这是根据您的要求的详细代码。请让如果你发现任何问题,我知道:
import android.app.Activity;
import android.content.Intent;
import android.net.Uri;
import android.os.Bundle;
import android.text.TextUtils;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.EditText;
import android.widget.Toast;
public class LinkActivity extends Activity {
EditText txtLink;
Button btnOpenLink;
String defaultLink;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_link);
defaultLink = "http://www.google.com";
txtLink = (EditText) findViewById(R.id.editTextLink);
btnOpenLink = (Button) findViewById(R.id.buttonOpenLink);
btnOpenLink.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View arg0) {
String page = txtLink.getText().toString();
if(!TextUtils.isEmpty(page)){
Uri uri = Uri.parse(defaultLink+"/"+page);
Intent intent = new Intent(Intent.ACTION_VIEW, uri);
startActivity(intent);
}else{
Toast.makeText(LinkActivity.this, "Please enter page on editText!!", Toast.LENGTH_LONG).show();
}
}
});
}
}
你的XML:
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="fill_parent"
android:layout_height="fill_parent"
android:orientation="vertical"
android:gravity="center" >
<EditText
android:id="@+id/editTextLink"
android:layout_width="300dp"
android:layout_height="wrap_content"
android:ems="10" >
<requestFocus />
</EditText>
<Button
android:id="@+id/buttonOpenLink"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="Open Link" />
</LinearLayout>
@LoopyLoo:我准备了一个详细的代码给你。租约检查是否有用。 – sUndeep 2015-02-12 07:01:11