Php 如何使用Android Studio将带有文本参数的图像上载到服务器

我是Android的新手。最近，我正在开发一个应用程序，使用Android Studio上传带有文本的图像

我在开发应用程序时遇到了一些问题。我可以把图像上传到服务器上。但是，我在上传带有图像的文本时遇到了一个问题。我真的不知道哪里出了问题。我希望我能从这里得到一些指导，你可以自由地修改代码

非常感谢你的帮助。

MainActivity.java

private void uploadVideo(final String name) {

    class UploadVideo extends AsyncTask<Void, Void, String> {

        ProgressDialog uploading;

        @Override
        protected void onPreExecute() {
            super.onPreExecute();
            uploading = ProgressDialog.show(MainActivityUpload.this, "Uploading File", "Please wait...", false, false);
        }

        @Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
            uploading.dismiss();
            textViewResponse.setText(Html.fromHtml("<b>Uploaded at <a href='" + s + "'>" + s + "</a></b>"));
            textViewResponse.setMovementMethod(LinkMovementMethod.getInstance());
        }

        @Override
        protected String doInBackground(Void... params) {
            upload u = new upload();
            String msg = u.uploadVideo(selectedPath , name); //the name is passed uploadVideo function in upload.java
            return msg;
        }
    }
    UploadVideo uv = new UploadVideo();
    uv.execute();
}

public class upload {

public static final String UPLOAD_URL= "http://192.168.1.10/VideoUpload/upload.php";

private int serverResponseCode;

public String uploadVideo(String file , String name) {

    String fileName = file;
    HttpURLConnection conn = null;
    DataOutputStream dos = null;
    String lineEnd = "\r\n";
    String twoHyphens = "--";
    String boundary = "*****";
    int bytesRead, bytesAvailable, bufferSize;
    byte[] buffer;
    int maxBufferSize = 1 * 1024 * 1024;

    File sourceFile = new File(file);
    if (!sourceFile.isFile()) {
        Log.e("Huzza", "Source File Does not exist");
        return null;
    }

    try {
        FileInputStream fileInputStream = new FileInputStream(sourceFile);
        URL url = new URL(UPLOAD_URL);
        conn = (HttpURLConnection) url.openConnection();
        conn.setDoInput(true);
        conn.setDoOutput(true);
        conn.setUseCaches(false);
        conn.setRequestMethod("POST");
        conn.setRequestProperty("Connection", "Keep-Alive");
        conn.setRequestProperty("ENCTYPE", "multipart/form-data");
        conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
        conn.setRequestProperty("myFile", fileName);
        conn.setRequestProperty("name", name); //the name of the image is passed here
        dos = new DataOutputStream(conn.getOutputStream());

        dos.writeBytes(twoHyphens + boundary + lineEnd);
        dos.writeBytes("Content-Disposition: form-data; name=\"myFile\";filename=\"" + fileName + "\"" + lineEnd);
        dos.writeBytes(lineEnd);

        bytesAvailable = fileInputStream.available();
        Log.i("Huzza", "Initial .available : " + bytesAvailable);

        bufferSize = Math.min(bytesAvailable, maxBufferSize);
        buffer = new byte[bufferSize];

        bytesRead = fileInputStream.read(buffer, 0, bufferSize);

        while (bytesRead > 0) {
            dos.write(buffer, 0, bufferSize);
            bytesAvailable = fileInputStream.available();
            bufferSize = Math.min(bytesAvailable, maxBufferSize);
            bytesRead = fileInputStream.read(buffer, 0, bufferSize);
        }

        dos.writeBytes(lineEnd);
        dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);

        dos.writeBytes("Content-Disposition: form-data; name=\"name\"" + lineEnd);
        dos.writeBytes(lineEnd);
        dos.writeBytes(name + lineEnd);
        dos.writeBytes(twoHyphens + boundary + lineEnd);

        serverResponseCode = conn.getResponseCode();

        fileInputStream.close();
        dos.flush();
        dos.close();

    } catch (MalformedURLException ex) {
        ex.printStackTrace();
    } catch (Exception e) {
        e.printStackTrace();
    }

    if (serverResponseCode == 200) {
        StringBuilder sb = new StringBuilder();
        try {
            BufferedReader rd = new BufferedReader(new InputStreamReader(conn
                    .getInputStream()));
            String line;
            while ((line = rd.readLine()) != null) {
                sb.append(line);
            }
            rd.close();
        } catch (IOException ioex) {
        }
        return sb.toString();
    }else {
        return "Could not upload";
    }
}
}

upload.php

<?php

 if($_SERVER['REQUEST_METHOD']=='POST'){
 $file_name = $_FILES['myFile']['name'];
 $file_size = $_FILES['myFile']['size'];
 $file_type = $_FILES['myFile']['type'];
 $temp_name = $_FILES['myFile']['tmp_name'];
 $name      = $_POST['name'];

 require_once('dbConnect.php');

        $sql ="SELECT id FROM uploads ORDER BY id ASC";

        $res = mysqli_query($con,$sql);

        $id = 0;

        while($row = mysqli_fetch_array($res)){
                $id = $row['id'];
        }

 $location = "uploads/";

 $actualpath = "http://192.168.1.9/VideoUpload/$location$file_name";

 $sql = "INSERT INTO uploads (image,name) VALUES ('$actualpath','$name')";

     if(mysqli_query($con,$sql)){
                move_uploaded_file($temp_name, $location.$file_name);
                echo "Successfully Uploaded";
     }
     mysqli_close($con);
 }else{
 echo "Error";
 }
?>

---

我也有同样的问题，我意识到如果POST参数只是一个数字，那就没问题了，但其他方面也有问题。如果你得到了答案，请在这里解释一下。非常感谢。