Question

根据这个回答"Android upload video to remote server using HTTP multipart form data"我做了所有步骤。

但我不知道我如何为服务器端编码！我的意思是一个PHP简单的页面，为我最糟糕的上传提供服务。

另一个问题是：YOUR_URL（以下代码段的第3行）必须是该PHP页面的地址吗？

private void uploadVideo(String videoPath) throws ParseException, IOException {

    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost(YOUR_URL);

    FileBody filebodyVideo = new FileBody(new File(videoPath));
    StringBody title = new StringBody("Filename: " + videoPath);
    StringBody description = new StringBody("This is a description of the video");

    MultipartEntity reqEntity = new MultipartEntity();
    reqEntity.addPart("videoFile", filebodyVideo);
    reqEntity.addPart("title", title);
    reqEntity.addPart("description", description);
    httppost.setEntity(reqEntity);

    // DEBUG
    System.out.println( "executing request " + httppost.getRequestLine( ) );
    HttpResponse response = httpclient.execute( httppost );
    HttpEntity resEntity = response.getEntity( );

    // DEBUG
    System.out.println( response.getStatusLine( ) );
    if (resEntity != null) {
      System.out.println( EntityUtils.toString( resEntity ) );
    } // end if

    if (resEntity != null) {
      resEntity.consumeContent( );
    } // end if

    httpclient.getConnectionManager( ).shutdown( );
}

Answer 1

这段代码工作正常，我应该使用的PHP代码就像这样简单：

<?php

    $file_path = "uploads/";

    $file_path = $file_path . basename( $_FILES['videoFile']['name']);
    if(move_uploaded_file($_FILES['videoFile']['tmp_name'], $file_path)) {
        echo "success";
    } else{
        echo "upload_fail_php_file";
    }
 ?>

注意，videoFile必须与

完全匹配

reqEntity.addPart("videoFile", filebodyVideo);

您可能遇到的最重要的问题是服务器配置中的post_max_size和upload_max_filesize的默认值！由于默认值太小，当您尝试上传大文件时，PHP脚本返回：＆＃34; upload_fail_php_file＆＃34; ，没有错误或异常抛出。 请记住将这些值设置得足够大......

享受编码。

Android：使用MultipartEntity上传大文件

1 个答案: