本文介绍了iOS解析数据时出错的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
在我的应用中,我使用JSON解析数据
In my app, I am parsing the data using JSON
NSString * urlString=[NSString stringWithFormat:@"http://[email protected]&latitude=59.34324&longitude=23.359257"];
NSURL * url=[NSURL URLWithString:urlString];
NSMutableURLRequest * request=[NSMutableURLRequest requestWithURL:url];
NSError * error;
NSURLResponse * response;
NSData *data=[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
NSString * outputData=[[NSString alloc]initWithData:data encoding:NSASCIIStringEncoding];
NSLog(@"%@",outputData);
SBJsonParser *jsonParser = [SBJsonParser new];
NSDictionary *jsonData = (NSDictionary *) [jsonParser objectWithString:outputData error:nil];
NSLog(@"%@",jsonData);
NSInteger success = [(NSNumber *) [jsonData objectForKey:@"success"] integerValue];
执行此代码后,在我的日志中将其打印为
After this code executes, In my log it is printed as
({
latitude = "0.000000000000000";
longitude = "0.000000000000000";
username = sunil;
},
{
latitude = "80.000000000000000";
longitude = "30.000000000000000";
username = arun;
})
但是在运行时,该应用程序崩溃了,如
But while running, the app crashes, as
'NSInvalidArgumentException', reason: '-[__NSArrayM objectForKey:]: unrecognized selector sent to instance 0x910d8d0'
推荐答案
我认为您的问题是,jsonParser objectWithString
返回一个包含字典的数组,而不是字典本身.
I think your problem is, that jsonParser objectWithString
returns an array with dictionaries in it not dictionaries itself.
请尝试以下操作:
NSArray *jsonData = (NSArray *) [jsonParser objectWithString:outputData error:nil];
for(NSDictionary *dict in jsonData) {
NSLog(@"%@",dict);
}
这对您有用吗?
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