我想把php放入TextView中,但我不知道怎么做。我无法将TextView的文本设置为php的行。这是我的密码。
public class ItemMenu extends ActionBarActivity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_item_menu);
final TextView gold=(TextView)findViewById(R.id.Gold);
Button back= (Button)findViewById(R.id.Back);
back.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
input in=new input();
in.execute();
}
});
}
}
这是获取php的代码。我应该换一下里面的东西吗?
public class input extends AsyncTask<String, Void, String>{
public String HttpIn(String urls){
HttpClient client=new DefaultHttpClient();
HttpGet in=new HttpGet(urls);
try{
HttpResponse response=client.execute(in);
if(response!=null){
String line="";
InputStream input=response.getEntity().getContent();
line=input.toString();
return line;
}else{
return"Unable to get Data";
}
}catch(ClientProtocolException Cie){
return "exception";
}catch(IOException io){
return "exception";
}
}
@Override
protected String doInBackground(String... urls) {
return HttpIn("http://worlddomination.pe.hu/login.php");
}
使用onPostExecute
方法更新textView:
在input
中创建构造函数以获取TextView:
TextView txtView;
public input(TextView txtView){
this.txtView=txtView;
}
重写input
类中的onPostExecute方法:
@Override
protected void onPostExecute(String result) {
txtView.setText(result);
}
并通过在按钮点击上传递TextView对象来创建input
类的对象:
public void onClick(View v) {
input in=new input(gold);
in.execute();
}