如何读取响应标头 +使用 scala 的调度库从单个 POST 获取正文
我一直在尝试使用 dispatch 库通过 http POST 请求下载文件。服务器返回一个“content-disposition”标头，建议其返回的数据文件的文件名。
我已成功将整个响应正文作为字符串读取，
http(r >~ { (x) => println(x.getLines.mkString("","\n","")) })
自行读取响应标头
http(r >:> { (x) => println(x) })
并获取响应正文的 BufferedReader
http(r >> { (x,c) => (new BufferedReader(new InputStreamReader(x,c))).readLine })
我将如何使用调度库一次性获取响应标头和响应正文？文档非常稀疏，而且我是 Scala 新手。
蒂亚
·迈克尔
I have been trying to use the dispatch library to download a file via an http POST request. The server returns a "content-disposition" header suggesting a filename for the data file it returns.
I have succeeded reading the entire response body as a string,
http(r >~ { (x) => println(x.getLines.mkString("","\n","")) })
reading the response headers on their own
http(r >:> { (x) => println(x) })
and getting a BufferedReader for the response body
http(r >> { (x,c) => (new BufferedReader(new InputStreamReader(x,c))).readLine })
How would I go about getting the response headers AND the response body in one go using the dispatch lib? The docs are very sparse and I am new to Scala.
TIA
Michael
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Dispatch 使用 Handler 来处理来自请求的 HTTP 响应。它提供了一些用于执行例行任务的便捷快捷方式，例如您在问题中概述的快捷方式，例如生成 InputStream、以字符串形式返回内容或查看响应的标头。还有一个方法，
>+，它组成两个独立的 Handler 并在同一个请求上执行它们。以下是使用该处理程序解决问题的方法：返回值是一个 Tuple2，在本例中，它包含一个字符串（网页内容）和另一个字符串（Content-Disposition 标头的值）。
Dispatch uses Handlers to handle HTTP responses from a request. It provides several handy shortcuts for performing routine tasks, like the ones that you outlined in the question such as generating an InputStream, returning the content as a string, or looking at the headers of the response. There is also a method,
>+, which composes two separate Handlers and executes them on the same request. Here's how you could solve your problem using that handler:The return value is a Tuple2 which contains, in this case, a string which is the content of the web page and another string which is the value of the Content-Disposition header.