Question

我有一个Android应用程序，它从我的mysql数据库请求数据。这样可以正常工作，但是当我尝试发送一个参数来检索要检索的项目时，我什么都没得到。

这是我的java代码：

        result = "";
        client = new DefaultHttpClient();
        post = new HttpPost("http://www.XXX.XXXX.XX/XX.php");

        nameValuePairs = new ArrayList<NameValuePair>();
        nameValuePairs.add(new BasicNameValuePair("itemsToGet", itemsToGet));

        try {
            post.setEntity((new UrlEncodedFormEntity(nameValuePairs)));
            response = client.execute(post);
            entity = response.getEntity();
        inputStream = entity.getContent();
    } catch (UnsupportedEncodingException e) {
            Log.e(TAG, e.toString());
        } catch (ClientProtocolException e) {
            Log.e(TAG, e.toString());
        } catch (IOException e) {
            Log.e(TAG, e.toString());
        }

        try {
            bufferedReader = new BufferedReader(new InputStreamReader(inputStream ,"iso-8859-1"), 8);
            stringBuilder = new StringBuilder();
            String line = null;

            while((line = bufferedReader.readLine()) != null) {
                stringBuilder.append(line + "\n");
            }
            inputStream.close();

            result = stringBuilder.toString();

            } catch (UnsupportedEncodingException e) {
            Log.e(TAG, e.toString());
        } catch (IOException e) {
            Log.e(TAG, e.toString());
        }

        try {
            JSONArray jsonArray = new JSONArray(result);

            for(int i = 0; i < jsonArray.length(); i++) {
                JSONObject object = jsonArray.getJSONObject(i);
                Log.i(TAG, object.getString("namn") + " " + i);
        }   
        } catch (JSONException e) {
            Log.e(TAG, e.toString());
    }

这是我的服务器端PHP代码：

<?php
mysql_connect("XXX.XXX.com", "XXX", "XXX") or die(mysql_error());
mysql_select_db("XXXXXXX") or die(mysql_error());

$data = mysql_query("SELECT * FROM artikel") or die(mysql_error());
$itemsToGet = intval($_POST['itemsToGet']);
$counter = 0;

while($info = mysql_fetch_array( $data ) && $counter < $itemsToGet)
{
    $databaseInfo[] = $info;
    $counter++;
}

print(json_encode($databaseInfo));

＆GT？;

Answer 1

我解决了。 while循环中的条件不正确。

while($info = mysql_fetch_array( $data ) && $counter < $itemsToGet)

应该是

while(($info = mysql_fetch_array( $data )) && ($counter < $itemsToGet))

从android应用程序发送参数到服务器

1 个答案: