Question

我有一个Android应用程序，用户可以在其中生成SQLite表的转储文件，zip文件并上传。

转储和压缩工作。我能够生成zip文件，将其解压缩并获得原始的转储文件。

但是，上传无法正常工作。

这是我的代码。我有一个启动Runnable的函数，该函数先启动转储文件，然后启动zip，最后到达uploadFile函数：

new Thread(new Runnable() {
    public void run() {
        createDumpFile();
    }
}).start();


public int uploadFile(String sourceFileUri) {

    Log.e(TAG, "uploading file");
    Log.e(TAG, "sourceFileUri = " + sourceFileUri);

    String fileName = sourceFileUri;

    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(sourceFileUri);

    if (!sourceFile.isFile()) {

        dialog.dismiss();

        Log.e(TAG, "Source File not exist :"
                + sourceFileUri);

        return 0;
    }
    else {
        try {

            Log.e(TAG, "file exists");

            // open a URL connection to the Servlet
            FileInputStream fileInputStream = new FileInputStream(sourceFile);
            URL url = new URL("http://www.myurl.com/subfolder/uploadToServer.php");

            // Open a HTTP connection to the URL
            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);

            // create a buffer of maximum size
            bytesAvailable = fileInputStream.available();

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

            // read file and write it into form...
            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);

            }

            // send multipart form data necesssary after file data...
            dos.writeBytes(lineEnd);
            dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);

            // Responses from the server (code and message)
            serverResponseCode = conn.getResponseCode();
            String serverResponseMessage = conn.getResponseMessage();

            Log.e(TAG, "HTTP Response is : "
                    + serverResponseMessage + ": " + serverResponseCode);

            if(serverResponseCode == 200){

                Log.e(TAG, "Upload success!");

                InputStream input = conn.getInputStream();
                BufferedReader reader = new BufferedReader(new InputStreamReader(input));
                StringBuilder result = new StringBuilder();
                String line;

                while ((line = reader.readLine()) != null) {
                    result.append(line);
                }

                Log.e(TAG, "result is = " + result.toString()); 
            }

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

        } catch (MalformedURLException ex) {

            dialog.dismiss();
            ex.printStackTrace();

            Log.e(TAG, "error: " + ex.getMessage(), ex);
        } catch (Exception e) {

            dialog.dismiss();
            e.printStackTrace();

            Log.e(TAG, "Exception : " + e.getMessage(), e);
        }
        dialog.dismiss();
        return serverResponseCode;

    } // End else block
}

这是接收文件的PHP代码：

<?php

    $file_path = "database/";

    $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";
    }
 ?>

返回的响应代码为200，大多数在线指南认为该指标足以说明文件已上传。但是，当我使用FileZilla检查服务器中的“数据库”文件夹时，发现文件没有上传。

这促使我检查PHP的echo语句。很快，我看到echo语句为“ fail”。

我尝试查找其他指南，但是它们大致相同，除了传递给uploadFile函数的String和对mainUIThread的少量更新之外。

此代码在哪里出错？

Answer 1

代码有效。只要确保您对文件所在的文件夹具有适当的读写权限即可。否则，您将无法在其中进行写入。

如果您拥有对服务器的ssh访问权限，请执行以下一种方法：

sudo chmod 777 /path/to/your/folder/here

在Android中将zip文件上传到服务器失败

1 个答案: