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我想讓我的android應用程序從在線數據庫獲取數據。這裏有兩種情況：來自sql server的數據
- 當我創建我的XAMPP分貝，我使用我的本地計算機的IP作爲參數
httpost功能我看到的輸出什麼，我希望看到（數據庫在logcat）。
我的問題是：如果我從我的手機運行應用程序，它會連接到我的本地機器服務器嗎？
- 我也有一個網站（可以說mysite.com），爲了不購買另一臺服務器，我將PHP文件和數據庫放在該服務器上。但是，然後我的Android應用程序連接（或者我認爲）到服務器，但它打印出整個HTML網站
logcat。我在想，這是因爲服務器需要用戶名和密碼，我不知道我是否提供了它們？
那麼，你有什麼建議嗎？我希望我的數據庫被髮送到我的應用程序（以便稍後使用它）。
我的代碼如下所示（我在評論只有情景變化2之間）位於任c:\xampp\htdocs或mysite的服務器上
public class MainActivity extends Activity {
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
setImageClickListener();
}
private void setImageClickListener() {
ImageView map_image=(ImageView)findViewById(R.id.map_icon);
map_image.setOnTouchListener(new ImageView.OnTouchListener() {
//OnTouchListener listener = new OnTouchListener() {
public boolean onTouch(View v, MotionEvent event) {
if(!(event.getAction() == MotionEvent.ACTION_DOWN))
return false; //If the touch event was not putting the finger down on the screen, return false(Actions may be move, up, and so on)
final float x = event.getX();
final float y = event.getY();
//System.out.println("Coordinates of button pressed are: X is %d"+x+" and Y is %d"+ y);
if(x>335 && x<395 && y>225 && y< 235)
DoFirst();
return true;
}
});
}
@SuppressWarnings("null")
private void DoFirst() {
Log.d("SnowReportApp","Do first thing");
setContentView(R.layout.layout_1);
String result = "";
InputStream is = null;
StringBuilder sb=null;
//the year data to send
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();//() before
nameValuePairs.add(new BasicNameValuePair("year","1980"));
//http post
try{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("192.168.1.67/test.php"); // only this changes to my server url : mysite.com/httpdocs/test.php
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
is = entity.getContent();
}catch(Exception e){
Log.e("log_tag", "Error in http connection "+e.toString());
}
//convert response to string
try{
BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
result=sb.toString();
}catch(Exception e){
Log.e("log_tag", "Error converting result "+e.toString());
}
//parse JSON data
try{
//JSONObject json_data_1 = new JSONObject(result);
JSONArray jArray = new JSONArray(result);
for(int i=0;i<jArray.length();i++){
JSONObject json_data = jArray.getJSONObject(i);
Log.i("log_tag","id: "+json_data.getInt("id")+
", name: "+json_data.getString("name")+
", sex: "+json_data.getInt("sex")+
", birthyear: "+json_data.getInt("birthyear")
);
}
}catch(JSONException e){
Log.e("log_tag", "Error parsing data "+e.toString());
}
}
我的php文件是這樣的：
<?php
mysql_connect("127.0.0.1","root","");
mysql_select_db("peopledata");
$q=mysql_query("SELECT * FROM people WHERE birthyear>'".$_REQUEST['year']."'");
while($e=mysql_fetch_assoc($q))
$output[]=$e;
print(json_encode($output));
mysql_close();?>