Intent to open Instagram user profile on Android

问题:

I'm developing a social networking app and our users can connect their Instagram account to our service. I'd like to open Instagram profiles directly in their official Android app (if it is installed) but I can't find any way to do that. However, there is a page on their developer site about the exact same feature on iOS but this doesn't seem to work on Android at all. Everything I found on the web only suggests various ways to open links in a browser. Any suggestions?

回答1:

I solved this problem using the following code.

    Uri uri = Uri.parse("http://instagram.com/_u/xxx");
    Intent likeIng = new Intent(Intent.ACTION_VIEW, uri);

    likeIng.setPackage("com.instagram.android");

    try {
        startActivity(likeIng);
    } catch (ActivityNotFoundException e) {
        startActivity(new Intent(Intent.ACTION_VIEW,
                Uri.parse("http://instagram.com/xxx")));
    }

回答2:

Although @jhondge's solution works and is correct. This is a more cleaner way to do this:

Uri uri = Uri.parse("http://instagram.com/_u/xxx");
    Intent insta = new Intent(Intent.ACTION_VIEW, uri);
    insta.setPackage("com.instagram.android");

    if (isIntentAvailable(mContext, insta)){
        startActivity(insta);
    } else{
        startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse("http://instagram.com/xxx")));
    }

private boolean isIntentAvailable(Context ctx, Intent intent) {
    final PackageManager packageManager = ctx.getPackageManager();
    List<ResolveInfo> list = packageManager.queryIntentActivities(intent, PackageManager.MATCH_DEFAULT_ONLY);
    return list.size() > 0;
}

回答3:

To open directly instagram app to a user profile :

String scheme = "http://instagram.com/_u/USER";
String path = "https://instagram.com/USER";
String nomPackageInfo ="com.instagram.android";
    try {
        activite.getPackageManager().getPackageInfo(nomPackageInfo, 0);
        intentAiguilleur = new Intent(Intent.ACTION_VIEW, Uri.parse(scheme));
        } catch (Exception e) {
            intentAiguilleur = new Intent(Intent.ACTION_VIEW, Uri.parse(path));
        }
        activite.startActivity(intentAiguilleur); 

// Use this link to open directly a picture
  String scheme = "http://instagram.com/_p/PICTURE";

回答4:

I tried this way and it worked for me..

instabtn.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {


                Intent instaintent = getActivity().getPackageManager().getLaunchIntentForPackage("com.instagram.android");

                instaintent.setComponent(new ComponentName( "com.instagram.android", "com.instagram.android.activity.UrlHandlerActivity"));
                instaintent.setData( Uri.parse( "https://www.instagram.com/_u/bitter_truth_lol") );

                startActivity(instaintent);

            }
        });

回答5:

I implemented this using fragment in webview but I have one issue, the instagram pop up comes three times :

webView.setWebViewClient(new WebViewClient()
        {
 public boolean shouldOverrideUrlLoading(WebView viewx, String urlx)
            {
 if(Uri.parse(urlx).getHost().endsWith("instagram.com")) {

                    gotoinstagram();

                  return false;
                }

                Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(urlx));
                viewx.getContext().startActivity(intent);
                return true;
            }



        });

outside of onCreateView

//instagram

public void gotoinstagram()
{

    Uri uri = Uri.parse("http://instagram.com/_u/XXXX");
    Intent likeIng = new Intent(Intent.ACTION_VIEW, uri);

    likeIng.setPackage("com.instagram.android");

    try {
        startActivity(likeIng);
    } catch (ActivityNotFoundException e) {
        startActivity(new Intent(Intent.ACTION_VIEW,
                Uri.parse("http://instagram.com/XXXX")));
    }

}