URLWithString: 返回 nil [英] URLWithString: returns nil
本文介绍了URLWithString: 返回 nil的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
这可能很容易，但我似乎不明白为什么 URLWithString: 在这里返回 nil.
it may be very easy, but I don't seems to find out why is URLWithString: returning nil here.
//localisationName is a arbitrary string here
NSString* webName = [localisationName stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSString* stringURL = [NSString stringWithFormat:@"http://maps.google.com/maps/geo?q=%@,Montréal,Communauté-Urbaine-de-Montréal,Québec,Canadae&output=csv&oe=utf8&sensor=false&key=", webName];
NSURL* url = [NSURL URLWithString:stringURL];
推荐答案
您还需要对硬编码 URL 中的非 ASCII 字符进行转义:
You need to escape the non-ASCII characters in your hardcoded URL as well:
//localisationName is a arbitrary string here
NSString* webName = [localisationName stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSString* stringURL = [NSString stringWithFormat:@"http://maps.google.com/maps/geo?q=%@,Montréal,Communauté-Urbaine-de-Montréal,Québec,Canadae&output=csv&oe=utf8&sensor=false", webName];
NSString* webStringURL = [stringURL stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL* url = [NSURL URLWithString:webStringURL];
您可能可以删除 localisationName 的转义，因为它将通过整个字符串的转义来处理.
You can probably remove the escaping of the localisationName since it will be handled by the escaping of the whole string.
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