Question

我正在尝试使用php脚本连接MySQL DB。但我没有得到任何输出只有异常代码。我无法弄清楚问题出在哪里。我使用了教程代码。

private EditText outputStream;

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    String result = null;
    InputStream input = null;
    StringBuilder sbuilder = null;
    outputStream = (EditText)findViewById(R.id.output);
    ArrayList <NameValuePair> nameValuePairs = new ArrayList <NameValuePair>();

    try{
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("http://ik.su.lt/~jbarzelis/Bandymas/index.php");
        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        HttpResponse response = httpclient.execute(httppost);
        HttpEntity entity = response.getEntity();
        input = entity.getContent();
    }
    catch(Exception e){
        Log.e("log_tag","Error in internet connection"+e.toString());
    }
    try{
        BufferedReader reader = new BufferedReader(new InputStreamReader(input,"iso-8859-1"),8);
        sbuilder = new StringBuilder();

        String line = null;

        while((line = reader.readLine()) != null){
            sbuilder.append(line + "\n");
            System.out.println(line);
        }
        input.close();
        result = sbuilder.toString();
    }
    catch(Exception e){
        Log.e("log_tag", "Error converting result "+e.toString());          
    }
    int fd_id;
    String fd_name;
    try{
        JSONArray jArray = new JSONArray(result);
        JSONObject json_data = null;
        for(int i=0;i<jArray.length();i++){
            json_data = jArray.getJSONObject(i);
            fd_id = json_data.getInt("FOOD_ID");
            fd_name = json_data.getString("FOOD_NAME");
            outputStream.append(fd_id +" " + fd_name + "\n");
        }


        }
    catch(JSONException e1){
        Toast.makeText(getBaseContext(), "No food found", Toast.LENGTH_LONG).show();
    }
    catch(ParseException e1){
        e1.printStackTrace();
    }
}

PHP脚本：

<?php
mysql_connect("localhost","**********","******");
mysql_select_db("test");
$sql = mysql_query("select FOOD_NAME as 'Maistas' from FOOD where FOOD_NAME like 'A%'");
while($row = mysql_fetch_assoc($sql)) $output[]=$row;
print(json_encode($output));
mysql_close;

＆GT;

任何想法如何解决？

Answer 1

首先，不要使用Exception.toString（），使用Exception.printStackTrace（）：

catch (Exception e) {
    e.printStackTrace();
}

其次，在您的PHP代码中，您没有检查任何错误。如果发生任何错误，我建议您在Android代码中发出不同的HTTP状态代码（如400）：

if (response.getStatusLine().getStatusCode() != 200) {
    Log.d("MyApp", "Server encountered an error.);
}

这样您就可以知道服务器上是否发生了某些事情。

希望这有帮助

Android，使用PHP连接MySQL

1 个答案: