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TypeScript antlr4ts侦听器类型引发错误

antlr4 typescript angularjs

Vinayak S · 技术社区 · 7 年前

我正在尝试在角度项目中使用antlr4解析器。

有一个dataservice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个方法(enterNonGraphScope)导出的接口。

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);
  };

}

正在粘贴scopeListener中的某些代码段。ts用于说明接口

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typescript代码。我出错了

src/app/rule解析器出错。服务ts(31,17):错误TS2559:“ScopeEvaluator”类型没有与“ParseTreeListener”类型相同的属性。

下面是在typescript中生成的侦听器。(已删除内容和antlr生成的注释)

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typescript解释或键入有关。我是javascript/typescript的新手。你能帮帮我吗?

网页包/生成的javascript代码功能正常,但此错误阻止生成。

非常感谢!

-维纳亚克

1 回复 | 直到 7 年前

Vinayak S 7 年前

我在谷歌上搜索了一下,发现这与弱类型处理更改有关。有关以下链接的更多详细信息。

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

在此基础上,在 After upgrading TypeScript, Angular controller registration now fails to compile

下面是对实现接口的类的修改。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-涉及重新澄清材料?

            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);
                };
            }

如果你觉得这样做不对,请发表评论。我对typescript的理解还不完整。