小编典典

如何为iPhone应用程序发出HTTP请求以获取JSON对象作为响应?

json

我在服务器上运行了一个Web服务,该服务以XML格式或JSON格式返回数据。我想请求JSON格式,但使用HTTP Post方法。

任何帮助,不胜感激。

提前致谢。


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2020-07-27

共1个答案

小编典典

这是适用于JSON发布请求的代码,TouchJSON Framework用于解析JSON,谢谢“ schwa”。

NSArray *keys = [NSArray arrayWithObjects:@"username", @"password", @"preference", @"uid", nil];
NSArray *objects = [NSArray arrayWithObjects:@"accuser", @"accpass", @"abc_region", @"100", nil];
NSDictionary *theRequestDictionary = [NSDictionary dictionaryWithObjects:objects forKeys:keys];

NSURL *theURL = [NSURL URLWithString:@"http://url.com/request.php"];
NSMutableURLRequest *theRequest = [NSMutableURLRequest requestWithURL:theURL cachePolicy:NSURLRequestReloadIgnoringCacheData timeoutInterval:10.0f];
[theRequest setHTTPMethod:@"POST"];

[theRequest setValue:@"application/json-rpc" forHTTPHeaderField:@"Content-Type"];
NSString *theBodyString = [[CJSONSerializer serializer] serializeDictionary:theRequestDictionary];
NSLog(@"%@", theBodyString);
NSData *theBodyData = [theBodyString dataUsingEncoding:NSUTF8StringEncoding];
// NSLog(@"%@", theBodyData);
[theRequest setHTTPBody:theBodyData];

NSURLResponse *theResponse = NULL;
NSError *theError = NULL;
NSData *theResponseData = [NSURLConnection sendSynchronousRequest:theRequest returningResponse:&theResponse error:&theError];
NSString *theResponseString = [[[NSString alloc] initWithData:theResponseData encoding:NSUTF8StringEncoding] autorelease];
NSLog(theResponseString);
NSDictionary *theResponseDictionary = [[CJSONDeserializer deserializer] deserialize:theResponseString];
NSLog(@"%@", theResponseDictionary);
NSString *theGreeting = [theResponseDictionary objectForKey:@"greeting"];
[self setValue:theGreeting forKey:@"greeting"];
2020-07-27