Xcode openURL 不读取链接

本文介绍了Xcode openURL 不读取链接的处理方法，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

我有一个动物列表和特殊按钮.当我按下按钮时，我想去维基百科阅读更多关于这种动物的信息.所以我写了这段代码:

I have an animal list and special button. When I press the button, I would like to go to Wikipedia and read about this animal more. So I wrote this code:

-(IBAction)goWiki:(id)sender
{
    NSString *wikiUrl = "http://ru.wikipedia.org/wiki/";
    NSString *url = [NSString stringWithFormat:@"%@%@",wikiUrl,animalTitle];
    [[UIApplication sharedApplication]openURL:[NSURL URLWithString:url]];
    NSLog(@"%@",url);
}

NSLog 显示 url 写入正确，但是没有任何反应.我 99.9% 肯定是因为 animalTitle.我的母语是俄语，animalTitle 也是俄语中的动物名称.因此，如果链接类似于 http://ru.wikipedia.org/wiki/Frog，它可以正常工作，但如果它像http://ru.wikipedia.org/wiki/Лягушка 没有任何反应.任何想法，我怎样才能转到俄语文章?谢谢！

NSLog shows that url was written correctly, however, nothing happened. I am 99,9% sure its because of animalTitle. My native language is russian and animalTitle is also an animal name in russian.So if link is like http://ru.wikipedia.org/wiki/Frog its fine and it works but if its likehttp://ru.wikipedia.org/wiki/Лягушка nothing happens.Any ideas, how can I move to a russian article?Thanks!

推荐答案

使用 stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding 如下 -

-(IBAction)goWiki:(id)sender
{
    NSString *wikiUrl = @"http://ru.wikipedia.org/wiki/";
    NSString *url = [NSString stringWithFormat:@"%@%@",wikiUrl,animalTitle];


    url = [url stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

    [[UIApplication sharedApplication]openURL:[NSURL URLWithString:url]];
    NSLog(@"%@",url);
}

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Russian