Question

我正在玩我发现here的代码。我只能让我的数据库返回一个空的json对象。这是我的代码。

public class TableTalkSplash extends Activity {
    private InputStream is;

    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

      String result = "";
      InputStream is = null;
      //the year data to send
      ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
      nameValuePairs.add(new BasicNameValuePair("row_id","1"));

      //http post
      try{
              HttpClient httpclient = new DefaultHttpClient();
              HttpPost httppost = new HttpPost("http://mywebsite.com/myscript.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);
              StringBuilder sb = new StringBuilder();
              String line = null;
              while ((line = reader.readLine()) != null) {
                      sb.append(line + "\n");
              }
              is.close();

              result=sb.toString();
              System.out.println(result); // this is printing "null" to the console
      }catch(Exception e){
              Log.e("log_tag", "Error converting result "+e.toString());
      }

      //parse json data
      try{
              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()); //exception is being caught here
      }
    }
}

这是我的php脚本。

<?php
mysql_connect("host","username","password");
mysql_select_db("mydb");

$q=mysql_query("SELECT * FROM dinner_info WHERE row_id ='".$_REQUEST['row_id']."'");
while($e=mysql_fetch_assoc($q))
        $output[]=$e;

print(json_encode($output));

mysql_close();
?>

我有一个当前正在运行网站的apache webserver，而.php文件与index.html文件位于同一目录中。我有一个mySQL数据库，它运行在与apache服务器相同的盒子上。我对数据库知之甚少，但我知道当我运行SELECT * FROM dinner_info时WHERE row_id = 1;它返回正确的信息。所以我猜我的错误是在PHP中。我尝试用localhost替换“host”，用我的用户名替换“username”，用我的密码替换“password”。有任何想法吗？如果您需要更多信息，请告诉我们！谢谢大家。

Answer 1

我猜测数据库连接或查询失败尝试这个来帮助调试：

<?php

error_reporting(-1);
ini_set("display_errors", 1);

mysql_connect("host","username","password") or die('Could not connect: ' . mysql_error());;
mysql_select_db("mydb");

$query = "SELECT * FROM dinner_info WHERE row_id ='".mysql_real_escape_string($_REQUEST['row_id'])."'"

$q=mysql_query($query);

if (!$q) {
    $message  = 'Invalid query: ' . mysql_error() . "\n";
    $message .= 'Whole query: ' . $query;
    die($message);
}

while($e=mysql_fetch_assoc($q))
        $output[]=$e;

print(json_encode($output));

mysql_close();

Php webservice总是返回null值

1 个答案: