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我想使用 php 将文件从 android 设备上传到服务器。此外，我想使用 WHERE 语句将文件名发送到数据库。我可以上传我的文件，但我无法发送 WHERE 语句，我不知道在哪里以及如何将它放在 android 源中。

这是我的php代码：

<form action="FileUp.php" method="post" enctype="multipart/form-data">
</form>

<?php

mysql_connect('localhost','root','root');
mysql_select_db('db');

$target_path  = "./";
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);
$file = basename( $_FILES['uploadedfile']['name']);
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)){
echo "The file ".  basename( $_FILES['uploadedfile']['name']).
" has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
$ident=addslashes($_POST['ident']);

mysql_query ("UPDATE table SET column = CONCAT(column,'$file') WHERE id = '$ident'");

?>

这是Java代码：

@Override
        public void onClick(View v) {
            // TODO Auto-generated method stub

            String exsistingFileName = file.getText().toString();
            String lineEnd = "\r\n";
            String twoHyphens = "--";
            String boundary = "*****";
            try {
                // ------------------ CLIENT REQUEST

                Log.e(Tag, "Inside second Method");

                FileInputStream fileInputStream = new FileInputStream(new File(
                        exsistingFileName));

                // open a URL connection to the Servlet

                URL url = new URL(urlString);

                // Open a HTTP connection to the URL

                conn = (HttpURLConnection) url.openConnection();

                // Allow Inputs
                conn.setDoInput(true);

                // Allow Outputs
                conn.setDoOutput(true);

                // Don't use a cached copy.
                conn.setUseCaches(false);

                // Use a post method.
                conn.setRequestMethod("POST");

                conn.setRequestProperty("Connection", "Keep-Alive");

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

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

                dos.writeBytes(twoHyphens + boundary + lineEnd);
                dos
                        .writeBytes("Content-Disposition: post-data; name=uploadedfile;filename="
                                + exsistingFileName + "" + lineEnd);
                dos.writeBytes(lineEnd);

                Log.e(Tag, "Headers are written");

                // create a buffer of maximum size

                int bytesAvailable = fileInputStream.available();
                int maxBufferSize = 1000;
                // int bufferSize = Math.min(bytesAvailable, maxBufferSize);
                byte[] buffer = new byte[bytesAvailable];

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

                int bytesRead = fileInputStream.read(buffer, 0, bytesAvailable);

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

                // send multipart form data necesssary after file data...

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

                // close streams
                Log.e(Tag, "File is written");
                fileInputStream.close();
                dos.flush();
                dos.close();

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

            catch (IOException ioe) {
                Log.e(Tag, "error: " + ioe.getMessage(), ioe);
            }

            try {
                BufferedReader rd = new BufferedReader(new InputStreamReader(conn
                        .getInputStream()));
                String line;
                while ((line = rd.readLine()) != null) {
                    Log.e("Dialoge Box", "Message: " + line);
                }
                rd.close();

            } catch (IOException ioex) {
                Log.e("MediaPlayer", "error: " + ioex.getMessage(), ioex);
            }

请帮助我需要将身份字符串放在哪里才能发送到php端！非常感谢！

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1 回答 1

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您可以通过MultipartEntity.

MultipartEntity，部分HttpMime 4.0和以后。允许您将由边界字符串分隔并使用给定字符集编码的多个部分放入 httppost 请求中。

有关更多信息以及如何使用 Multipart，请参阅this和this。

于 2012-10-13T08:47:14.487 回答