how to open instagram app on button click

问题

Hi all am trying to open instagram app on a button click but am not able to i set Url scheme as instagram in plist also

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];

i did like this but app not opening am new to iOS any help can be appreciated

回答1:

You can open instagram app by username like,

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

You can refer iPhone Hooks of Instagram for more ways and details about api!!!

Update :

Replace below line,

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

with

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

You are assigning directly string as url!!

回答2:

Add key Value In Info.plist File

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

来源：https://stackoverflow.com/questions/38803801/how-to-open-instagram-app-on-button-click

标签

ios

instagram