Question

我正在编写一个代码，允许我的用户上传任何类型的文件，这是我的代码：

Intent fileIntent = new Intent(Intent.ACTION_GET_CONTENT);
            fileIntent.setType("*/*"); // intent type to filter application based on your requirement
            startActivityForResult(fileIntent, 1);

用户可以选择他们的文件，然后通过以下代码获取路径和文件名：

@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    if (requestCode == 1) {
        if (resultCode == Activity.RESULT_OK) {
            String result = data.getStringExtra("result");
            Uri u = data.getData();
                filepath = u.getPath();
                filename = u.toString();
                chooseFile=true;



        }
    }
}

这是我将所选文件发送到服务器的代码，我使用assyctask。这是代码;

@Override
    protected String doInBackground(String... params) {
        if (filepath != null) {//file entekhab shode bud
            int serverResponseCode = 0;
            HttpURLConnection connection = null;
            DataOutputStream outputStream = null;
            DataInputStream inputStream = null;
            String urlServer = getString(R.string.url)+"/getFile.php?id="+sefareshId;
            String lineEnd = "\r\n";
            String twoHyphens = "--";
            String boundary = "*****";

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

            try {
                FileInputStream fileInputStream = new FileInputStream(new File(filepath));

                URL url = new URL(urlServer);
                connection = (HttpURLConnection) url.openConnection();

                // Allow Inputs &amp; Outputs.
                connection.setDoInput(true);
                connection.setDoOutput(true);
                connection.setUseCaches(false);

                // Set HTTP method to POST.
                connection.setRequestMethod("POST");

                connection.setRequestProperty("Connection", "Keep-Alive");
                connection.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);

                outputStream = new DataOutputStream(connection.getOutputStream());
                outputStream.writeBytes(twoHyphens + boundary + lineEnd);
                outputStream.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\"" + filepath + "\"" + lineEnd);
                outputStream.writeBytes(lineEnd);

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

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

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

                outputStream.writeBytes(lineEnd);
                outputStream.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
                serverResponseCode = connection.getResponseCode();

                fileInputStream.close();
                outputStream.flush();
                outputStream.close();
            } catch (Exception ex) {
                Log.v("this", ex.getMessage());
            }
            return serverResponseCode + "";
        } else
            return "5";

    }

通过运行此代码，我收到此错误：

/document/image:134342: open failed: ENOENT (No such file or directory)

我认为此文件的url路径错误，如何获取正确的文件路径然后上传？

谢谢

没有找到onActivityResult上的android-文件来选择文件

0 个答案: