Question

即使commonHint有值，此Meteor服务器递归方法result也会向控制台返回finalRes未定义。有关如何将finalRes返回给来电者的任何建议？ THX

  //call the recursive method
  let result = this.commonHint(myCollection.findOne({age: 44}), shortMatches);
        console.log('got most common hint: ' + result); // <=== undefined ====

  'commonHint': function (doc, shortMatches, hinters, results = []) {
      // first call only first 2 args are defined,
      if (!hinters) {
        hinters = [...lib.getCombinations(['arg1', 'arg2', 'arg3'], 2, 3)];
        this.commonHint(doc, shortMatches, hinters, results);  // hinters is an array of length 3 with 2 elements each
        return;
      }

      // get an element from hinters, use its 2 hinters and remove that element from the hinters
      let hintersToUse = hinters.pop();
      let hinter1 = this.cleanMatchItem(hintersToUse[0]);
      let hinter2 = this.cleanMatchItem(hintersToUse[1]);
      let intersect = _.intersection(hinter1, hinter2);

      // which item of the shortMatches best matches with the intersect
      let tempCol = new Meteor.Collection();
      for (let i = 0; i < shortMatches.length; i++) { tempCol.insert({match: shortMatches[i]}); }
      results.push(mostSimilarString(tempCol.find({}), 'match', intersect.join(' ')));

      if (hinters.length > 0) {
        this.commonHint(doc, shortMatches, hinters, results);
      } else {
        let finalRes = lib.mostCommon(results);
        console.log(finalRes);  //<==== has a value
        return finalRes;        //<==== so return it to caller
      }
    },

Answer 1

您呼叫commonHint的任何地方都需要返回通话值。

  ... 

  if (!hinters) {
    hinters = [...lib.getCombinations(['arg1', 'arg2', 'arg3'], 2, 3)];
    return this.commonHint(doc, shortMatches, hinters, results);  // hinters is an array of length 3 with 2 elements each
  }

  ...

  if (hinters.length > 0) {
    return this.commonHint(doc, shortMatches, hinters, results);

Answer 2

返回结果的递归函数中的每个路径必须返回结果。在您的网站中，您的路径不是：未提供hinters时，hinters.length > 0为真。

你应该返回递归调用的结果：

  if (!hinters) {
    hinters = [...lib.getCombinations(['arg1', 'arg2', 'arg3'], 2, 3)];
    return this.commonHint(doc, shortMatches, hinters, results);  // hinters is an array of length 3 with 2 elements each
//  ^^^^^^
  }

  // ...

  if (hinters.length > 0) {
    return this.commonHint(doc, shortMatches, hinters, results);
//  ^^^^^^
  } else {
    let finalRes = lib.mostCommon(results);
    console.log(finalRes);  //<==== has a value
    return finalRes;        //<==== so return it to caller
  }

从重复功能返回值

2 个答案: