我在用 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,并在战后解析它。
Log.d("RESULT", result);
该行发布了正确的结果:
2[{"longtitude":"32.32","latitude":"33.12"}]
但是
Log.d("JSON","Finished");
从来没有接到电话, 所以问题似乎出在这一行
JSONObject json_data = new JSONObject(result);
这件事取自教程,我在互联网和本网站上看到了很多示例,其中有一些指出的错误,但不是这个。
任何帮助都会很棒! 谢谢
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();
?>
最佳答案
您的 JSON 数据 2[{"longtitude":"32.32","latitude":"33.12"}]
无效(数字 2
无效) t 正确的 JSON 语法)。
我可以建议你实际上的意思
[{"longtitude":"32.32","latitude":"33.12"}]`
(即开头没有 2
)
您可以使用http://jsonlint.com/处的 validator 检查您的 JSON 代码。
关于java - 尝试从 MySQL 查询解析 JSON 时出现异常,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10721663/