无法在Android的Mysql中插入数据

时间:2015-03-10 18:27:21

标签: php android mysql

我正在尝试连接Android,PHP和MySql。 为此,使用Android作为前端我试图将数据插入MySql Db。 我已经使用logcat输出将信息显示为适当的位置。 响应将在logcat中显示,但MySql中没有数据。 请帮帮我!!

这是我的代码:

public JSONObject getJsonData(String url_create_product, List<NameValuePair> parameters)
{

    try 
    {
        Log.e("url is:"," "+url_create_product.toString());

        DefaultHttpClient httpClient= new DefaultHttpClient();
        Log.e("httpClient is:"," "+httpClient.toString());

        HttpPost httpPost=new HttpPost(url_create_product);
        Log.e("httpPost is:"," "+httpPost.toString());

        httpPost.setEntity(new UrlEncodedFormEntity(parameters));

        Log.e("encoded parameters are: "," "+parameters.toString());

            try 
            {
                httpResponse=httpClient.execute(httpPost);
                Log.e("httpResponse is:"," "+httpResponse.toString());
            } 
            catch (Exception e) 
            {
                e.printStackTrace();
            }


        int statusCode=httpResponse.getStatusLine().getStatusCode();

        if (statusCode==200) 
        {
            Log.e("statusCode is:","200");
            HttpEntity httpEntity=httpResponse.getEntity();
            inputStream=httpEntity.getContent();

            Log.e("inputStream is:"," "+inputStream.toString());

            BufferedReader reader = new BufferedReader(new InputStreamReader(
                    inputStream, "iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) 
            {
                sb.append(line + "\n");
            }
            inputStream.close();
            json = sb.toString();
            Log.e("json is:"," "+json.toString());

            jsonObject=new JSONObject(json);
            Log.e("jsonObject is:"," "+jsonObject.toString());

        }
    } 

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


}

0 个答案:

没有答案