ios 7中的openURL后UIButton不起作用

Question

我有多个调用 openURL 的共享按钮。按下第一个按钮后，它会打开 Safari 进行共享，当我返回我的应用程序时，所有按钮都不起作用。 （仅 iOS 7.1，但 iOS 8 工作正常）

下面是我的代码：

- (IBAction)btnFacebook_TouchUpInSide:(id)sender
{
    NSLog(@"Touch FaceBook");

    NSString *shareUrl = @"some url";

    NSString *shareText = @"some text";

    NSString *shareDescription = @"some text";

    NSString *sharePictureUrl = @"some text";

    NSString *sharingURL = [self encodeURL:[NSString stringWithFormat:@"https://www.facebook.com/dialog/feed?app_id=388711667988471&display=page&caption=%@&description=%@&picture=%@&link=%@&redirect_uri=%@", shareText, shareDescription, sharePictureUrl, shareUrl, shareUrl]];

    if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:sharingURL]])
    {
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:sharingURL]];
    }
}

- (IBAction)btnTwitter_TouchUpInSide:(id)sender
{
    NSLog(@"Touch Twitter");

    NSString *shareUrl = @"some url";

    NSString *shareText = @"some text";

    NSString *sharingURL = [self encodeURL:[NSString stringWithFormat:@"twitter://post?message=%@", [self encodeURL:[NSString stringWithFormat:@"%@ %@", shareUrl, shareText]]]];

    if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:sharingURL]])
    {
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:sharingURL]];
    }
    else
    {
        NSLog(@"Open web to share twitter!");

        sharingURL = [NSString stringWithFormat:@"https://twitter.com/intent/tweet?url=%@&text=%@&count=none/",[self encodeURL:shareUrl],[self encodeURL:shareText]];

        if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:sharingURL]])
        {
            [[UIApplication sharedApplication] openURL:[NSURL URLWithString:sharingURL]];
        }
    }
}

- (IBAction)btnPinterest_TouchUpInSide:(id)sender
{
    NSLog(@"Touch Pinterest");

    NSString *shareUrl = @"some url";

    NSString *shareText = @"some text";

    NSString *shareMedia = @"some text";

    NSString *sharingURL = [self encodeURL:[NSString stringWithFormat:@"pinit12://pinterest.com/pin/create/link/?url=%@&media=%@&description=%@", shareUrl, shareMedia, shareText]];

    if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:sharingURL]])
    {
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:sharingURL]];
    }
    else
    {
        NSLog(@"Open web to share Pinterest!");

        sharingURL = [self encodeURL:[NSString stringWithFormat:@"https://www.pinterest.com/pin/create/link/?url=%@&media=%@&description=%@", shareUrl, shareMedia, shareText]];

        if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:sharingURL]])
        {
            [[UIApplication sharedApplication] openURL:[NSURL URLWithString:sharingURL]];
        }
    }
}

- (IBAction)btnLinkedIn_TouchUpInSide:(id)sender
{
    NSLog(@"Touch LinkedIn");

    NSString *shareUrl = @"some url";

    NSString *shareTitle = @"some text";

    NSString *shareSummary = @"some text";

    NSString *shareSource = @"some text";

    NSString *sharingURL = [NSString stringWithFormat:@"https://www.linkedin.com/shareArticle?mini=true&url=%@&title=%@&summary=%@&source=%@",[self encodeURL:shareUrl],[self encodeURL:shareTitle],[self encodeURL:shareSummary],[self encodeURL:shareSource]];

    if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:sharingURL]])
    {
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:sharingURL]];
    }
}

谢谢。

Answer 1

试试这个：

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:sharingURL]];

而不是这个：

if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:sharingURL]])
    {
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:sharingURL]];
    }

您将 openURL 移出 if 块 canOpenURL