將文件上傳到服務器Android

Question

我想將一些mp3文件上傳到我的本地主機服務器，但是我不確定我要去哪里。 我認為這是php文件，但這是我的php文件：

<?php
    $file_path = "/storage/emulated/0/sample.txt";
    $file_path = $file_path . basename( $_FILES['uploaded_file']['name']);
    if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $file_path)) {
        echo "success";
    } else{
        echo "fail";
    }
 ?>

這是我的java代碼：

public class UploadFile extends Activity {
    int serverResponseCode = 0;
    String serverResponseMessage;
    String uploadFileName;
    String uploadFilePath;
    String upLoadServerUri = null;
    ProgressBar loading;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.uploads);
        uploadFilePath = "/storage/emulated/0/";
        uploadFileName = "sample.txt";

        upLoadServerUri = "http://10.0.3.2/files/uploadfile.php";
        new Thread(new Runnable() {

            public void run() {
                uploadFile(uploadFilePath + "" + uploadFileName);
            }
        }).start();
    }

    public void uploadFile(String sourceFileUri) {
        String fileName = sourceFileUri;
        HttpURLConnection conn = null;
        DataOutputStream dos = null;
        String lineEnd = "\r\n";
        String twoHyphens = "--";
        String boundary = "*****";
        int code;

        int bytesRead, bytesAvailable, bufferSize;
        byte[] buffer;
        int maxBufferSize = 1 * 1024 * 1024;

        File sourceFile = new File(sourceFileUri);

        if (!sourceFile.isFile()) {
            code = 0;
            this.runOnUiThread(new Runnable() {
                public void run() {

                }
            });
        }
        else {
            try {
                FileInputStream fileInputStream = new FileInputStream(sourceFile);
                URL url = new URL(upLoadServerUri);

                conn = (HttpURLConnection) url.openConnection();
                conn.setDoInput(true); // Allow Inputs
                conn.setDoOutput(true); // Allow Outputs

                conn.setUseCaches(false); // Don't use a Cached Copy
                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("uploaded_file", fileName);
                dos = new DataOutputStream(conn.getOutputStream());
                dos.writeBytes(twoHyphens + boundary + lineEnd);
                dos.writeBytes("Content-Disposition: form-data; name='uploaded_file';filename='"+fileName+"'"+lineEnd);
                dos.writeBytes(lineEnd);
                bytesAvailable = fileInputStream.available();
                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);
                serverResponseCode = conn.getResponseCode();
                code = serverResponseCode;
                if (code==200) {
                    this.runOnUiThread(new Runnable() {
                        public void run() {

                        }
                    });
                }
                else {
                    this.runOnUiThread(new Runnable() {
                        public void run() {

                        }
                    });
                }
                serverResponseMessage = conn.getResponseMessage();
                fileInputStream.close();
                dos.flush();
                dos.close();
            }
            catch (MalformedURLException ex) {
                this.runOnUiThread(new Runnable() {
                    public void run() {

                    }
                });
            }
            catch (Exception e) {
                this.runOnUiThread(new Runnable() {
                    public void run() {

                    }
                });
            }
        }
    }
}

我將此代碼在線獲取並出於我的目的對其進行了修改-但它無法正常工作，但正如我所提到的，它可能是php文件

任何幫助都將真正起到很大的作用（即使對此進行修復，也可以讓我正常飲食和睡眠）

謝謝

Answer 1

使用以下php代碼更改您的代碼，即可正常運行。

<?php
$file_path = "uploads/";
    $file_path = $file_path . basename( $_FILES['uploaded_file']['name']);
    if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $file_path)) {
        echo "success";
    } else{
        echo "fail";
    }
 ?>

我已經測試過它工作完美

Answer 2

我認為您應該為此嘗試使用json解析器，將您的PHP腳本設置為輸出json，以獲取成功和失敗的信息。 您需要將Java代碼中的圖像文件名字符串解析為Web服務，並獲取它來執行操作。 您還可以使用$ _GET從Java代碼獲取文件