TypeScript antlr4ts 监听器类型抛出错误

Question

我正在尝试在 Angular 项目中使用 antlr4 解析器。

有一个数据服务类，其中调用的函数如下所示

parseRule() {
  const ruleString = ' STRING TO PARSE';

  const inputStream = new ANTLRInputStream(ruleString);
  const lexObject = new lexer.scopeLexer(inputStream);
  const tokenStream = new CommonTokenStream(lexObject);
  const parseObject = new parser.scopeParser(tokenStream);
  const result = parseObject.file();
  const evaluator = new ScopeEvaluator();
  const walker = new ParseTreeWalker();

  walker.walk(evaluator, result);
  console.log(' result :', result.text);
}

客户侦听器正在实现仅使用一种方法导出的接口，即 enterNonGraphScope。

import { scopeListener } from './antlrscope/scopeListener';
import { NonGraphScopeContext, NamesetSimpleContext } from './antlrscope/scopeParser';

export class ScopeEvaluator implements scopeListener {

  constructor() {
    console.log('constructed the asdfasf');
  }

  enterNonGraphScope = function (ctx: NonGraphScopeContext) {
    console.log('Tis ', ctx.text);
  };

}

从 scopeListener.ts 中粘贴一些 sn-p 用于说明界面

export interface scopeListener extends ParseTreeListener {

/**
 * Enter a parse tree produced by the `NonGraphScope`
 * labeled alternative in `scopeParser.scope`.
 * @param ctx the parse tree
*/
enterNonGraphScope?: (ctx: NonGraphScopeContext) => void;

当我运行 angular ng serve 时，它​​正在编译打字稿代码。 我收到一个错误

src/app/rule-parser.service.ts(31,17) 中的错误：错误 TS2559：“ScopeEvaluator”类型与“ParseTreeListener”类型没有共同的属性。

下面是打字稿中生成的监听器。 （删除内容和antlr生成的cmets）

export interface scopeListener extends ParseTreeListener {
    enterNonGraphScope?: (ctx: NonGraphScopeContext) => void;
    exitNonGraphScope?: (ctx: NonGraphScopeContext) => void;
    enterBlockScope?: (ctx: BlockScopeContext) => void;
    exitBlockScope?: (ctx: BlockScopeContext) => void;
    enterNamesetSimple?: (ctx: NamesetSimpleContext) => void;
    exitNamesetSimple?: (ctx: NamesetSimpleContext) => void;
    enterGrainExpression?: (ctx: GrainExpressionContext) => void;
    exitGrainExpression?: (ctx: GrainExpressionContext) => void;
    enterGrainSimple?: (ctx: GrainSimpleContext) => void;
    exitGrainSimple?: (ctx: GrainSimpleContext) => void;
    enterNamesetExpression?: (ctx: NamesetExpressionContext) => void;
}

这似乎与打字稿解释或打字有关。 我是 javascript/typescript 的新手。你能帮帮我吗？

webpack / 生成的 javascript 代码运行良好，但此错误阻止生成构建。

非常感谢！

-维纳亚克

Answer 1

我对此进行了一些谷歌搜索，发现这可以通过弱类型处理更改来做一些事情。更多详情请点击以下链接。

https://blogs.msdn.microsoft.com/typescript/2017/06/12/announcing-typescript-2-4-rc/

基于此，另一个 * 讨论位于 After upgrading TypeScript, Angular controller registration now fails to compile

对实现接口的类的以下修改有效。 选项 1 更好，因为它只是一个类型断言。

选项 1

    parseRule() {
      const ruleString = ' Scope: (Dim1.Attr1 * &Namset1);' +
        ' Measure.[asdfa] = ddd ;' +
        'end scOPe;';

      const inputStream = new ANTLRInputStream(ruleString);
      const lexObject = new lexer.scopeLexer(inputStream);
      const tokenStream = new CommonTokenStream(lexObject);
      const parseObject = new parser.scopeParser(tokenStream);
      const result = parseObject.file();
      const evaluator = new ScopeEvaluator();
      const walker = new ParseTreeWalker();
      console.log(' type of ' , evaluator);
      walker.walk(evaluator as ParseTreeListener, result);
        // Parse and execute the code.
        console.log(' p :', parseObject);
      console.log(' result :', result.text);
    }

选项 2 - 涉及重新声明内容？

            import { scopeListener } from './antlrscope/scopeListener';
            import { NonGraphScopeContext, NamesetSimpleContext } from './antlrscope/scopeParser';
            import { ParseTreeListener } from 'antlr4ts/tree/ParseTreeListener';


            export class ScopeEvaluator implements scopeListener {

                visitTerminal = () => { };
                visitErrorNode = () => { };
                enterEveryRule = () => { };
                exitEveryRule = () => { };
                enterNonGraphScope = function (ctx: NonGraphScopeContext) {
                    console.log('Tis ', ctx.text);
                };
            }

如果您觉得这不是正确的做法，请发表评论。我对 typescript 的理解还不完整。