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fix(linter/no-control-regex): false negative for flags in template literals (#6531)

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Updates regex flag extraction logic to handle no-substitution template literals, allowing cases like this to be correctly reported:
```js
let r = new RegExp('\\u{0}', `u`);
```
This commit is contained in:
DonIsaac 2024-10-14 00:59:28 +00:00
parent ecce5c53f8
commit 2b86de9e74
2 changed files with 9 additions and 3 deletions
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10
crates/oxc_linter/src/ast_util.rs
View file

@ -287,11 +287,15 @@ pub fn extract_regex_flags<'a>(
if args.len() <= 1 {
return None;
}
let Argument::StringLiteral(flag_arg) = &args[1] else {
return None;
let flag_arg = match &args[1] {
Argument::StringLiteral(flag_arg) => flag_arg.value.clone(),
Argument::TemplateLiteral(template) if template.is_no_substitution_template() => {
template.quasi().expect("no-substitution templates always have a quasi")
}
_ => return None,
};
let mut flags = RegExpFlags::empty();
for ch in flag_arg.value.chars() {
for ch in flag_arg.chars() {
let flag = RegExpFlags::try_from(ch).ok()?;
flags |= flag;
}

2
crates/oxc_linter/src/rules/eslint/no_control_regex.rs
View file

@ -278,6 +278,8 @@ mod tests {
],
vec![
r"let r = /\u{0}/u",
r"let r = new RegExp('\\u{0}', 'u');",
r"let r = new RegExp('\\u{0}', `u`);",
r"let r = /\u{c}/u",
r"let r = /\u{1F}/u",
r"let r = new RegExp('\\u{1F}', 'u');", // flags are known & contain u