【问题标题】:Upload a file plus string from android to server not working将文件和字符串从android上传到服务器不起作用
【发布时间】:2012-10-04 00:01:24
【问题描述】:

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,但我不能发送 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'");

?>

这里是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以发送到 php 端! 非常感谢!

【问题讨论】:

    标签: php android file upload webserver


    【解决方案1】:

    您可以通过MultipartEntity上传图片。

    MultipartEntityHttpMime 4.0 和更高版本的一部分。允许您将多个部分(由边界字符串分隔并使用给定字符集编码)放入 httppost 请求中。

    有关详细信息以及如何使用 Multipart,请参阅 thisthis

    【讨论】:

    • 感谢您的回复!这是一个很好的解决方案,但对我来说它不起作用。每次我得到 Dalvik 错误,我已经尝试过建议的技巧......所以我提出请求,它并不优雅,但有效。谢谢!
    • 我认为那个 jar 文件导致了 Dalvik 问题,但没有。每次都会出现,不知道怎么解决。
    • 我已经解决了这个问题,现在它适用于您的解决方案。很抱歉,感谢您的帮助!
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