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我想创建一个应用程序，它将从HTTP请求中检索JSON数据，并将其作为JSON对象发送给应用程序主控制器，然后再从中进行进一步处理。iPhone JSON对象
我坚持的地方实际上是创建一个类，它将用作一个JSON类，在该类中将获取URL，获取数据并返回该对象。
独自一人，即使能够使这个类的工作，但我不能让类来存储我的主控制器的对象来检索它。
因为IM相当新的Objective-C的本身，我的想法是，即时通讯搞乱了我的init调用内：
-initWithURL:(NSString *) value
{
responseData = [[NSMutableData data] retain];
NSURL *theURL = [NSURL URLWithString:value];
NSURLRequest *request = [NSURLRequest requestWithURL:theURL];
[[NSURLConnection alloc] initWithRequest:request delegate:self];
return self;
}
JSON对象的处理发生在这里：
- (void)connectionDidFinishLoading:(NSURLConnection *)connection {
[connection release];
NSString *responseString = [[NSString alloc] initWithData:responseData encoding:NSUTF8StringEncoding];
[responseData release];
NSError *jsonError;
SBJSON *json = [[SBJSON new] autorelease];
NSDictionary *parsedJSON = [json objectWithString:responseString error:&jsonError];
// NSArray object.
listings = [parsedJSON objectForKey:@"posts"];
NSEnumerator *enumerator = [listings objectEnumerator];
NSDictionary* item;
// to prove that it does work.
while (item = (NSDictionary*)[enumerator nextObject]) {
NSLog(@"posts:id = %@", [item objectForKey:@"id"]);
NSLog(@"posts:address = %@", [item objectForKey:@"address"]);
NSLog(@"posts:lat = %@", [item objectForKey:@"lat"]);
NSLog(@"posts:lng = %@", [item objectForKey:@"lng"]);
}
[responseString release];
}
现在当调用主控制器内的对象时，我在viewDidLoad方法调用中使用了这一点代码：
- (void)viewDidLoad {
[super viewDidLoad];
JSON_model *jsonObj = [[JSON_model alloc] initWithURL:@"http://localhost/json/faith_json.php?user=1&format=json"];
NSEnumerator *enumerator = [[jsonObj listings] objectEnumerator];
NSDictionary* item;
//
while (item = (NSDictionary*)[enumerator nextObject]) {
NSLog(@"posts:id = %@", [item objectForKey:@"id"]);
NSLog(@"posts:address = %@", [item objectForKey:@"address"]);
NSLog(@"posts:lat = %@", [item objectForKey:@"lat"]);
NSLog(@"posts:lng = %@", [item objectForKey:@"lng"]);
}
}