Php 如何将数据从android发送到mysql？

我想从android发送数据到php文件，该文件正在使用POST方法将数据添加到mysql数据库中，我不知道我是如何做到这一点的。。有什么帮助吗

我已经通过方法post连接了http。。不知道现在发生了什么

public URL makeUrl(String stringUrl) {
        URL url = null;
        try {
            url = new URL(stringUrl);
        } catch (MalformedURLException e) {
            Log.e(LOG_TAG, "Problem building the URL ", e);
        }
        return url;
    }

public static void HttpConnect(URL url) {

        HttpURLConnection urlConnection = null;
        InputStream inputStream = null;
        try {
            urlConnection = (HttpURLConnection) url.openConnection();
            urlConnection.setConnectTimeout(15000 /* milliseconds */);
            urlConnection.setRequestMethod("POST");
            urlConnection.connect();

            // If the request was successful (response code 200),
            // then read the input stream and parse the response.
            if (urlConnection.getResponseCode() == 200) {
                //to do
            } else {
                Log.e(LOG_TAG, "Error response code: " + urlConnection.getResponseCode());
            }
        } catch (IOException e) {
            Log.e(LOG_TAG, "Problem retrieving the JSON results.", e);
        } finally {
            if (urlConnection != null) {
                urlConnection.disconnect();
            }

        }

    }

用POST方法添加的PHP文件

<?php 
if ($_SERVER['REQUEST_METHOD'] == 'POST'){

    require 'connect.php';
    create();
}

function create(){
    global $connect;

    $name = $_POST["nazwa"];
    $ingredients = $_POST["skladniki"];
    $price = $_POST["cena"];
    $type = $_POST["typ"];
    $photo = $_POST["zdjecie"];

    $query = "INSERT INTO dania('nazwa','skladniki','cena','typ','zdjecie') VALUES ('$name','$ingredients','$price','$type','$photo');";

    mysqli_query($connect, $query) or die (mysqli_error($connect));
    mysqli_close($connect);
}



?>

---

使用Uri.Builder附加POST有效负载并将其写入OutputStream

在下面的代码中，将您的_值1、_值2、_值3、_值4、_值5替换为您希望发布到服务器的合适值

我还建议将这段代码包装在AsyncTask中

例如

---

对于完整的示例，我已经在我的

中写了关于AsyncTask的内容。这是临时的，我想先添加come内容，以检查所有内容是否一起工作。

public static void HttpConnect(URL url) {

    StringBuilder stringBuilder = new StringBuilder();

        try {

            HttpURLConnection httpURLConnection = (HttpURLConnection)url.openConnection();
            httpURLConnection.setRequestMethod("POST");
            httpURLConnection.setConnectTimeout(10000);
            httpURLConnection.setReadTimeout(15000);
            httpURLConnection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
            httpURLConnection.setDoInput(true);
            httpURLConnection.setDoOutput(true);

            Uri.Builder builder = new Uri.Builder();
            builder.appendQueryParameter("nazwa",YOUR_VALUE1);
            builder.appendQueryParameter("skladniki",YOUR_VALUE2);
            builder.appendQueryParameter("cena",YOUR_VALUE3);
            builder.appendQueryParameter("typ",YOUR_VALUE4);
            builder.appendQueryParameter("zdjecie",YOUR_VALUE5);

            String urlQuery = builder.build().getEncodedQuery();

            OutputStream outputStream = httpURLConnection.getOutputStream();

            BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(outputStream,"UTf-8"));
            bufferedWriter.write(urlQuery);
            bufferedWriter.flush();
            bufferedWriter.close();
            outputStream.close();

            httpURLConnection.connect();

            InputStream inputStream = httpURLConnection.getInputStream();

            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream,"UTF-8"));

            String line = "";

            while ((line = bufferedReader.readLine()) != null){
                stringBuilder.append(line).append("\n");
            }

            // your json response output is here
            Log.d("RESULT",stringBuilder.toString());

        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (ProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }


}