如何在打字稿中获得联合类型的完整 AST 表示？

Question

我知道 ts-ast-viewer，但我不知道他们如何从联合中提取元素列表。

我已经尝试了几种现有的解决方案，包括，但似乎大多数都已过时。一些 ts.[methods] 现在已弃用。

这是启动调试编译器的初始代码API:

import * as ts from "typescript";

const code = "type Foo = 'one'|'two'|'three'";
const sourceFile = ts.createSourceFile(
  "foo.ts",
  code,
  ts.ScriptTarget.Latest,
  true
);

function print(node: ts.Node, name: string) {
  console.log({ node });
}

print(sourceFile, "Foo");

我知道 TS wiki 中的示例。这是我稍微修改过的版本：

//@ts-ignore
import * as ts from "typescript";

const code = "type Foo = 'one'|'two'|'three'";
const sourceFile = ts.createSourceFile(
  "foo.ts",
  code,
  ts.ScriptTarget.Latest,
  true
);

function extract(identifiers: string[]): void {
  //@ts-ignore
  const unfoundNodes = [];
  //@ts-ignore
  const foundNodes = [];
  //@ts-ignore

  ts.forEachChild(sourceFile, (node) => {
    let name = "";
    //@ts-ignore
    if (ts.isFunctionDeclaration(node)) {
      //@ts-ignore
      name = node.name.text;
      //@ts-ignore
      node.body = undefined;
      //@ts-ignore
    } else if (ts.isVariableStatement(node)) {
      //@ts-ignore
      name = node.declarationList.declarations[0].name.getText(sourceFile);
      //@ts-ignore
    } else if (ts.isInterfaceDeclaration(node)) {
      name = node.name.text;
    }
    //@ts-ignore
    const container = identifiers.includes(name) ? foundNodes : unfoundNodes;
    //@ts-ignore
    container.push([name, node]);
  });
  //@ts-ignore
  return (unfoundNodes[0][1] as any).type.types.map(
    (elem: any) => elem.literal.text
  );
}

// Run the extract function with the script's arguments
console.log(extract(["Foo"]));

并且有效。但是，如果我将源代码从 "type Foo = 'one'|'two'|'three'" 更改为 "type Foo = keyof Array<any>" - 它不会。

我知道我的版本不对

此外，我试过 this 示例，但似乎 forEachDescendant 在 node 上不存在。

如何写一个函数来获取联合数组的元素？

预期函数：

import * as ts from "typescript";

const sourceCode = "type Foo = keyof Array<number>";


union(sourceCode, "Foo") // ["forEach", "reduce", "map" ...]

我需要这个用于调试目的。

Answer 1

您不能为此使用 AST，因为 AST 仅包含有关文本在文件中的外观的信息，而不包含代码的某些部分与其他部分的关系。为此，您需要使用类型检查器。这是一个独立的例子：

// available as @ts-morph/bootstrap on npm
import { ts, createProjectSync } from "https://deno.land/x/ts_morph@13.0.3/bootstrap/mod.ts";

// setup code... you can use the vanilla ts compiler for this, but it's a lot of work
const project = createProjectSync();
const sourceCode = "type Foo = keyof Array<number>";
const sourceFile = project.createSourceFile("file.ts", sourceCode);
const typeChecker = project.createProgram().getTypeChecker();

// example use with the compiler API
const fooTypeAlias = sourceFile.statements[0] as ts.TypeAliasDeclaration;

const fooType = typeChecker.getTypeAtLocation(fooTypeAlias.name);
if (fooType.isUnion()) {
  for (const type of fooType.types) {
    console.log(typeChecker.typeToString(type, fooTypeAlias.name));
  }
}

输出：

number
"length"
"toString"
"toLocaleString"
"pop"
"push"
"concat"
"join"
"reverse"
"shift"
"slice"
"sort"
"splice"
"unshift"
"indexOf"
"lastIndexOf"
"every"
"some"
"forEach"
"map"
"filter"
"reduce"
"reduceRight"

如何在打字稿中获得联合类型的完整 AST 表示？

How to get full AST representation of union type in typescript?

typescript

typescript-compiler-api