How to find Instagram developer support?

Question

I am thinking on a project that uses insta APIs but when I signup for instagramdeveloper account I have some kinda issue with it. I cannot find a button to create new Client and when I hit Manage client button this is what I got:

when I hit Registration Disabled button Nothing Happens. Is that mean am I ban from an Instagram developer account? please note I haven't created any kind of Client ID since I created an Instagram developer account.

OR was this some kinda bug? How can I report my issue to the Instagram support team? your suggestions are appreciated:)

Answer 1

I dont know why my registration button is disabled too. Maybe Instagram api update. But I realize this guide and It works for me. https://developers.facebook.com/docs/instagram-basic-display-api/getting-started

Updated :

In my case, i am using webview in android. So, below is the example code : (Ignore the Dialog, you can implement only webview and its onpagefinished method)

    public class AuthenticationDialog extends Dialog {
        private String TAG = AuthenticationDialog.class.getSimpleName();
        private AuthenticationListener listener;
        private Context context;
        private WebView webView;

        private final String url = "https://api.instagram.com/" + "oauth/authorize/?app_id=" +
                getResources().getString(R.string.app_id)
                + "&redirect_uri="
                + getResources().getString(R.string.redirect_url)
                + "&response_type=code"
                + "&scope=user_profile,user_media";

        public AuthenticationDialog(@NonNull Context context, AuthenticationListener listener) {
            super(context, android.R.style.Theme_Black_NoTitleBar_Fullscreen);

            this.context = context;
            this.listener = listener;
        }

        @Override
        protected void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            this.setContentView(R.layout.auth_dialog);
            this.getWindow().setLayout(ViewGroup.LayoutParams.MATCH_PARENT, ViewGroup.LayoutParams.WRAP_CONTENT);
            initializeWebView();
        }

        private void initializeWebView() {
            webView = (WebView) findViewById(R.id.webView);
            webView.getSettings().setUseWideViewPort(true);
            webView.getSettings().setLoadWithOverviewMode(true);

            webView.getSettings().setJavaScriptEnabled(true);
            webView.loadUrl(url);
            Log.d(TAG, "url: " + url);
            webView.setWebViewClient(new WebViewClient() {

                String access_token;
                boolean authComplete;

                @Override
                public void onPageStarted(WebView view, String url, Bitmap favicon) {
                    super.onPageStarted(view, url, favicon);
                    Log.d(TAG, "onPageStarted called");
                }

                @Override
                public void onPageFinished(WebView view, String url) {
                    super.onPageFinished(view, url);
                    Log.d(TAG, "onPageFinished called " + url);
                    if (url.contains("?code=") && !authComplete) {
                        Log.d(TAG, " inside access_token");
                        access_token = url;
                        //get the whole token after "=" sign
                        access_token = access_token.replace("https://www.instagram.com/?code=","");
                        access_token = access_token.replace("#_","");
                        Log.d(TAG, "token: " + access_token);
                        authComplete = true;
                        listener.onTokenReceived(access_token);
                        webView.loadUrl("https://instagram.com/accounts/logout/");

                        dismiss();
                    } else if (url.contains("?error")) {
                        Log.d(TAG, "getting error fetching access token");
                        dismiss();
                    } else {
                        Log.d(TAG, "outside both" + url.toString());
                    }
                }
            });
        }
    }