java - Android NameValuePair 发送值

Question

我有连接到 MySql 并将数据编码为 JSON 的 PHP 代码。稍后我会过滤它并获得特定的 JSON 对象。当我使用一个 NameValuePair 对象时它工作正常，但现在我想使用用户名和密码等变量。现在我在 logcat 中收到此警报 Error parsing data .org.json.JSONException: Value null of type org.json.JSONObject$1 cannot be converted to JSONArray.

我应该如何更改可以正常工作的代码？

$q=mysql_query("SELECT username, firstname, lastname, email, phone1, skype, city, description FROM mdl_user WHERE username LIKE '$username' AND password LIKE '$password'");
while($e=mysql_fetch_assoc($q))
    $output[]=$e;
print(json_encode($output));

发送请求的代码:

ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
        nameValuePairs.add(new BasicNameValuePair("usern",""+usr));
        nameValuePairs.add(new BasicNameValuePair("passw",""+psw));
        InputStream is = null; 
        String result = "";
        //http post
        try{
                HttpClient httpclient = new DefaultHttpClient();
                HttpPost httppost = new HttpPost("http://ik.su.lt/~jbarzelis/Bdarbas/getUserInfo.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());
        }

已编辑

$username = mysql_real_escape_string($_REQUEST['usern']);
$password = mysql_real_escape_string($_REQUEST['passw']);
$q=mysql_query("SELECT username, firstname, lastname, email, phone1, skype, city, description
FROM mdl_user WHERE username LIKE '$username' AND password LIKE '$password'");
while($e=mysql_fetch_assoc($q))
    $output[]=$e;
print(json_encode($output));

Answer 1

如果这是您放在服务器端的全部代码，那么我想您需要先使用 将值输入 $username 和 $password >$_POST[] 这样的方法

$username=$_POST["usern"];
$password=$_POST["passw"];

密码也是如此。

因为现在变量中没有值，您的 SELECT 语句返回 null 值，该值以 JSON 格式发送给客户端给出一个 null 值错误。