我已经阅读了与TouchJSON序列化相关的问题和答案,但我仍然没有让它开始工作。
我用示例数据创建了一个NSDictionary,并使用JSONTouch序列化程序将NSDictionary转换为JSON。但是,当我记录NSData对象'theJSONData‘时,结果是这样的:
<7b223131 31353535 34343434 223a2250 . 65227d>
此外,当我将这个'theJSONData‘数据发送到web服务(这是期望的JSON)时,这就是我得到的:
2011-07-31 18:48:46.572街灯7169:207序列化错误:(null)
2011-07-31 18:48:46.804街灯7169:207 returnData:(空)
2011-07-31 18:48:46.805街灯7169:207错误:错误Domain=kJSONScannerErrorDomain代码=-201“无法扫描数组。数组没有由'[‘字符启动。” UserInfo=0x4d51ab0 {snippet=!HERE>!xml version="1.0“、location=0、NSLocalizedDescription=Could not扫描数组。数组未由“[”字符.、character=0、line=0}启动。
我做错什么了?在我将JSON NSData对象'theJSONData‘发送到web服务之前,是否需要将它转换为另一个类型?我又错过了一步吗?
// Create the dictionary
NSDictionary *outage = [[NSDictionary alloc] initWithObjectsAndKeys:
@"YCoord", @"12678967.543233",
@"XCoord", @"12678967.543233",
@"StreetLightID", @"666",
@"StreetLightCondition", @"Let's just say 'BAD'",
@"PhoneNumber", @"1115554444",
@"LastName", @"Smith",
@"Image",@"",
@"FirstName", @"Dawn",
@"Comments", @"Pole knocked down",
nil];
NSError *error = NULL;
// Serialize the data
NSData *theJSONData = [[CJSONSerializer serializer] serializeDictionary:outage error:&error];
NSLog(@"theJSONData: %@", theJSONData);
NSLog(@"Serialization Error: %@", error);
// Set up the request and send it
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL: [NSURL URLWithString: @"http://24.52.35.127:81/StreetLight/StreetlightService/CreateStreetLightOutage"]];
[request setHTTPMethod: @"POST"];
[request setHTTPBody: theJSONData];
// Deserialize the response
NSData *returnData = [ NSURLConnection sendSynchronousRequest: request returningResponse: nil error:&error];
NSString *returnString = [[NSString alloc] initWithData:returnData encoding: NSUTF8StringEncoding];
NSData *theReturnData = [returnString dataUsingEncoding:NSUTF8StringEncoding];
id theObject = [[CJSONDeserializer deserializer] deserializeAsArray:theReturnData error:&error];
NSLog(@"returnData: %@",theObject);
NSLog(@"Error: %@", error);发布于 2011-08-06 21:49:03
谢谢大家的帮助。最后,我使用Fiddler跟踪需要用JSON发送到服务的内容,然后发现我没有正确地格式化标头。这是最后为我工作的代码。
// Create the NSDictionary
NSDictionary *outage = [[NSDictionary alloc] initWithObjectsAndKeys:
@"12.543233",@"YCoord",
@"12.543233",@"XCoord",
@"111",@"StreetLightID",
@"Dented pole",@"StreetLightCondition",
@"1115554444",@"PhoneNumber",
@"Black",@"LastName",
[NSNull null],@"Image",
@"White",@"FirstName",
@"Hit by a car",@"Comments",
nil];
// Serialize the data
NSError *error = NULL;
NSData *theJSONData = [[CJSONSerializer serializer] serializeDictionary:outage error:&error];
NSLog(@"Serialization Error: %@", error);
// Change the data back to a string
NSString* theStringObject = [[NSString alloc] initWithData:theJSONData encoding:NSUTF8StringEncoding];
// Determine the length of the data
NSData *requestData = [NSData dataWithBytes: [theStringObject UTF8String] length: [theStringObject length]];
NSString* requestDataLengthString = [[NSString alloc] initWithFormat:@"%d", [requestData length]];
// Create request to send to web service
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL: [NSURL URLWithString: @"http://11.22.33.444:55/StreetLight/StreetlightService/CreateStreetLightOutage"]];
[request setHTTPMethod:@"POST"];
[request setHTTPBody:requestData];
[request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[request setValue:requestDataLengthString forHTTPHeaderField:@"Content-Length"];
[request setTimeoutInterval:30.0];
// Deserialize the response
NSData *returnData = [ NSURLConnection sendSynchronousRequest: request returningResponse: nil error:&error];
NSString *returnString = [[NSString alloc] initWithData:returnData encoding: NSUTF8StringEncoding];
NSData *theReturnData = [returnString dataUsingEncoding:NSUTF8StringEncoding];
id theObject = [[CJSONDeserializer deserializer] deserializeAsArray:theReturnData error:&error];
NSLog(@"returnData: %@",returnString);
NSLog(@"Error: %@", error);发布于 2011-08-01 02:26:05
首先,您已经在NSDictionary中反转了对象和键。
我对TouchJSON的了解还不足以帮助完成这部分代码。
发布于 2011-08-03 01:51:03
在解析一个很好的v3 api响应时,我遇到了类似的问题。离解决我的问题还差得很远,但我发现,如果您使用的是deserializerAsArray,那么JSON响应必须包含在"“和”deserializerAsDictionary“中,如果您是deserializerAsDictionary,那么JSON响应必须包含在"{”和"}“中。
由于google响应采用"{“}”格式,所以我需要使用v3方法。
我怀疑您已经知道这一点,但是在查看了Jonathan的代码之后,我已经解决了自己的问题,因为Jonathan的代码在解析JSON响应时专门检查上面的内容。
谢谢,
时间
https://stackoverflow.com/questions/6893665
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