Ankita Sethi
Ankita Sethi

Reputation: 43

open a link within android app instead of open a webview

I am trying to open a social link within my app on a button click listener.i want to open it within app i do not want to open a webview for it. I dont find any way to open that link.the url for this link is come from server also i dont understand how to load url on a button click.Please provide any solution.

this.linkedin.setOnClickListener(new View.OnClickListener() {
       @Override
       public void onClick(View v) {
           String URL=sbobj.getLinkdin_url();
           Intent intent=new Intent(getApplicationContext(),LinkedinProfile.class);
           intent.putExtra("URL",sbobj.getLinkdin_url());
           startActivity(intent);
       }
});

sbobj is the object of class from which I am trying to get url.

"linkdin_url": "https://in.linkedin.com/in/kumaraguru"

this is the url,which i have to get from server.

Upvotes: 1

Views: 1849

Answers (3)

RATHI
RATHI

Reputation: 5299

If you are ok with adding new activity then you better use this. I will beautiful Webview out of box and very easy to use but yes you need to add one more activity. Here is the link: FinestWebView. See the beautiful UI of this:

enter image description here

Upvotes: 0

Anjali
Anjali

Reputation: 2535

Custom Tab layout will work for you.

https://developer.chrome.com/multidevice/android/customtabs

this.linkedin.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                String URL = sbobj.getLinkdin_url();
                CustomTabsIntent.Builder builder = new CustomTabsIntent.Builder();
                CustomTabsIntent customTabsIntent = builder.build();
                customTabsIntent.launchUrl(this, Uri.parse(URL));

            }
        });

Upvotes: 4

Vishal Sojitra
Vishal Sojitra

Reputation: 195

You can try this

this.linkedin.setOnClickListener(new View.OnClickListener() {
       @Override
       public void onClick(View v) {
           String url = sbobj.getLinkdin_url();
                            if (url.contains("https://in.linkedin.com")) {
                                Intent i = new Intent(Intent.ACTION_VIEW);
                                i.setData(Uri.parse(url));
                                startActivity(i);
                            } else {
                                Utils.ShowToast(context, "Url not Valid!");
                            }
       }
});

Upvotes: 0

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