[typescript] parsing template literal as type #9748
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tanhauhau
commented
Mar 22, 2019
tanhauhau
commented
| Q | A |
|---|---|
| Fixed Issues? | Fixes #9706 |
| Patch: Bug Fix? | |
| Major: Breaking Change? | |
| Minor: New Feature? | |
| Tests Added + Pass? | Yes |
| Documentation PR Link | |
| Any Dependency Changes? | |
| License | MIT |
|
Build successful! You can test your changes in the REPL here: https://babeljs.io/repl/build/10593/ |
danez
added
pkg: parser
area: typescript
PR: Bug Fix 🐛
| node.literal = this.finishNode( | ||
| noSubstitutionTemplateLiteralNode, | ||
| "NoSubstitutionTemplateLiteral", | ||
| ); |
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I don't see why we need a new node type here. We can just re-use TemplateLiteral: if node.expressions.length === 0, then it doesn't have any substitution.
| "NoSubstitutionTemplateLiteral", | ||
| ); | ||
| } else { | ||
| throw this.unexpected(startPosition); |
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Suggested change
| throw this.unexpected(startPosition); | |
| throw this.raise(templateNode.expressions[0].start, "Template literal types cannot have any substitution"); |
nicolo-ribaudo
approved these changes
Mar 25, 2019
existentialism
approved these changes
Mar 25, 2019
|
Thanks! |
This was referenced Aug 17, 2019
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