在JAVA的HttpURLConnection中在一个post Request中发送多个文件

Question

我使用HttpURLConnection通过post方法发送文件。我用文件发送一个参数，这是'student_id'。当在每个post请求中发送一个文件时，代码工作正常。但是，我如何更新下面的代码在一个post请求中发送多个文件，其中所有文件都属于相同的'student_id'?

       try{  
            File textFile2 = new File("info.txt");  
         URL url = new URL("htttp://wwww.students.com");
         HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();         
         urlConnection.setChunkedStreamingMode(0);
         urlConnection.setDoOutput(true);   

        String boundary = Long.toHexString(System.currentTimeMillis()); // Just generate some unique random value.
        urlConnection.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + boundary);          
        OutputStream output = urlConnection.getOutputStream();
        PrintWriter writer = new PrintWriter(new OutputStreamWriter(output, charset), true);
        writer.append("--" + boundary).append(CRLF);
        writer.append("Content-Disposition: form-data; name="student_id"").append(CRLF);
        writer.append("Content-Type: text/plain; charset=" + charset).append(CRLF);
        writer.append(CRLF).append("25").append(CRLF).flush();

        writer.append("--" + boundary).append(CRLF);
        writer.append("Content-Disposition: form-data; name="newfile"; filename="" + textFile2.getName() + """).append(CRLF);
        writer.append("Content-Type: text/plain; charset=" + charset).append(CRLF); // Text file itself must be saved in this charset!
        writer.append(CRLF).flush();                                                            
        Files.copy(textFile2.toPath(), output);//copies all bytes in  a file to the output stream 
        output.flush(); // Important before continuing with writer!
        writer.append(CRLF).flush();
        writer.append("--" + boundary + "--").append(CRLF).flush();
        InputStream responseStream; 
        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

我试图在'newfile'中添加'multiple'参数，但它不起作用

        writer.append("--" + boundary).append(CRLF);
    writer.append("Content-Disposition: form-data; name="newfile" multiple; filename="" + textFile2.getName() + """).append(CRLF);
    writer.append("Content-Type: text/plain; charset=" + charset).append(CRLF); // Text file itself must be saved in this charset!
    writer.append(CRLF).flush();                                                            
    Files.copy(textFile2.toPath(), output);//copies all bytes in  a file to the output stream 
    output.flush(); // Important before continuing with writer!
    writer.append(CRLF).flush();
    writer.append("--" + boundary + "--").append(CRLF).flush();

Answer 1

似乎您正试图发布具有1 表单字段的multipart/form-data请求与student_id的name参数和多个文件部分来上传文件。

您可以通过在单独的部分中提供每个文件来发送多个文件，但所有文件都具有相同的name参数。

例如，上传textFile1的文件，发送第一个文件部分， newfile参数name:

writer.append("--" + boundary).append(CRLF);
writer.append("Content-Disposition: form-data; name="newfile"; filename="" + textFile1.getName()+ """).append(CRLF);
writer.append("Content-Type: text/plain; charset=" + charset).append(CRLF);
writer.append("Content-Transfer-Encoding: binary").append(CRLF);
writer.append(CRLF);
writer.flush();
FileInputStream inputStream = new FileInputStream(textFile1);
byte[] buffer = new byte[4096];
int bytesRead = -1;
while ((bytesRead = inputStream.read(buffer)) != -1) {
    outputStream.write(buffer, 0, bytesRead);
}
outputStream.flush();
inputStream.close();
writer.append(CRLF);
writer.flush();

然后，通过发送newfile的相同name参数的文件部分，可以上传textFile2的另一个文件:

writer.append("--" + boundary).append(CRLF);
writer.append("Content-Disposition: form-data; name="newfile"; filename="" + textFile2.getName()+ """).append(CRLF);
writer.append("Content-Type: text/plain; charset=" + charset).append(CRLF);
writer.append("Content-Transfer-Encoding: binary").append(CRLF);
writer.append(CRLF);
writer.flush();
FileInputStream inputStream = new FileInputStream(textFile2);
byte[] buffer = new byte[4096];
int bytesRead = -1;
while ((bytesRead = inputStream.read(buffer)) != -1) {
    outputStream.write(buffer, 0, bytesRead);
}
outputStream.flush();
inputStream.close();
writer.append(CRLF);
writer.flush();

正如您所看到的，除了要上传的文件之外，代码几乎是相同的。建议将代码放入一个方法中并调用它来发送每个文件部分。