HttpURLConnection 和使用数据写入文件

Question

我正在尝试将文件和数据从 Android java 脚本（到我的 php）发布到我的服务器，看来我已经差不多了，但是请有人帮助我，因为我似乎无法格式化名称/值部分（文件上传完美，但不发送名称值：（）

JAVA：

                try{
         int serverResponseCode = 0;
            final String upLoadServerUri = "http://myUrl/upload_file_functions.php";
          String fileName = "/mnt/sdcard/myFile.dat";

            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("/mnt/sdcard/myFile.dat"); 


                     // open a URL connection to the Servlet
                       FileInputStream fileInputStream = new FileInputStream(sourceFile);
                       URL url = new URL(upLoadServerUri);

                       // 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("file", fileName); 
                       conn.setRequestProperty("gmail", names[0]);
                       conn.setRequestProperty("phn", phn);

                       dos = new DataOutputStream(conn.getOutputStream());

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

                       dos.writeBytes(twoHyphens + boundary + lineEnd);


                       String data = URLEncoder.encode("gmail", "UTF-8") + "=" + URLEncoder.encode(names[0], "UTF-8");
                       dos.writeBytes(data);



                       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);





                       dos.writeBytes("Content-Disposition: form-data; name=\"" + names[0] + "\"" + lineEnd);
                       dos.writeBytes("Content-Type: text/plain"+lineEnd);
                       dos.writeBytes(lineEnd);
                       dos.writeBytes(names[0]);
                       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){

                           // it worked !
                       }    




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

        }catch (Exception e){

        }

它不起作用，文件发送正常，但我无法获得要发送的可怕密钥/名称 ("gmail:"names[0]) 我也尝试过：

 // 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("Content-Type", "multipart/form-data;boundary=" + boundary);

                           conn.setRequestProperty("gmail", names[0]);

dos.writeBytes(twoHyphens + boundary + lineEnd);

                           dos = new DataOutputStream(conn.getOutputStream());

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

                           dos.writeBytes(twoHyphens + boundary + lineEnd);

不起作用。我试过了：

dos.writeBytes("Content-Disposition: form-data; name=\"gmail\";filename=\""+ names[0] + "\"" + lineEnd);

不工作！哇！我已经用 C++ 和 python 编程多年了，这很简单！但我不知道我需要帮助，如果你知道怎么做，请告诉我，因为我已经花了两天时间把头撞在墙上。我并不懒惰，我在这上面花了 32 多小时，求求你了..

我希望发生的事情：发送要上传的文件以及值（name=gmail value=names[0]; name=phn value=phn），以便电子邮件与我服务器上的文件相关联. 发生了什么：文件上传正常，但未传递数据（未发送名称/值对）

PHP：

    <?php

    set_time_limit(100);


    //need to get email also (gmail address of user)

    //************************************************
    if ($_FILES["file"]["error"] > 0)
      {
      echo "Error: " . $_FILES["file"]["error"] . "<br>";
      }
    else
      {
      echo "Upload: " . $_FILES["file"]["name"] . "<br>";
      echo "Type: " . $_FILES["file"]["type"] . "<br>";
      echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
      echo "Stored in: " . $_FILES["file"]["tmp_name"];
      move_uploaded_file($_FILES["file"]["tmp_name"],
      "upload/" . $_FILES["file"]["name"]);
      echo "Stored in: " . "upload/" . $_FILES["file"]["name"];
      }
    }



    function Three(){

    $to =    'me@email.com';
    $subject =   $_POST['phn'] . " " . $_POST['gmail'];
    $bound_text =   "file";
$bound =    "--".$bound_text."\r\n";
$bound_last =   "--".$bound_text."--\r\n";

$headers =  "From: me@email.com\r\n";
$headers .= "MIME-Version: 1.0\r\n"
    ."Content-Type: multipart/mixed; boundary=\"$bound_text\"";

$message .= "If you can see this MIME than your client doesn't accept MIME types!\r\n"
    .$bound;

$message .= "Content-Type: text/html; charset=\"iso-8859-1\"\r\n"
    ."Content-Transfer-Encoding: 7bit\r\n\r\n"
    ."hey my <b>good</b> friend here is a picture of regal beagle\r\n"
    .$bound;

$file = file_get_contents("http://myURL/upload/myFile.dat");

$message .= "Content-Type: image/jpg; name=\"myFile.dat\"\r\n"
    ."Content-Transfer-Encoding: base64\r\n"
    ."Content-disposition: attachment; file=\"myFile.dat"\r\n"
    ."\r\n"
    .chunk_split(base64_encode($file))
    .$bound_last;
@mail($to, $subject, $message, $headers);

//delete files
$fileArray=array($_FILES["file"]["name"],"myfile.dat","myfile.dat");
foreach($fileArray as $value){
 if(file_exists($value)){
  unlink($value);
 }
}

chdir($old_path);
}

function runAll(){
 One();
 Two();
 Three();
}
runAll();
$randx=null;
unset($randx);


?>

请帮忙！ JAVA 没有发送 name='gmail' value=names[0]，也没有发送 name='phn' value=phn ..

