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2a3f5192be
This happens whenever we load standard library information from export data, and prevents the editor from understanding the file names Change-Id: If5c04176a3e669ba396f322275993616e51ec097 Reviewed-on: https://go-review.googlesource.com/c/149612 Run-TryBot: Ian Cottrell <iancottrell@google.com> Reviewed-by: Rebecca Stambler <rstambler@golang.org> TryBot-Result: Gobot Gobot <gobot@golang.org>
48 lines
1.2 KiB
Go
48 lines
1.2 KiB
Go
// Copyright 2018 The Go Authors. All rights reserved.
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// Use of this source code is governed by a BSD-style
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// license that can be found in the LICENSE file.
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package source
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import (
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"fmt"
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"net/url"
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"path/filepath"
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"runtime"
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"strings"
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)
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const fileSchemePrefix = "file://"
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// URI represents the full uri for a file.
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type URI string
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// Filename gets the file path for the URI.
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// It will return an error if the uri is not valid, or if the URI was not
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// a file URI
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func (uri URI) Filename() (string, error) {
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s := string(uri)
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if !strings.HasPrefix(s, fileSchemePrefix) {
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return "", fmt.Errorf("only file URI's are supported, got %v", uri)
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}
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s = s[len(fileSchemePrefix):]
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s, err := url.PathUnescape(s)
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if err != nil {
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return s, err
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}
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s = filepath.FromSlash(s)
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return s, nil
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}
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// ToURI returns a protocol URI for the supplied path.
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// It will always have the file scheme.
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func ToURI(path string) URI {
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const prefix = "$GOROOT"
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if strings.EqualFold(prefix, path[:len(prefix)]) {
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suffix := path[len(prefix):]
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//TODO: we need a better way to get the GOROOT that uses the packages api
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path = runtime.GOROOT() + suffix
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}
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return URI(fileSchemePrefix + filepath.ToSlash(path))
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}
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