How to open a website when a Button is clicked in

问题:

I am designing an app, with several button for users to click on. Once button is clicked, user is directed to appropriate website. How do I accomplish this?

回答1:

If you are talking about an RCP app, then what you need is the SWT link widget.

Here is the official link event handler snippet.

Update

Here is minimalist android application to connect to either superuser or stackoverflow with 2 buttons.

package ap.android;

import android.app.Activity;
import android.content.Intent;
import android.net.Uri;
import android.os.Bundle;
import android.view.View;

public class LinkButtons extends Activity {

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
    }

    public void goToSo (View view) {
        goToUrl ( "http://stackoverflow.com/");
    }

    public void goToSu (View view) {
        goToUrl ( "http://superuser.com/");
    }

    private void goToUrl (String url) {
        Uri uriUrl = Uri.parse(url);
        Intent launchBrowser = new Intent(Intent.ACTION_VIEW, uriUrl);
        startActivity(launchBrowser);
    }

}

And here is the layout.

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:orientation="vertical" android:layout_width="fill_parent" android:layout_height="fill_parent">
    <TextView android:layout_width="fill_parent" android:layout_height="wrap_content" android:text="@string/select" />
    <Button android:layout_height="wrap_content" android:clickable="true" android:autoLink="web" android:cursorVisible="true" android:layout_width="match_parent" android:id="@+id/button_so" android:text="StackOverflow" android:linksClickable="true" android:onClick="goToSo"></Button>
    <Button android:layout_height="wrap_content" android:layout_width="match_parent" android:text="SuperUser" android:autoLink="web" android:clickable="true" android:id="@+id/button_su" android:onClick="goToSu"></Button>
</LinearLayout>

Time permitting, I'll add the site logos to the buttons to make it a bit sexier ;-)

回答2:

In your Java file write the following piece of code...

ImageView Button = (ImageView)findViewById(R.id.yourButtonsId);

Button.setOnClickListener(new OnClickListener() {
    public void onClick(View v) {
        Intent intent = new Intent();
        intent.setAction(Intent.ACTION_VIEW);
        intent.addCategory(Intent.CATEGORY_BROWSABLE);
        intent.setData(Uri.parse("http://www.yourURL.com"));
        startActivity(intent);
    }
});

回答3:

Here is a workable answer.

Manifest.xml

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
  package="com.tutorial.todolist"
  android:versionCode="1"
  android:versionName="1.0">
  <uses-sdk android:minSdkVersion="3"></uses-sdk>
  <application android:icon="@drawable/icon" android:label="@string/app_name">
    <activity android:name=".todolist"
              android:label="@string/app_name">
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />
            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>
  </application>
</manifest>

todolist.java

package com.tutorial.todolist;
import android.app.Activity; 
import android.content.Intent;
import android.net.Uri;
import android.os.Bundle; 
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;

public class todolist extends Activity {    
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

        Button btn = (Button) findViewById(R.id.btn_clickme);
        btn.setOnClickListener(new OnClickListener() {
            public void onClick(View v) {
                Intent myWebLink = new Intent(android.content.Intent.ACTION_VIEW);
                myWebLink.setData(Uri.parse("http://www.anddev.org"));
                    startActivity(myWebLink);
             }
        });
    }
}

main.xml

<?xml version="1.0" encoding="utf-8"?> 
<LinearLayout xmlns:android="schemas.android.com/apk/res/android"
  android:orientation="vertical" 
  android:layout_width="fill_parent" 
  android:layout_height="fill_parent" > 

    <Button android:id="@+id/btn_clickme" 
      android:text="Click me..." 
      android:layout_width="fill_parent" 
      android:layout_height="wrap_content" /> 
</LinearLayout>

回答4:

Import import android.net.Uri;

Intent openURL = new Intent(android.content.Intent.ACTION_VIEW);
openURL.setData(Uri.parse("http://www.example.com"));
startActivity(openURL);

or it can be done using,

TextView textView = (TextView)findViewById(R.id.yourID);

textView.setOnClickListener(new OnClickListener() {
    public void onClick(View v) {
        Intent intent = new Intent();
        intent.setAction(Intent.ACTION_VIEW);
        intent.addCategory(Intent.CATEGORY_BROWSABLE);
        intent.setData(Uri.parse("http://www.typeyourURL.com"));
        startActivity(intent);
    } });

回答5:

ImageView Button = (ImageView)findViewById(R.id.button);

Button.setOnClickListener(new OnClickListener() {

    public void onClick(View v) {
        Uri uri = Uri.parse("http://google.com/");

        Intent intent = new Intent(Intent.ACTION_VIEW, uri);
        startActivity(intent);
    }
});

回答6:

you can use this on your button click activity

Intent webOpen = new Intent(android.content.Intent.ACTION_VIEW);
            WebOpen.setData(Uri.parse("http://www.google.com"));
                startActivity(myWebLink);

and import this on your code

import android.net.Uri;

回答7:

Add this to your button's click listener:

  Intent intent = new Intent(android.content.Intent.ACTION_VIEW);
    try {
        intent.setData(Uri.parse(url));
        startActivity(intent);
    } catch (ActivityNotFoundException exception) {
        Toast.makeText(getContext(), "Error text", Toast.LENGTH_SHORT).show();
    }

If you have a website url as a variable instead of hardcoded string then don't forget to handle an ActivityNotFoundException and show error. Or you may receive invalid url and app will simply crash. (Pass random string instead of url variable and see for youself )

回答8:

You can wrap the buttons in anchors that href to the appropriate website.

<a href="http://www.stackoverflow.com">
  <input type="button" value="Button" />
</a>
<a href="http://www.stackoverflow.com">
  <input type="button" value="Button" />
</a>
<a href="http://www.stackoverflow.com">
  <input type="button" value="Button" />
</a>

When the user clicks the button (input) they are directed to the destination specified in the href property of the anchor.

Edit: Oops, I didn't read "Eclipse" in the question title. My mistake.

回答9:

public class MainActivity extends Activity {

private WebView webView1;
Button google; 


@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    google = (Button) findViewById(R.id.google);
    google.setOnClickListener(new OnClickListener() {

        public void onClick(View arg0) {
            webView1 = (WebView) findViewById(R.id.webView);
            webView1.getSettings().setJavaScriptEnabled(true);
            webView1.loadUrl("http://www.google.co.in/");

      }
});        
    }

@Override
public boolean onCreateOptionsMenu(Menu menu) {
    // Inflate the menu; this adds items to the action bar if it is present.
    getMenuInflater().inflate(R.menu.main, menu);
    return true;
}

}

回答10:

I just need one line to show a website in my app:

startActivity(new Intent(Intent.ACTION_VIEW, Uri.parse("http://match4app.com")));