android 并将文件发送到HTTP服务器

发布于 2024-11-25 04:38:37 字数 3077 浏览 4 评论 0原文

我尝试编写将文件发送到 HTTP 服务器的应用程序。这是我的 android 端代码:

public void send(View view)
{
    HttpURLConnection connection = null;
    DataOutputStream outputStream = null;
    DataInputStream inputStream = null;

    String pathToOurFile = "/data/file.txt";
    String urlServer = "http://localhost/zad1.php";
    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(pathToOurFile) );

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

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

        // Enable POST method
        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=\"" + pathToOurFile +"\"" + 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);

        // Responses from the server (code and message)
        int serverResponseCode = connection.getResponseCode();
        String serverResponseMessage = connection.getResponseMessage();

        fileInputStream.close();
        outputStream.flush();
        outputStream.close();

        Log.i("ODPOWIEDZ", serverResponseMessage);
    }
    catch (Exception ex)
    {
        Log.i("WYJATEK", ex.getMessage());
    }
}

}

这是我的 php 脚本

<?php
    $target_path  = "./";
    $target_path = $target_path . 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!";
    }
?>

我的问题是,当我尝试将文件发送到服务器时,我收到“localhost/127.0.0.1:80 - 连接被拒绝”异常。你们有人知道我做错了什么吗?感谢您的每一个帮助和提示。

I try to write application which will be send file to HTTP server. Here is my android-side code:

public void send(View view)
{
    HttpURLConnection connection = null;
    DataOutputStream outputStream = null;
    DataInputStream inputStream = null;

    String pathToOurFile = "/data/file.txt";
    String urlServer = "http://localhost/zad1.php";
    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(pathToOurFile) );

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

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

        // Enable POST method
        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=\"" + pathToOurFile +"\"" + 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);

        // Responses from the server (code and message)
        int serverResponseCode = connection.getResponseCode();
        String serverResponseMessage = connection.getResponseMessage();

        fileInputStream.close();
        outputStream.flush();
        outputStream.close();

        Log.i("ODPOWIEDZ", serverResponseMessage);
    }
    catch (Exception ex)
    {
        Log.i("WYJATEK", ex.getMessage());
    }
}

}

and here is my php script

<?php
    $target_path  = "./";
    $target_path = $target_path . 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!";
    }
?>

My problem is that when I try to send file to server I get "localhost/127.0.0.1:80 - Connection refused" exception. Do someone of you know what I doing wrong? Thanks for every help and tip.

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。

评论(1

丑丑阿 2024-12-02 04:38:37

使用
导入 org.apache.http.entity.mime.MultipartEntity;

Use
import org.apache.http.entity.mime.MultipartEntity;

~没有更多了~
我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
原文