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go/internal/lsp/source/uri.go
Ian Cottrell d0600fd9f1 internal/lsp: make source independent of protocol
I realized this was a mistake, we should try to keep the source
directory independent of the LSP protocol itself, and adapt in
the outer layer.
This will keep us honest about capabilities, let us add the
caching and conversion layers easily, and also allow for a future
where we expose the source directory as a supported API for other
tools.
The outer lsp package then becomes the adapter from the core
features to the specifics of the LSP protocol.

Change-Id: I68fd089f1b9f2fd38decc1cbc13c6f0f86157b94
Reviewed-on: https://go-review.googlesource.com/c/148157
Reviewed-by: Rebecca Stambler <rstambler@golang.org>
2018-11-07 18:42:35 +00:00

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Go

// Copyright 2018 The Go Authors. All rights reserved.
// Use of this source code is governed by a BSD-style
// license that can be found in the LICENSE file.
package source
import (
"fmt"
"net/url"
"path/filepath"
"strings"
)
const fileSchemePrefix = "file://"
// URI represents the full uri for a file.
type URI string
// Filename gets the file path for the URI.
// It will return an error if the uri is not valid, or if the URI was not
// a file URI
func (uri URI) Filename() (string, error) {
s := string(uri)
if !strings.HasPrefix(s, fileSchemePrefix) {
return "", fmt.Errorf("only file URI's are supported, got %v", uri)
}
s = s[len(fileSchemePrefix):]
s, err := url.PathUnescape(s)
if err != nil {
return s, err
}
s = filepath.FromSlash(s)
return s, nil
}
// ToURI returns a protocol URI for the supplied path.
// It will always have the file scheme.
func ToURI(path string) URI {
return URI(fileSchemePrefix + filepath.ToSlash(path))
}