Question

请帮我在下面的代码中找到错误。

我已将Android文件中的值通过namevaluepair发布到PHP文件。

我收到以下错误：

注意：未定义索引：城市{＆＃34;结果＆＃34;：[]}

<?php
$host = "localhost";
$user = "root";
$password = "";
$dbname = "sampledb";
$con = new mysqli($host, $user, $password, $dbname);
$city = $_POST['city'];
$con->query("SET NAMES 'utf8'");
$stmtsearch = "SELECT a.title, a.phone, a.description, a.email, a.post_date FROM `ad_master` a INNER JOIN tbl_city_master b ON a.city_id = b.fld_city_id WHERE b.fld_city_name = '$city'";
$resultsearch = $con->query($stmtsearch);
$json = array();
while ($row = mysqli_fetch_array($resultsearch)) {
    array_push($json, array(
        'title' => $row[0],
        'phone' => $row[1],
        'description' => $row[2],
        'email' => $row[3],
        'post_date' => $row[4]
    ));
}
echo json_encode(array("result" => $json));
$con->close();
?>

Android代码：

public void getData() {
    class GetDataJSON extends AsyncTask<String, Void, String> {
        @Override
        protected String doInBackground(String... params) {
            InputStream inputStream = null;
            String result = null;
            Intent in = getIntent();
            String city = in.getStringExtra(("city"));
            List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
            nameValuePairs.add(new BasicNameValuePair("city", city));
            DefaultHttpClient httpClient = new DefaultHttpClient(new BasicHttpParams());
            HttpPost httppost = new HttpPost("http://10.0.2.2/qsearchdetail.php");
            try {
                HttpResponse response = httpClient.execute(httppost);
                HttpEntity entity = response.getEntity();
                inputStream = entity.getContent();
                BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream, "utf-8"), 8);
                StringBuilder sb = new StringBuilder();
                String line = null;
                while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
                }
                result = sb.toString();
            } catch (Exception e) {
                e.printStackTrace();
            } finally {
                try {
                    if (inputStream != null) inputStream.close();
                } catch (Exception squish) {
                }
                return result;
            }
        }

        @Override
        protected void onPostExecute(String result) {
            myJSON = result;
            showList();
        }
    }
    GetDataJSON g=new GetDataJSON();
    g.execute();
}

public void showList() {
    try {
        JSONObject jsonObj = new JSONObject(myJSON);
        search = jsonObj.getJSONArray(TAG_RESULT);
        for (int i = 0; i < search.length(); i++) {
            JSONObject c = search.getJSONObject(i);
            String title = c.getString(TAG_TITLE);
            String phone = c.getString(TAG_PHONE);
            String email = c.getString(TAG_EMAIL);
            String description = c.getString(TAG_DESCRIPTION);
            String postDate = c.getString(TAG_DATE);
            HashMap<String, String> search = new HashMap<String, String>();

            search.put(TAG_TITLE, title);
            search.put(TAG_PHONE, phone);
            search.put(TAG_EMAIL, email);
            search.put(TAG_DESCRIPTION, description);
            search.put(TAG_DATE, postDate);
            searchList.add(search);
        }
        ListAdapter adapter = new SimpleAdapter(
            QResultDetail.this,searchList, R.layout.search_result,
            new String[]{TAG_TITLE, TAG_PHONE, TAG_EMAIL, TAG_DESCRIPTION,TAG_DATE},
            new int[] { R.id.tvTitle, R.id.tvMobile, R.id.tvEmail, R.id.tvDesp, R.id.tvDate }
        );
        listView.setAdapter(adapter);
    } catch (JSONException e) {
        e.printStackTrace();
    }
}

Answer 1

您没有发送nameValuePairs。

PHP文件中未定义的索引错误（从Android文件传递值）

1 个答案: