我正在解析JSON
并将value.Store
值获取为NSString
,并且当我想在UILabel
上显示时,应用崩溃
unrecognized selector sent to instance 0x7b070470`
NSString *idvalue = [jsonResponse objectForKey:@"id"];`
labl_id.text =idvalue;
NSLog(@"id value%@",idvalue);
output: id value 6
请让我知道如何解决此问题
谢谢
最佳答案
mb你得到NSNumber
NSString *idvalue = [[jsonResponse valueForKey:@"id"] stringValue];