Question

大家好我想点击按钮打开Instagram应用，但不是能够将url方案设置为plist中的instagram

NSString *instagramURL = @"instagram://app"; 
NSURL *ourURL = [NSURLURLWithString:instagramURL]; 
if ([[UIApplication sharedApplication]canOpenURL:ourURL]) {
    [[UIApplication sharedApplication]openURL:ourURL];

} else {
    //The App is not installed. It must be installed from iTunes code.
    NSString *iTunesLink = @"//Some other Url goes here";
    [[UIApplication sharedApplication] openURL:[NSURL URLWithString:iTunesLink]];

    UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"URL error"
          message:[NSString stringWithFormat: @"No custom URL defined for %@", ourURL]
          delegate:self cancelButtonTitle:@"Ok" otherButtonTitles:nil];
    [alert show];

我确实喜欢这个，但应用程序没有打开iOS新的任何帮助都可以理解

Answer 1

您可以按用户名

打开Instagram应用

 NSURL *instagramURL = [NSURL URLWithString:@"instagram://user?username=USERNAME"];
if ([[UIApplication sharedApplication] canOpenURL:instagramURL]) {
    [[UIApplication sharedApplication] openURL:instagramURL];
}

您可以参考iPhone Hooks of Instagram了解有关api的更多方式和详细信息!!!

更新：

替换下面的行，

 NSString *instagramURL = @"instagram://app";

与

   NSURL *instagramURL = [NSURL URLWithString:@"instagram://app"];

您直接将字符串指定为url !!

Answer 2

在Info.plist文件中添加键值

<key>LSApplicationQueriesSchemes</key>
    <array>
        <string>instagram</string>
        <string>twitter</string>
    </array>

如何在按钮点击时打开Instagram应用程序

2 个答案: