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我正在尝试在 MYSQL 数据库中插入一个新行（以在该数据库中发布数据）

这是我的主要课程：

try {
            JSONObject json = new JSONObject();
            json.put("latitude", "4332314");
            json.put("longitude", "1234567");
            json.put("description", "Hahalela");
            HttpParams httpParams = new BasicHttpParams();
            HttpConnectionParams.setConnectionTimeout(httpParams,TIMEOUT_MILLISEC);
            HttpConnectionParams.setSoTimeout(httpParams, TIMEOUT_MILLISEC);
            HttpClient client = new DefaultHttpClient(httpParams);
            //
            //String url = "http://10.0.2.2:8080/sample1/webservice2.php?" + 
            //             "json={\"UserName\":1,\"FullName\":2}";
            String url = "http://localhost/writemysql.php";

            HttpPost request = new HttpPost(url);
            request.setEntity(new ByteArrayEntity(json.toString().getBytes(
                    "UTF8")));
            request.setHeader("json", json.toString());
            HttpResponse response = client.execute(request);
            HttpEntity entity = response.getEntity();
            // If the response does not enclose an entity, there is no need
            if (entity != null) {
                InputStream instream = entity.getContent();

                String result = RestClient.convertStreamToString(instream);
                Log.i("Read from server", result);
              //  Toast.makeText(this,  result,
                      //  Toast.LENGTH_LONG).show();
            }
        } catch (Throwable t) {
           // Toast.makeText(this, "Request failed: " + t.toString(),
             //       Toast.LENGTH_LONG).show();
        }

这是我的 RestClient 数据：

public class RestClient {

public static String convertStreamToString(InputStream is) {

BufferedReader reader = new BufferedReader(new InputStreamReader(is));
StringBuilder sb = new StringBuilder();

String line = null;
try {
    while ((line = reader.readLine()) != null) {
        sb.append(line + "\n");
    }
} catch (IOException e) {
    e.printStackTrace();
} finally {
    try {
        is.close();
    } catch (IOException e) {
        e.printStackTrace();
    }
}
return sb.toString();
}


  public static void connect(String url)
{

HttpClient httpclient = new DefaultHttpClient();

// Prepare a request object
HttpGet httpget = new HttpGet(url); 

// Execute the request
HttpResponse response;
try {
    response = httpclient.execute(httpget);
    // Examine the response status
    Log.i("Praeda",response.getStatusLine().toString());

    // Get hold of the response entity
    HttpEntity entity = response.getEntity();
    // If the response does not enclose an entity, there is no need
    // to worry about connection release

    if (entity != null) {

        // A Simple JSON Response Read
        InputStream instream = entity.getContent();
        String result= convertStreamToString(instream);
        Log.i("Praeda",result);

        // A Simple JSONObject Creation
        JSONObject json=new JSONObject(result);
        Log.i("Praeda","<jsonobject>\n"+json.toString()+"\n</jsonobject>");

        // A Simple JSONObject Parsing
        JSONArray nameArray=json.names();
        JSONArray valArray=json.toJSONArray(nameArray);
        for(int i=0;i<valArray.length();i++)
        {
            Log.i("Praeda","<jsonname"+i+">\n"+nameArray.getString(i)+"\n</jsonname"+i+">\n"
                    +"<jsonvalue"+i+">\n"+valArray.getString(i)+"\n</jsonvalue"+i+">");
        }

        // A Simple JSONObject Value Pushing
        json.put("sample key", "sample value");
        Log.i("Praeda","<jsonobject>\n"+json.toString()+"\n</jsonobject>");

        // Closing the input stream will trigger connection release
        instream.close();
    }


} catch (ClientProtocolException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
} catch (IOException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
} catch (JSONException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}
  }}

这是我的 php 脚本：

<?php

 $con = mysql_connect("localhost","root","mypassword");


 $d = mysql_select_db("radars");
 if ($json = file_get_contents('php://input')) {
// Never mind. We'll do it ourselves.
$a = json_decode($json, true); // Now we have a nice PHP array.
$insert = '';
foreach ($a as $v) {
    if ($insert) $insert .= ',';
    $insert .= ' ("' . mysql_real_escape_string($v['latitude']) . '", "' . mysql_real_escape_string($v['longitude']) . '", "' . mysql_real_escape_string($v['description']) . '")';

$sql = 'INSERT INTO `radars` (`latitude`, `longitude`, `description`) VALUES' . $insert;
mysql_query($sql);
}
}
?>

运行应用程序后，我在 logcat 中收到这两个：

 05-11 19:52:29.273: E/Buffer Error(532): Error converting result java.lang.NullPointerException
 05-11 19:52:29.294: E/JSON Parser(532): Error parsing data org.json.JSONException: End of input at character 0 of

有谁知道问题是什么，或者我应该改变什么才能通过 JSON 向 MYSQL 发送数据？欢迎任何提示或建议。谢谢 ！

4

1 回答 1

0

我认为 instream 是 null 因为 entity.getContent() 返回 null

...
    InputStream instream = entity.getContent();
    String result= convertStreamToString(instream);
...

你能提供更多的调试信息吗？

我们这样做的原因：

...
        request.setEntity(new ByteArrayEntity(json.toString().getBytes(
                "UTF8")));

        request.setHeader("json", json.toString());//<---that
...

尝试添加检查服务器答案（因为服务器不响应正文）：

...
HttpPost request = new HttpPost(url);
            request.setEntity(new ByteArrayEntity(json.toString().getBytes(
                    "UTF8")));
            request.setHeader("json", json.toString());
            HttpResponse response = client.execute(request);
            HttpEntity entity = response.getEntity();
            // If the response does not enclose an entity, there is no need
            if (entity != null) {
                InputStream instream = entity.getContent();
                //check instream for null
                if(insream!=null)
                {

                String result = RestClient.convertStreamToString(instream);
                Log.i("Read from server", result);
              //  Toast.makeText(this,  result,
                      //  Toast.LENGTH_LONG).show();
                }
...
于 2012-05-12T07:38:23.767 回答