iphone - 如何从 URL 中捕获所需的值

Question

我需要从字符串中提取变量的值，该字符串恰好是一个 URL。字符串/url 作为单独的 php 查询的一部分加载，而不是浏览器中的 url。

网址看起来像:

http://gmail.com?access_token=ab8w4azq2xv3dr4ab37vvzmh&token_type=bearer&expires_in=3600

如何获取 access_token 的值，在此示例中为 ab8w4azq2xv3dr4ab37vvzmh？

Answer 1

这段代码应该做到这一点:

- (NSString *)extractToken:(NSURL *)URL
{
    NSString *urlString = [URL absoluteString];
    NSRange start = [urlString rangeOfString:@"access_token="];
    if (start.location != NSNotFound)
    {
        NSString *token = [urlString substringFromIndex:start.location+start.length];
        NSRange end = [token rangeOfString:@"&"];
        if (end.location != NSNotFound)
        {
            //trim off other parameters
            token = [token substringToIndex:end.location];
        }
        return token;
    }
    //not found
    return nil;
}

或者，这里有一个更通用的解决方案，它将所有查询参数提取到一个字典中:

- (NSDictionary *)URLQueryParameters:(NSURL *)URL
{
    NSString *queryString = [URL query];
    NSMutableDictionary *result = [NSMutableDictionary dictionary];
    NSArray *parameters = [queryString componentsSeparatedByString:@"&"];
    for (NSString *parameter in parameters)
    {
        NSArray *parts = [parameter componentsSeparatedByString:@"="];
        NSString *key = [[parts objectAtIndex:0] stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
        if ([parts count] > 1)
        {
            id value = [[parts objectAtIndex:1] stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
            [result setObject:value forKey:key];
        }
    }
    return result;
}