php - 如何在android中将postparameters声明为php

我想问一些关于android编程的问题..so，我从一个名为list_movie的类中得到了一个值，这是代码>我只张贴了一半

public void adapter_listview() {
    ListAdapter adapter = new SimpleAdapter(this, dftr,R.layout.list_item,new String[] { "title", "id"}, new int[] {R.id.tangk, R.id.kodeangk});

    setListAdapter(adapter);
    ListView lv = getListView();


    lv.setOnItemClickListener(new OnItemClickListener() {
        public void onItemClick(AdapterView<?> parent, View view,int position, long id) {
            String kode_ang = ((TextView) view.findViewById(R.id.kodeangk)).getText().toString();
            Intent in = new Intent(getApplicationContext(), studio.class);
            in.putExtra("kode_intent", kode_ang);

            startActivity(in);
        }
    });
}

因此，我有一个包含id的值“ kode_intent”（此id是从表中的id_movie字段获取的）。接下来，在名为studio的类中使用它。

public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    requestWindowFeature(Window.FEATURE_NO_TITLE);
    getWindow().setFlags(WindowManager.LayoutParams.FLAG_FULLSCREEN, WindowManager.LayoutParams.FLAG_FULLSCREEN);
    setContentView(R.layout.studio);


    Bundle b = getIntent().getExtras();
    kode = b.getString("kode_intent");

    ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>();
    postParameters.add(new BasicNameValuePair("id_movie", kode.getText().toString())); >> this line "kode" is getting an error

所以我的问题是，如何将kode = b.getString("kode_intent")的值传递给postParameters.add(new BasicNameValuePair("id_movie", kode.getText().toString()));

这是我的PHP代码：

$db_host  = "127.0.0.1";

$db_uid  = "root";

$db_pass = "";

$db_name  = "platinum";

$db_con = mysql_connect($db_host,$db_uid,$db_pass) or die('could not connect');
mysql_select_db($db_name);

$sql = "Select studio_name, date from cinema where id_movie = '".   $_POST["id_movie"]."'";

$result = mysql_query($sql);
while($row=mysql_fetch_assoc($result))
$output[]=$row;
print(json_encode($output));
mysql_close();

由于我是android编程的新手，所以任何帮助对我来说都是如此。
谢谢

我试图将postParameters.add(new BasicNameValuePair("id_film", kode.getText().toString()));更改为此一个postParameters.add(new BasicNameValuePair("id_movie", kode));

看起来不错..但是现在我又遇到了一个错误。代码如下：

public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    requestWindowFeature(Window.FEATURE_NO_TITLE);
    getWindow().setFlags(WindowManager.LayoutParams.FLAG_FULLSCREEN, WindowManager.LayoutParams.FLAG_FULLSCREEN);
    setContentView(R.layout.studio);

    Bundle b = getIntent().getExtras();
    kode = b.getString("kode_intent");

    ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>();
    postParameters.add(new BasicNameValuePair("id_movie", kode));

    try
    {
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("http://10.0.2.2/cinemainfo/movie_time.php");
            HttpResponse response = httpclient.execute(httppost,postParameters);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();
            Log.e("Pass 1", "success!");
    }

    catch(Exception e)
    {
        Log.e("Fail 1", e.toString());
        Toast.makeText(getApplicationContext(), "Invalid IP Address",
        Toast.LENGTH_LONG).show();
    }

我在此行HttpResponse response = httpclient.execute(httppost,postParameters); >>上出现错误，执行错误消息：“类型为HttpClient的方法execute（HttpUriRequest，HttpContext）不适用于参数（HttpPost，
 数组列表）”

我做错什么了吗？？？

最佳答案

postParameters.add(new BasicNameValuePair("id_movie", kode.getText().toString()));

  “未为类型String定义方法getText（）”

postParameters.add(new BasicNameValuePair("id_movie", kode);

修复后，您将遇到另一个错误NetworkOnMainThreadException

您有2个解决方案：


StrictMode选中此answer
AsyncTask


建议使用第二个选项。