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This repository has been archived by the owner on Nov 22, 2020. It is now read-only.
When creating a URI from a file path, always use vscode.URI.file(path), e.g vscode.URI.file('c:\\test')
Use vscode.URI.parse(uriString) when the input string is in the form scheme://authority/path
I saw that it works right there are four times only that he used Uri and it's in its place.
private showToolbarForActiveDocumentIfNeeded(documentUri: vscode.Uri) {
if (this.isMicropythonProject(documentUri))
'use strict';
import Base from '../base';
import * as vscode from 'vscode';
import * as fs from 'fs';
import * as path from 'path';
import * as _ from 'lodash';
In 1.26 we discovered that several extensions call the 'vscode.openFolder' command with invalid file URIs:
When creating a URI from a file path, always use
vscode.URI.file(path)
, e.gvscode.URI.file('c:\\test')
Use
vscode.URI.parse(uriString)
when the input string is in the formscheme://authority/path
See microsoft/vscode#57267 for more information or for questions.
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