Question

这是我的标题部分：

@interface RootViewController : UIViewController
{
    NSString *status_id;
}

在控制器文件中，我正在分配变量：

- (void)updateStatus
{
    NSURL *url = [NSURL URLWithString:@"http://localhost/RightNow/API/status.json"];
    ASIHTTPRequest *request = [ASIHTTPRequest requestWithURL:url];
    [request startSynchronous];
    NSError *error = [request error];
    NSString *response = [NSString alloc];

    NSError *error2;
    NSData* data = [response dataUsingEncoding:NSUTF8StringEncoding];
    NSMutableDictionary *json = [NSJSONSerialization JSONObjectWithData:data      options:kNilOptions error:&error2];
    status_id = [json objectForKey:@"id"];
}

现在，当我再次尝试使用status_id时，我收到错误

- (IBAction)likeClick:(id)sender
{
    NSURL *url = [NSURL URLWithString:@"http://localhost/RightNow/API/vote"];
    ASIFormDataRequest *request = [ASIFormDataRequest requestWithURL:url];
    [request setRequestMethod:@"POST"];
    [request setPostValue:status_id forKey:@"id"]; //The error comes here
    [request setPostValue:@"like" forKey:@"vote"];
    [request startSynchronous];
}

抱歉我的英语不好。请帮帮我，谢谢！

Answer 1

[json objectForKey：@“id”];将返回autorelease池中的对象。您需要向其发送复制消息，如

status_id = [[json objectForKey:@"id"] copy];

并在适当时释放它（如果不使用ARC）

使用stringWithFormat时出现EXC_BAD_ACCESS错误

1 个答案: