小编Chr*_*nen的帖子

目前我正在编写一个应用程序(目标iOS 6,启用了ARC),它使用JSON进行数据传输,使用Core Data进行持久存储.JSON数据由PHP脚本通过json_encode从MySQL数据库生成.

我的问题是,对于某些表中的数据,以下代码失败:

- (NSDictionary *)executeFetch:(NSString *)query
{
    NSURL *requesturl = [NSURL URLWithString:[query stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];

    NSError *dataError = nil;
    self.jsonData = [NSData dataWithContentsOfURL:requesturl options:kNilOptions error:&dataError];

    NSError *error = nil;
    self.jsonSerializationResult = [NSJSONSerialization JSONObjectWithData:self.jsonData options:NSJSONReadingMutableContainers|NSJSONReadingMutableLeaves error:&error];

    return self.jsonSerializationResult;

}

该程序总是在EXC_BAD_ACCESS错误的位置崩溃,它表示self.jsonSerializationResult和Instruments说有一个Zombie被检测到.我知道这意味着我发送消息的一些对象是零,但我无法找到如何解决它...这就是乐器所说的:

#   Address Category    Event Type  RefCt   Timestamp   Size    Responsible Library Responsible Caller
0   0xa1b8a70   CFString (mutable)  Malloc  1   00:01.603.081   32  Foundation  -[NSPlaceholderMutableString initWithBytesNoCopy:length:encoding:freeWhenDone:]
1   0xa1b8a70   CFString (mutable)  Release 0   00:01.603.137   0   Foundation  newJSONValue
2   0xa1b8a70   CFString (mutable)  Zombie  -1  00:01.603.259   0 …