lantianyiming 2013-01-17 08:24 采纳率: 0%
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在lable中显示string时应用崩溃

在用NSString更新UILabel时应用崩溃了,在console中显示同样的NSString就没问题:

-(void)connectionDidFinishLoading:(NSURLConnection *)connection
{
    if (self.connectionData)
    {
        NSError *error;
        self.dict = [NSJSONSerialization JSONObjectWithData:self.connectionData options:kNilOptions error:&error];
        self.matchesArray = self.dict[@"matches"];

        NSString *title = [self.matchesArray valueForKey:@"title"];
        NSLog(@"NSString TITLE contains: %@", title);
        self.titleLabel.text = title;
    }
}

CONSOLE OUTPUT:

2013-01-16 13:54:08.550 ZEITreisen[3168:c07] NSString TITLE contains: (
    "Mark und Dollar"
)
2013-01-16 13:54:08.552 ZEITreisen[3168:c07] -[__NSArrayI isEqualToString:]: unrecognized selector sent to instance 0xde93850
(lldb) 
  • 写回答

2条回答 默认 最新

  • g989_1314125 2013-01-17 08:45
    关注

    试试:

    -(void)connectionDidFinishLoading:(NSURLConnection *)connection
    {
        if (self.connectionData)
        {
            NSError *error;
            self.dict = [NSJSONSerialization JSONObjectWithData:self.connectionData options:kNilOptions error:&error];
            self.matchesArray = self.dict[@"matches"];
    
            NSString *title = [self.matchesArray valueForKey:@"title"];
            NSLog(@"NSString TITLE contains: %@", title);
            self.titleLabel.text = [NSString stringWithFormat:@"%@",[title objectAtIndex:0]];
        }
    }
    
    评论

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