Intent to open Instagram user profile on Android

微笑、不失礼 提交于 2019-11-28 03:56:38
jhondge

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")));
    }

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;
}

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";

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);

            }
        });
vb217

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")));
    }

}
fun getOpenInstagram(context: Context, url: String) {
    val likeIng = Intent(Intent.ACTION_VIEW, Uri.parse("http://instagram.com/_u/$url"))
    likeIng.setPackage("com.instagram.android")
    try {
        context.startActivity(likeIng)
    } catch (e: ActivityNotFoundException) {
        context.startActivity(Intent(Intent.ACTION_VIEW, Uri.parse("http://instagram.com/$url")))
    }
}
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