android app开发 - 传递参数到数据库

时间:2014-08-13 07:20:51

标签: database android

我正在尝试将我的应用程序连接到localhost服务器上的数据库。我可以连接到它,问题是如何将参数从app传递到php脚本。例如,我想要所有年龄小于10的名字,所以我会将参数传递给php.below是我连接数据库的代码。请提供良好的参考资料

/ * * / 在这里输入代码

public class TestExternalDatabaseActivity extends Activity {
/** Called when the activity is first created. */

TextView resultView;
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
   StrictMode.enableDefaults(); //STRICT MODE ENABLED
   resultView = (TextView) findViewById(R.id.result);

    getData();
}
public void getData(){
    String result = "";
    InputStream isr = null;
    try{
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("http://localhost/tahseen0amin/php/getAllCustomers.php"); //YOUR PHP SCRIPT ADDRESS 
        HttpResponse response = httpclient.execute(httppost);
        HttpEntity entity = response.getEntity();
        isr = entity.getContent();
}
catch(Exception e){
        Log.e("log_tag", "Error in http connection "+e.toString());
        resultView.setText("Couldnt connect to database");
}
//convert response to string
try{
        BufferedReader reader = new BufferedReader(new InputStreamReader(isr,"iso-8859-1"),8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
                sb.append(line + "\n");
        }
        isr.close();

        result=sb.toString();
}
catch(Exception e){
        Log.e("log_tag", "Error  converting result "+e.toString());
}

//parse json data

   try {
   String s = "";
   JSONArray jArray = new JSONArray(result);

   for(int i=0; i<jArray.length();i++){
       JSONObject json = jArray.getJSONObject(i);
       s = s + 
               "Name : "+json.getString("FirstName")+" "+json.getString("LastName")+"\n"+
               "Age : "+json.getInt("Age")+"\n"+
               "Mobile Using : "+json.getString("Mobile")+"\n\n";
   }

   resultView.setText(s);

   } catch (Exception e) {
// TODO: handle exception
   Log.e("log_tag", "Error Parsing Data "+e.toString());
}

}

}

0 个答案:

没有答案