php无法使用HttpURLConnection接收Android发送的数据

时间:2015-04-27 23:14:14

标签: php android database

我尝试使用OutputStreamWriter将一些数据从Android发送回服务器(数据库),但是我无法在数据库中看到任何结果并且Logcat很好。我试过在stackoverflow上发布的解决方案,但没有运气。这是我的代码: -

private class sentFeedback extends AsyncTask<String,Void,String>{
    @Override
    protected String doInBackground(String... params) {
        if(isNetworkAvailable()){
            try{
                URL feedback = new URL("http:.../setFeedback.php");
                HttpURLConnection conn = (HttpURLConnection) feedback.openConnection();
                conn.setDoOutput(true);
                conn.setRequestMethod("POST");
                OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
                String data = getData(params);
                wr.write(data);
                wr.flush();

            } catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }else{
            return "No internet connection";
        }
        return null;
    }

    public String getData(String... params) throws UnsupportedEncodingException {

        String data = "id" + "=" + URLEncoder.encode(params[0], "UTF-8");
        data += "&" + "rating" + "=" + URLEncoder.encode(params[1], "UTF-8");
        data += "&" + "message" + "=" + URLEncoder.encode(params[2], "UTF-8");
        return data;
    }
}

php代码: -

$id = $_POST['id'];
$rating = (double)$_POST['rating'];
$message = $_POST['message'];

$feedbackQuery = "INSERT INTO feedback(message, id, rating) VALUES ('$message', '$id', '$rating')";
$feedbackQueryResult = $mysqli->query($feedbackQuery);

$mysqli->close();

目前我正在使用免费的000webhost帐户。这是我可以发回数据的原因吗?

提前致谢!!

1 个答案:

答案 0 :(得分:0)

php文件中的

tablenName替换为查询字符串中的dbName.tableName,例如,如果db name为mydb

$feedbackQuery = "INSERT INTO mydb.feedback(message, id, rating) VALUES ('$message', '$id', '$rating')";

再次测试,在一个项目中,我的问题用这个解决了