Question

我尝试读取SQL并将其编码回android时，标题中有错误。

我尝试了以下内容：

通过打印到日志文件验证$ androidIMEI的值，按预期返回值。
验证了$ sql的输出，并且查询正常（包括来自android的$ _POST [＆＃39; myIMEI_toString＆＃39;]）值。
验证了$ json数组的值，当查询从SQL返回2行时，返回2个数组，OK。

更换

$androidIMEI = isset($_POST['myIMEI_toString']) ? $_POST['myIMEI_toString'] : '';

WITH

$androidIMEI = "000000000000000" //works fine but I want to get that programmatically.

代码：

Android（发送IMEI）：

TelephonyManager mngr = (TelephonyManager)getSystemService(Context.TELEPHONY_SERVICE); 
myIMEI = mngr.getDeviceId();

myIMEI_toString = myIMEI.toString();

...............

protected String doInBackground(String... arg0) {

              ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();         

            nameValuePairs.add(new BasicNameValuePair("myIMEI_toString",myIMEI_toString));

                try
                {
                HttpClient httpclient = new DefaultHttpClient();
                HttpPost httppost = new HttpPost("http://path_on_server/file.php");
                httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                HttpResponse response = httpclient.execute(httppost); 
                HttpEntity entity = response.getEntity();
                is = entity.getContent();
                InputStreamReader ireader = new InputStreamReader(is);
                BufferedReader bf = new BufferedReader(ireader);
                sb = new StringBuilder();
                String line = null;
                while ((line = bf.readLine()) != null) {
                    sb.append(line);
                }
                Log.e("pass 1", "connection success ");


            }
                catch(Exception e)

            {
                System.out.println("Error catch");

            }
                return id;     

        }

Android（JSON）：

      try {
   JSONObject jsonResponse = new JSONObject(jsonResult);
   JSONArray jsonMainNode = jsonResponse.optJSONArray("myarray");
   for (int i = 0; i < jsonMainNode.length(); i++) {
    JSONObject jsonChildNode = jsonMainNode.getJSONObject(i);
    name = jsonChildNode.getString("Request_Name");
    number = jsonChildNode.getString("Request_Number");
    username = jsonChildNode.getString("Request_Username");
    status = jsonChildNode.getString("Status");

    arrName.add(name);
    arrNumber.add(number);
    arrUsername.add(username);
    arrStatus.add(status);

    System.out.println("Name: "+name);
    System.out.println("Number: "+number);
    System.out.println("Username: "+username);
    System.out.println("Status: "+status);

   }
  } catch (JSONException e) {

      Log.i("Error Log: ", e.toString());
      System.out.println("Error: "+e.toString());


      Toast.makeText(getApplicationContext(), "Error" + e.toString(), Toast.LENGTH_SHORT).show();
  }

PHP：

<?php

include 'config.php';

//$androidIMEI = "000000000000000";
$androidIMEI = isset($_POST['myIMEI_toString']) ? $_POST['myIMEI_toString'] : '';

    //$f = fopen("log.txt", "w");
    //fwrite($f, print_r($androidIMEI, true));
    //fclose($f); 

$con=mysql_connect("$servername", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$dbname")or die("cannot select DB");

$sql = "SELECT * from users WHERE Request='0' AND IMEI = '$androidIMEI' "; 
$result = mysql_query($sql);

$json = array(); 
if(mysql_num_rows($result)){
    while($row=mysql_fetch_assoc($result)){
    $json['myarray'][]=$row;
    }

}
else
{
    //$error =  "Error selecting record: " . $conn->error " ";
    //$f = fopen("$error.txt", "w");
    //fwrite($f, print_r($error, true));
    //fclose($f); 
}

    $f = fopen("log.txt", "w");
    fwrite($f, print_r($sql, true));
    fclose($f); 

mysql_close($con);
echo json_encode($json); 
?>

Logcat错误：

org.json.JSONException: Value [] of type org.json.JSONarray cannot be converted to JSONObject

（我之前已经问过这个问题，但会再次尝试提供更多信息）

org.json.JSONarray类型的值[]无法转换为JSONObject

0 个答案: