在 iOS 中创建 JsonString

Question

我是 iOS 新手。我像这样创建了JSON NSDictionary：

NSArray *keys = [NSArray arrayWithObjects:@"User", @"Password", nil];
NSArray *objects = [NSArray arrayWithObjects:@"Ali", @"2020", nil];
NSDictionary *jsonDictionary = [NSDictionary dictionaryWithObjects:objects forKeys:keys];

然后我可以通过两种机制将其转换为NSString：

1)

NSError *error; 
NSData *jsonData = [NSJSONSerialization dataWithJSONObject:jsonDictionary options:0 error:&error];

NSString *jsonString = nil;
if (! jsonData) {
     NSLog(@"Got an error: %@", error);
} else {
     jsonString = [[NSString alloc] initWithData:jsonData encoding:NSUTF8StringEncoding];
}

2)

NSString *jsonString = [jsonDictionary JSONRepresentation];

第二种方式我得到这个warning：

Instance method '-JSONRepresentation' not found (return type defaults to 'id')

但是当我运行项目时，两种机制都可以正常工作：

NSLog(@"Val of json parse obj is %@",jsonString); 

您知道如何以第二种方式删除警告吗？

我的主要目标是 POST 使用 RESTful Web 服务将此 json 字符串发送到外部数据库。 基本上考虑到我的主要目标，哪种方式更好？

Answer 1

您应该使用NSJSONSerialization，因为它速度更快，并且直接与 iOS SDK 一起提供，只要您的“目标受众”是 iOS5+

要将数据发布到您的网络服务，您需要按照这些方式创建一个请求...

NSDictionary * postDictionary = [NSDictionary dictionaryWithObjects:[NSArray arrayWithObjects:@"value1", @"value2", nil]
                                                                  forKeys:[NSArray arrayWithObjects:@"key1", @"key2", nil]];

NSError * error = nil;
NSData * jsonData = [NSJSONSerialization dataWithJSONObject:postDictionary options:NSJSONReadingMutableContainers error:&error];

NSMutableURLRequest * urlRequest = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:@"your_webservice_post_url"]];
[urlRequest setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[urlRequest setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[urlRequest setHTTPMethod:@"POST"];
[urlRequest setHTTPBody:jsonData];

NSURLConnection * myConnection = [[NSURLConnection alloc] initWithRequest:urlRequest delegate:self startImmediately:YES];

请阅读NSURLConnectionDelegate 协议。

Answer 2

对于 iOS 5.0 >：

像这样使用NSJSONSerialization：

NSError *error;
NSData *jsonData = [NSJSONSerialization dataWithJSONObject:dictionary options:NSJSONWritingPrettyPrinted error:&error];
NSString *resultAsString = [[NSString alloc] initWithData:jsonData encoding:NSUTF8StringEncoding];
 NSLog(@"jsonData as string:\n%@ Error:%@", resultAsString,error);

对于 ：

使用json-framework第三方库，使用NSDictionary的类别来提供json字符串：

 NSString *jsonString = [dictionary JSONRepresentation];

 //with options
 NSString *jsonString = [dictionary JSONStringWithOptions:JKSerializeOptionNone error:nil]

Answer 3

这会帮助你...Convert NSDictionary to JSON with SBJson

Answer 4

用这个方法希望对你有帮助

 NSArray *keys = [NSArray arrayWithObjects:@"User", @"Password", nil];
 NSArray *objects = [NSArray arrayWithObjects:@"Ali", @"2020", nil];
 NSDictionary *jsonDictionary = [NSDictionary dictionaryWithObjects:objects forKeys:keys];

 NSError *error = nil;
 // NSError *error; 
 NSData *jsonData = [NSJSONSerialization dataWithJSONObject:jsonDictionary options:0 error:&error];


 id result = [NSJSONSerialization JSONObjectWithData:jsonData options:NSJSONReadingMutableContainers error:&error];

 NSLog(@"\n\n\n id for json==%@ \n\n\n\n\n",result);

Answer 5

JSON DEFAULT METHOD......

+(NSDictionary *)stringWithUrl:(NSURL *)url postData:(NSData *)postData httpMethod:(NSString *)method { 

NSDictionary *returnResponse=[[NSDictionary alloc]init];

@try {
 NSMutableURLRequest *urlRequest = [NSMutableURLRequest requestWithURL:url cachePolicy:NSURLRequestReloadIgnoringCacheData timeoutInterval:180]; [urlRequest setHTTPMethod:method];

if(postData != nil)
{
    [urlRequest setHTTPBody:postData];
}

[urlRequest setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[urlRequest setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[urlRequest setValue:@"text/html" forHTTPHeaderField:@"Accept"];

NSData *urlData;
NSURLResponse *response;
NSError *error;
urlData = [NSURLConnection sendSynchronousRequest:urlRequest
                                returningResponse:&response
                                            error:&error];
returnResponse = [NSJSONSerialization
                  JSONObjectWithData:urlData
                  options:kNilOptions
                  error:&error];
} @catch (NSException *exception) { returnResponse=nil; } @finally { return returnResponse; } }

Return Method :

+(NSDictionary )methodName:(NSString)string { 
NSDictionary *returnResponse; NSData *postData = [NSData dataWithBytes:[string UTF8String] length:[string length]]; NSString *urlString = @"https//:..url...."; returnResponse=[self stringWithUrl:[NSURL URLWithString:urlString] postData:postData httpMethod:@"POST"];
return returnResponse; 
}