mySQL到PHP到JSON：String无法转换为JSONObject

Question

我试图使用PHP从mySQL数据库中获取数据。 这是我第一次尝试远程获取数据并使用JSON。 php文件运行正常，因为它在浏览器中输出为JSON字符串，并使用JSONLint对其进行了标记 。 所以，我不确定我的错在哪里。 任何帮助将不胜感激

这就是LogCat抛出的内容：

Error parsing data org.json.JSONException: Value <?xml of type java.lang.String cannot be converted to JSONObject
threadid=9: thread exiting with uncaught exception (group=0x401dce20)
FATAL EXCEPTION: Thread-10
java.lang.NullPointerException
at com.andaero.test.JSON.JSONMain$1.run(JSONMain.java:39)
at java.lang.Thread.run(Thread.java:1020)

更新： 我从标记请求的php文件中删除了echo方法。 我认为它与“JSONArray a = json.getJSONArray（”监管“）有关。我也试过其他人的方法，但没有优势。

以下是课程：

public class JSONfunctions {

    public static JSONObject getJSONfromURL(String url) {
        InputStream is = null;
        String result = "regulatory";
        JSONObject jArray = null;

        // http post
        try {
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost(url);
            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());
        }

        // convert response to string
        try {
            BufferedReader reader = new BufferedReader(new InputStreamReader(
                    is, "iso-8859-1"), 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 {

            jArray = new JSONObject(result);
        } catch (JSONException e) {
            Log.e("log_tag", "Error parsing data " + e.toString());
        }

        return jArray;
    }
}

列表活动：

public class JSONMain extends ListActivity {
    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.listview);

        final ArrayList<HashMap<String, String>> mylist = new ArrayList<HashMap<String, String>>();

        new Thread(new Runnable() {
            public void run() {
                JSONObject json = JSONfunctions
                        .getJSONfromURL("http://192.168.1.34/andaero/regulatory_list_ASC.php");

                try {

                    JSONArray a = json.getJSONArray("regulatory");

                    for (int i = 0; i < a.length(); i++) {
                        HashMap<String, String> map = new HashMap<String, String>();
                        JSONObject e = a.getJSONObject(i);

                        map.put("id", String.valueOf(i));
                        map.put("label", e.getString("label"));
                        map.put("title", e.getString("title"));
                        map.put("caption", e.getString("description"));
                        map.put("dummy", e.getString("gotoURL"));
                        mylist.add(map);
                    }
                } catch (JSONException e) {
                    Log.e("log_tag", "Error parsing data " + e.toString());
                }
            }
        }).start();

        ListAdapter adapter = new SimpleAdapter(this, mylist,
                R.layout.list_item, new String[] { "label", "title", "caption",
                        "dummy" }, new int[] { R.id.label, R.id.listTitle,
                        R.id.caption, R.id.dummy });

        setListAdapter(adapter);

        final ListView lv = getListView();
        lv.setTextFilterEnabled(true);
        lv.setOnItemClickListener(new OnItemClickListener() {
            public void onItemClick(AdapterView<?> parent, View view,
                    int position, long id) {
                @SuppressWarnings("unchecked")
                HashMap<String, String> o = (HashMap<String, String>) lv
                        .getItemAtPosition(position);
                Toast.makeText(JSONMain.this,
                        "ID '" + o.get("id") + "' was clicked.",
                        Toast.LENGTH_SHORT).show();

            }
        });
    }
} 

PHP：

<?php
//MySQL Database Connect
include 'andaerologin.php';

mysql_select_db("andaero");
$sql=mysql_query("select * from regulatory_list");

$output = array();
while($row = mysql_fetch_assoc($sql)) {
    $output['regulatory'][] = $row;
}

exit (json_encode($output));

mysql_close();
?>

Answer 1

尝试将PHP更改为：

$output = new stdClass();
$output->regulatory = array();
while($row = mysql_fetch_assoc($sql)) {
    $output->regulatory[] = $row;
}

header('Content-type: application/json');
echo (json_encode($output));

Answer 2

尝试将PHP脚本更改为：

<?php

  // Hide errors to prevent data corruption
  ini_set('display_errors', 0);

  // For debugging, uncomment these lines to show errors
  //ini_set('display_errors', 0);
  //error_reporting(E_ALL);

  //MySQL Database Connect
  require 'andaerologin.php';

  if (!mysql_select_db("andaero")) {
    // Use trigger_error() so you can find out in the server logs if something
    // goes wrong
    trigger_error('Unable to select MySQL database');
    header('HTTP/1.1 500 Internal Server Error');
    exit;
  }

  $query = "SELECT *
            FROM regulatory_list";
  if (!$result = mysql_query($query)) {
    trigger_error('MySQL error: '.mysql_error());
    header('HTTP/1.1 500 Internal Server Error');
    exit;
  }

  if (!mysql_num_rows($query)) {
    trigger_error('MySQL returned no results');
    header('HTTP/1.1 500 Internal Server Error');
    exit;
  }

  // Build an array of the results
  $output = array();
  while ($row = mysql_fetch_assoc($result)) {
    $output[] = $row;
  }

  // Send the results back as JSON
  exit(json_encode($output));

  // Closing the database connection happens implicitly at the end of the
  // script. Also, you don't need to have a closing PHP tag at the end of the
  // file and omitting it is a good habit to get into as it can avoid problems

Answer 3

在您的PHP代码中，进行更改

json_encode($output)

至

json_encode($output, JSON_FORCE_OBJECT)

JSON_FORCE_OBJECT选项需要PHP版本> = 5.3.0

您的问题似乎是在jArray = new JSONObject(result);

我不知道JSONObject构造函数需要什么，但我知道你要向它发送一个JSON 数组 ，而不是一个对象 。

Answer 4

你真的需要桌子上的所有领域吗？

我曾经做过相同的SELECT * FROM表和json_encode（）所有的结果。 即使JSON结果看起来很好，jQuery似乎在读取数据时遇到了问题。

所以我试图限制数据并通过SELECT field1，field2 FROM table将所需字段仅发送到浏览器，而不是所有字段。

一切都很好。 我只能怀疑jQuery可以解析的JSON数据量有限制。

我知道你没有使用jQuery但我只是留下我的经验以防万一。