Question

我正在尝试将一些压缩数据和文本发布到Web服务器上的perl程序，但无法将标头从octet-stream更改为multipart / form-data。

我的代码是： -

HttpClient httpclient = new DefaultHttpClient();
String url = "http://webaddress/perl.pl";
HttpPost httppost = new HttpPost(url);
httppost.setHeader("Content-Type","multipart/form-data");
MultipartEntity entity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);

try {

    entity.addPart("mtdata",new ByteArrayBody(data, file.getLastPathSegment()));
    entity.addPart("email", new StringBody(strUserName));
    entity.addPart("mtfilename", new StringBody(file.getLastPathSegment()));
    httppost.setEntity(entity);

    HttpResponse httpresponse = httpclient.execute(httppost);
    HttpEntity resEntity = httpresponse.getEntity();

    response = EntityUtils.toString(resEntity);
}

收到的原始数据是： -

Buffer = --FfT4ZNRUCPw6yONkpSsXNkA3WA2l6fvy53
Content-Disposition: form-data; name="mtdata"; filename="filename"
Content-Type: application/octet-stream             <<=== cannot change this

... Some binary data ...

我做错了什么？

Answer 1

最后，我发现最好的方法是将其保存为文件，然后使用MultipartEntity发布文件，并使用FileBody明确说明Content-Type。

我使用的代码是： -

pairs.addPart("File", new FileBody(fout,"multipart/form-data"));

另外，我还发现我需要使用ZipOutputStream而不是GZIPOutputStream，以便我可以添加未压缩的文件名。

代码是： -

data = bos.toByteArray();

OutputStream fos = new FileOutputStream(extStorageDirectory +  newfilename);
ZipOutputStream zos = new ZipOutputStream(new BufferedOutputStream(fos));
try {
    ZipEntry entry = new ZipEntry(file.getLastPathSegment());
    zos.putNextEntry(entry);
    zos.write(data);
    zos.closeEntry();
}
finally {
    zos.close();
}
fos.close();

Answer 2

尝试插入

httppost.addHeader("Content-Type", "multipart/form-data");

并删除

MultipartEntity entity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);

无法将标头设置为HttpPost的multipart / form-data

2 个答案: