Java Android应用程序中从字符串到JSONArray的解析错误

我希望从MySQL服务器接收表中的所有数据。 因此，我编写了一个PHP脚本，用于获取完整的表并将其编码为JSON格式

PHP代码：

<?php
include ("connect.php"); 

mysql_query("SET NAMES 'utf8'");

$query = mysql_query('SELECT * FROM MyTable);

while ($row = mysql_fetch_assoc($query))
{
$buffer[] = $row;
}

print json_encode( $buffer );
mysql_close($dbconnect);
?>

---

这有什么奇怪的，在我看来像JSON？！你的错误是什么。请发布错误消息。这一行中缺少一个
'
：
$query=mysql\u query（'SELECT*FROM MyTable'）。如果是打字错误（你没有剪切/粘贴代码？），那就奇怪了；如果不是，那就更奇怪了：）嗨！对不起，是的，我在复制/粘贴过程中忘记了'
String result = "";
        ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();

        // HTTP Verbindung
        try
        {
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("http://mywebsite.com");
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse response = httpclient.execute(httppost);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();
        }
        catch(Exception e)
        {
            Log.e("log_tag", "Fehler bei der http Verbindung "+e.toString());
        }



        try
        {
            BufferedReader reader = new BufferedReader(new InputStreamReader(is,"utf-8"),8);
            StringBuilder 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());
        }



        try
        {
            JSONArray jArray = new JSONArray(result);

            for(int i=0;i<jArray.length();i++)
            {
                json_data = jArray.getJSONObject(i);


                    );
            }
        }
        catch(JSONException e)
        {
            Log.e("log_tag", "Error parsing data "+e.toString());
        }