Lesson 08 · The request line - Build an HTTP Client

The request line

With a URL parsed, you can write the first line an HTTP server reads - the request line naming the method, the target, and the protocol version. Today you build the Request type and produce that line exactly.

The goal

Produce the request line for a method and URL, ending in a carriage return and line feed.

Start here - the target

TO DO

Scenario: Forming the request line

Givena Request with method "GET" and the URL parsed from "http://example.com/a/b"

Whenits request line is produced

Thenit is exactly "GET /a/b HTTP/1.1\r\n"

Anda URL with a query, "http://example.com/s?q=1", gives "GET /s?q=1 HTTP/1.1\r\n" (the target carries the query)

Background

An HTTP/1.1 request opens with a request line: the method, a space, the request-target, a space, the protocol version, and a CRLF - the two bytes \r\n that end every line in an HTTP message. For a normal request the target is the origin form: the URL’s path, with ? and the query appended when there is one. So GET on http://example.com/a/b produces GET /a/b HTTP/1.1\r\n.

Two details matter for exactness. The version is literally HTTP/1.1 - that is the protocol this whole project speaks. And the line ends in \r\n, not a bare newline; HTTP is strict about carriage-return-line-feed, and every line you emit from here on carries it. The path already defaults to / from chapter one, so the target is never empty.

Make it work

// the request line is: METHOD SP request-target SP "HTTP/1.1" CRLF
// the request-target is the URL's path, plus "?"+rawQuery if a
// query is present. CRLF is the two bytes "\r\n".
type Request struct { Method string; URL *URL }
func (r *Request) requestLine() string {
  // target := path; if rawQuery != "" { target += "?" + rawQuery }
}

CheckpointDONE

A Request can produce its request line with the correct target and CRLF ending. Commit and stop here.