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在我的 iphone 应用程序上创建登录视图时，我遇到了这个错误：

'NSInvalidArgumentException', reason: '-[SBJsonParser objectWithString:error:]:
    unrecognized selector sent to instance 0x6695330'

它来自这条线：

NSDictionary *results = [parser objectWithString:json_string error:nil];

从这个方法：

+ (BOOL)loginWithUsername:(NSString *)username password:(NSString *)password
{
    NSString *urlString = [NSString stringWithFormat:@"%@login3", ROSE_ROOT_URL];
    NSURL *url = [NSURL URLWithString:urlString];


    NSString *requestString = [NSString stringWithFormat:@"&mobile=1&username=%@&password=%@", username, password];


    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:url];
    [request setHTTPMethod:@"POST"];
    [request setHTTPBody:[NSData dataWithBytes:[requestString UTF8String] length:[requestString length]]];

    NSData *response = [NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil];
    NSString *json_string = [[NSString alloc] initWithData:response encoding:NSUTF8StringEncoding];

    [request release];
    // parse the JSON response into an object
    // Here we're using NSArray since we're parsing an array of JSON product objects
    SBJsonParser *parser = [[SBJsonParser alloc] init];
    NSDictionary *results = [parser objectWithString:json_string error:nil];

    [json_string release];
    [parser release];


    //  NSDictionary *results = [RoseFetcher fetch:request]; 
    //  [request release];

    if ([[results objectForKey:@"password"] intValue] == 0)
        return NO;

    return YES;
}

任何帮助或解释将不胜感激。

4

1 回答 1

3

根据 的文档SBJsonParser，它没有一个名为的方法-objectWithString:error:- 您在运行时的发现证实了这一观点。尝试向它发送一条它会响应的消息，例如-objectWithString:.

于 2012-01-10T14:18:37.003 回答