Question

我在使用Java解析来自MySQL的JSON响应时遇到问题。

try {
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost("http://parkfinder.zxq.net/default.php");
    httppost.setEntity(new UrlEncodedFormEntity(coordinatesToSend));
    HttpResponse response = httpclient.execute(httppost);
    Log.d("HTTP Client", "HTTP Request made");

    HttpEntity entity = response.getEntity();
    inputStream = entity.getContent();
    BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream,
            "iso-8859-1"), 8);
    sb = new StringBuilder();
    sb.append(bufferedReader.readLine() + "\n");

    String line = "0";
    while ((line = bufferedReader.readLine()) != null) {
        sb.append(line + "\n");
    }
    inputStream.close();
    bufferedReader.close();
    result = sb.toString();
    Log.d("RESULT", result);
    JSONObject json_data = new JSONObject(result);
    Log.d("JSON","Finished");
    JSONArray nameArray = json_data.names();
    JSONArray valArray = json_data.toJSONArray(nameArray);
    for (int i = 0; i < nameArray.length(); i++) {
        Log.d("NAMES", nameArray.getString(i));
    }
    for (int i = 0; i < nameArray.length(); i++) {
        Log.d("NAMES", nameArray.getString(i));
    }

} catch (Exception e) {
    // TODO: handle exception
}

这是MySQL访问和检索信息，并在事后解析它。在

Log.d("RESULT", result);

发布正确的结果：

2[{"longtitude":"32.32","latitude":"33.12"}]

然而

Log.d("JSON","Finished");

永远不会被召唤，所以问题似乎就在这一行

JSONObject json_data = new JSONObject(result);

这件事情是从一个教程中得到的，我在互联网和本网站上看到了很多这样的例子，一些陈述的错误，但不是这个。

任何帮助都会很棒！感谢

编辑： printStackTrace（）输出：

0`5-14 21:38:18.639: WARN/System.err(665): org.json.JSONException: A JSONObject text must begin with '{' at character 1 of 2[{"longtitude":"32.32","latitude":"33.12"}]`

php代码：

<?php
$host = "localhost";
$user = "**MASKED**";
$password = "**MASKED**";
$database = "parkfinder_zxq_coordinates";
$connection = mysql_connect($host, $user, $password) or die("couldn't connect to server");
$db = mysql_select_db($database, $connection) or die("couldn't select database.");
//$request_parked = $_REQUEST['parked'];
$request_long = $_REQUEST['longtitude'];
$request_lat = $_REQUEST['latitude'];
//if ($request_parked == 'FIND') {
$q = mysql_query("SELECT * FROM Coordinates");
while ($e = mysql_fetch_assoc($q))
    $output[] = $e;

print (json_encode($output));
//}

mysql_close();
?>

Answer 1

您的JSON数据2[{"longtitude":"32.32","latitude":"33.12"}]无效（数字2不是正确的JSON语法）。

我可以建议你实际意味着

[{"longtitude":"32.32","latitude":"33.12"}]`

（即开头没有2）

您可以使用http://jsonlint.com/处的验证程序检查您的JSON代码。

尝试从MySQL查询解析JSON时出现异常

1 个答案: