将带有多部分/表单数据的curl请求(POST)转换为scala代码

Question

我有下面的curl请求，我正试图在Scala中实现。

curl-u“username”-x POST“https://post_url.com”-h“content-type：multipart/form-data”-f“xmlrequest=@/home/@file.xml；type=text/xml”

我已经尝试了下面的，但我得到一个坏的请求。

val client = new DefaultHttpClient()
val requestEntity = MultipartEntityBuilder.create().addBinaryBody("xmlRequest",
  new File("/home/@file.xml")).build()
val post = new HttpPost("https://post_url.com")
post.addHeader(BasicScheme.authenticate(new
      UsernamePasswordCredentials(username, password), "UTF-8", false))
post.addHeader("Content-Type", "multipart/form-data")
post.setEntity(requestEntity)
val response = client.execute(post)
println(response.getStatusLine)

Answer 1

我已经能够通过使用另一个addBinaryBody函数来解决我的问题。

  val client = new DefaultHttpClient()
  val requestEntity = MultipartEntityBuilder.create().addBinaryBody("xmlRequest",
  new File("/home/@file.xml"), ContentType.DEFAULT_BINARY, "").build()
  val post = new HttpPost("https://post_url.com")
  post.addHeader(BasicScheme.authenticate(new
  UsernamePasswordCredentials(username, password), "UTF-8", false))
  post.addHeader("Content-Type", "multipart/form-data")
  post.setEntity(requestEntity)
  val response = client.execute(post)
  println(response.getStatusLine)