name"];

    NSDictionary *tagData = [user objectForKey:@"tags"];
    NSArray *tagDataArray = [tagData objectForKey:@"data"];
    for(NSDictionary *tagData2 in tagDataArray)
      NSLog(@"tagged user id is: %@ \n", [tagData2 objectForKey:@"id"]);
        [checkinsA.taggedID addObject:[tagData2 objectForKey:@"id"]];
      NSLog(@"tagged user name is: %@\n ", [tagData2 objectForKey:@"name"]);
         [checkinsA.taggedName addObject:[tagData2 objectForKey:@"name"]];
    

    [checkinArray addObject:checkinsA];
    test = [checkinArray objectAtIndex:count2];
    NSLog(@"Check array: %@",test.name);
    count2++;

for(int i=0;i<count2;i++)

    test = [checkinArray objectAtIndex:i];
    NSLog(@"%@",test.name);

我的 NSobject 是 CheckinsA。我正在使用 NSDictionary 从响应字符串中收集数据。我在这里要做的是将每个 CheckinsA 保存在 checkinArray 中。这看起来不错：

`[checkinArray addObject:checkinsA];

test = [checkinArray objectAtIndex:count2];

NSLog(@"Check array: %@",test.name);`

并且正在打印正确的值。但是当循环完成并且我尝试使用循环打印数组中的值时，只打印出最后一个 CheckinsA!!我尝试用以下方式打印它：

test = [checkinArray objectAtIndex:0];
NSLog(@"%@",test.name);
test = [checkinArray objectAtIndex:1];
NSLog(@"%@",test.name);

（以防我的循环是问题，但它仍然只打印最后一个 CheckinsA）。 那么为什么我的整个数组只填充了最后一个 CheckinsA？好奇怪！

【参考方案1】：

您每次都将相同的对象添加到数组中。像这样移动代码：

Checkin *checkinsA
for(NSDictionary *user in users) 
    checkinsA = [[Checkin alloc] init];
    // loop code
    [checkinArray addObject:checkinsA];
    [checkinsA release];

【讨论】：

感谢 vakio，但 Deepak 先回答，所以我必须将正确答案归功于他。

【参考方案2】：

每次循环时都需要初始化一个新对象。否则，您将更改同一个对象。

将Checkin *checkinsA = [[Checkin alloc] init]; 设为循环中的第一行，在完成将checkinsA 添加到数组后的最后，执行[checkinsA release];

【讨论】：

这个怎么样？测试 = [checkinArray objectAtIndex:count2];?还是一样？

太棒了！！现在好了！只需等待 7 分钟即可将您的答案标记为正确的答案！！谢谢。

【参考方案3】：

通过检查您的数组 NSLog(@"%@",[checkinArray 描述]);