Question

下面的代码将图像成功发送到服务器。

但我需要的是将图像描述和图像日期等参数添加到数据库中。

我不知道如何在HttURLconnection中添加参数。

    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);


            FileInputStream fileInputStream = new FileInputStream(sourceFile_imagepath);
            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);

            // 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.i("uploadFile", "HTTP Response is : "
                    + serverResponseMessage + ": " + serverResponseCode);

            if (serverResponseCode == 200) {

                runOnUiThread(new Runnable() {
                    public void run() {
                        String msg = "File Upload Completed.\n\n See uploaded file here : \n\n"
                                + " c:/wamp/www/echo/uploads";
                        messageText.setText(msg);
                        Toast.makeText(MainActivity.this,
                                "File Upload Complete.", Toast.LENGTH_SHORT)
                                .show();
                    }
                });
            }

            // close the streams //
            fileInputStream.close();
            dos.flush();
            dos.close();

和PHPCODE：

<?php
  error_reporting(0);
   $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";
   }
?>

上面的代码适用于将代码上传到服务器。

必须在db。中存储图像描述。

如何在上面的代码中添加参数。

如何在php服务器端接收它。

任何相关的教程或文档。

提前致谢。

Answer 1

选中此link

Android代码：

String parameter =“value”;

        dos.writeBytes("Content-Disposition: form-data; name=\"parameter\"" + lineEnd); 
        //dos.writeBytes("Content-Type: text/plain; charset=UTF-8" + lineEnd);
        //dos.writeBytes("Content-Length: " + parameter.length() + lineEnd);
        dos.writeBytes(lineEnd);
        dos.writeBytes(parameter); // mobile_no is String variable
        dos.writeBytes(lineEnd);

Php代码：

$parameter =$_POST['parameter'];

如何在httpurlconnection android中添加参数

1 个答案: