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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
In 1.26 we discovered that several extensions call the 'vscode.openFolder' command with invalid file URIs:
136,51: yield vscode.commands.executeCommand('vscode.openFolder', vscode.Uri.parse(gitRepoFolder));
27,51: yield vscode.commands.executeCommand('vscode.openFolder', vscode.Uri.parse(gitRepoFolder));
27,51: yield vscode.commands.executeCommand('vscode.openFolder', vscode.Uri.parse(gitRepoFolder));
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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