UIApplication openUrl 不适用于格式化的 NSString

发布于 2024-08-26 21:23:21 字数 269 浏览 6 评论 0 原文

我有以下代码来打开谷歌地图：

NSString *urlString = [NSString stringWithFormat:@"http://maps.google.com/maps?q=%@, Anchorage, AK",addressString];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:urlString]];

但它不起作用，也没有错误。它就是打不开。

原文

I have the following code to open google maps:

NSString *urlString = [NSString stringWithFormat:@"http://maps.google.com/maps?q=%@, Anchorage, AK",addressString];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:urlString]];

But it doesn't work and there is no error. It just doesn't open.

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

醉梦枕江山 2024-09-02 21:23:21

URLWithString 需要一个百分比转义字符串。您的示例 url 包含空格，这会导致创建 nil NSURL。另外，addressString还可能包含需要转义的字符。首先尝试对 url 字符串进行百分比转义：

NSString *urlString = [NSString stringWithFormat:@"http://maps.google.com/maps?q=%@, Anchorage, AK",addressString];
NSString *escaped = [urlString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:escaped]];

URLWithString requires a percent-escaped string. Your sample url contains spaces which results in a nil NSURL being created. Additionally, the addressString may also contain characters that need to be escaped. Try percent-escaping the url string first:

NSString *urlString = [NSString stringWithFormat:@"http://maps.google.com/maps?q=%@, Anchorage, AK",addressString];
NSString *escaped = [urlString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:escaped]];

回复收藏 0 原文

挽清梦 2024-09-02 21:23:21

需要转义 urlString ，
否则 [NSURL URLWithString:urlString] 将返回 null。

NSString *urlString = [NSString stringWithFormat:@"http://maps.google.com/maps?q=%@, Anchorage, AK",addressString];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:[urlString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding] ]];

Need to escape the urlString ,
else [NSURL URLWithString:urlString] will return nill.

NSString *urlString = [NSString stringWithFormat:@"http://maps.google.com/maps?q=%@, Anchorage, AK",addressString];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:[urlString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding] ]];

回复收藏 0 原文

~没有更多了~