UIApplication openUrl不使用格式化的NSString [英] UIApplication openUrl not working with formatted NSString

问题描述

我有以下代码打开谷歌地图：

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]];

但它不起作用且没有错误。它只是没有打开。

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

推荐答案

URLWithString 需要一个百分比转义字符串。您的示例网址包含空格，这会导致创建无NSURL。此外，addressString还可能包含需要转义的字符。首先尝试以百分比形式转义url字符串：

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]];  

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