iphone - 在iOS的JSON中使用未声明的标识符

我在json中使用了未声明的标识符错误，但是我正在按照http://blog.zachwaugh.com/post/309924609/how-to-use-json-in-cocoaobjective-c中的示例进行操作

我该如何解决？是的，我对objective-c \ ios非常陌生：）谢谢

我将此代码放在我的viewcontroller.m文件中基于视图的应用程序中

问题在于“ SBJSON * parser = [[SBJSON alloc] init];”

 // Create new SBJSON parser object
   SBJSON *parser = [[SBJSON alloc] init];

// Prepare URL request to download statuses from Twitter
NSURLRequest *request = [NSURLRequest requestWithURL:[NSURL URLWithString:@"http://twitter.com/statuses/public_timeline.json"]];

// Perform request and get JSON back as a NSData object
NSData *response = [NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil];

// Get JSON as a NSString from NSData response
NSString *json_string = [[NSString alloc] initWithData:response encoding:NSUTF8StringEncoding];

// parse the JSON response into an object
// Here we're using NSArray since we're parsing an array of JSON status objects
NSArray *statuses = [parser objectWithString:json_string error:nil];

// Each element in statuses is a single status
// represented as a NSDictionary
for (NSDictionary *status in statuses)
{
    // You can retrieve individual values using objectForKey on the status NSDictionary
    // This will print the tweet and username to the console
    NSLog(@"%@ - %@", [status objectForKey:@"text"], [[status objectForKey:@"user"] objectForKey:@"screen_name"]);
}

最佳答案

尝试将行更改为SBJsonParser *parser = [[SBJsonParser alloc] init];

编辑：

一种更简单的方法是使用SBJSON提供的NSString类别：

NSArray *statuses = (NSArray *)[json_string JSONValue];