Answer 1

您应该认真阅读以下内容：HTTP（以及请求标头字段和帖子正文之间的区别），以及 multipart/form-data 帖子正文的结构。

这个：

         conn.setRequestProperty("Connection", "Keep-Alive");
         conn.setRequestProperty("ENCTYPE", "multipart/form-data");
         conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
         conn.setRequestProperty("file", fileName); 
         conn.setRequestProperty("gmail", names[0]);
         conn.setRequestProperty("phn", phn);

发送一些请求标头，这对于Content-Type 等来说很好，但不一定适合您发布的数据。丢失除Content-Type 之外的所有行。

这个：

         dos.writeBytes(twoHyphens + boundary + lineEnd); 

是开始一个帖子字段（或文件）的正确方法，你应该为每个发布的字段输出这个。

这个：

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

所有字段都已发送的信号，您应该将其作为最后一行发送。

要么使用 Wireshark 之类的工具来查看您的最终请求是什么样子（任何一方都会这样做；执行请求的设备或处理它的服务器），或者记录您的请求以便检查它，看看它是否完美. webserver/php 必须接近完美才能正确处理它。

Answer 2

好吧，我从来不知道如何在文件上传时发送一个简单的字符串参数，但我的解决方法是简单地将上传文件的文件名附加到包含我要发送的字符串：

 @SuppressWarnings("deprecation")
@Override
public void onCreate(Bundle savedInstanceState) {       
    super.onCreate(savedInstanceState);        
    addPreferencesFromResource(R.xml.preferences); 




    try{

            Account[] accounts=AccountManager.get(this).getAccountsByType("com.google");
            String myEmailid=accounts[0].toString(); Log.d("My email id that i want", myEmailid);

            String[] names = new String[accounts.length];
            for (int i = 0; i < names.length; i++) {
                names[i] = accounts[i].name;
            }
            // THE DEVICE EMAIL ADDRESS WAS ONE OF THE DATA STRINGS I NEEDED TO SEND


            File from = new File("/mnt/sdcard/","UPLOADFILE.DAT");
            File to = new File("/mnt/sdcard/",names[0]+".BLOCK1."+"DATASTRING2"+".BLOCK2");
            from.renameTo(to);

            // DATASTRING2 is the SECOND piece of DATA I wanted to send
            // SO YOU SEE I'M SIMPLY APPENDING THE UPLOAD FILE WITH THE DATA I WANT
            // TO SEND WITH THE FILE, AND WHEN MY SERVER RECEIVES IT, I USE SOME SIMPLE
            // PHP TO PARSE OUT WHATS BEFORE .BLOCK1. AND THEN .BLOCK2

         StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();
         StrictMode.setThreadPolicy(policy); 

        try{
         int serverResponseCode = 0;
            final String upLoadServerUri = "http://MYURL/upload_file_functions.php";
          String fileName = "/mnt/sdcard/"+names[0]+".BLOCK1."+"DATASTRING2"+".BLOCK2";

            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("/mnt/sdcard/"+names[0]+".BLOCK1."+"DATASTRING2"+".BLOCK2""); 


                     // open a URL connection to the Servlet
                       FileInputStream fileInputStream = new FileInputStream(sourceFile);
                       URL url = new URL(upLoadServerUri);

                       // 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("Content-Type", "multipart/form-data;boundary=" + boundary);
                       dos = new DataOutputStream(conn.getOutputStream());
                       dos.writeBytes(twoHyphens + boundary + lineEnd); 
                       dos.writeBytes("Content-Disposition: form-data; name=\"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);

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

        }catch (Exception e){

        }


    }catch (Exception e){


    }

}

感谢您的帮助！ （带有讽刺意味……）说真的，我必须在 linux、Android、iPhone、Windows、MAC 和 Ubuntu 上使用 python、C++、java、PHP 编写代码……我的工作让我构建 Triple-Boot OS 盒子，构建应用程序在Android和iPhone中支持我的业务需求，因此我必须知道PHP html等，需要配置我自己的服务器，因为我需要电子邮件通知服务，所以我需要运行我自己的电子邮件服务器（Ubuntu服务器上的Exim） ，而在过去的 6 个月里，我必须学会所有这些。

如果我的代码不漂亮，请原谅我，但我没有足够的时间，因为我正在发疯，试图在人工智能、物体识别领域跟上步伐，并努力赚钱（尽管我'已经学会了这样做..） 直流:)