Question

尤其是我想访问答案的数值，以便稍后对其求和。

以前尝试过的长度，不适用于对象。

这是我的数据：

const qData = [
    {
        id: 0,
        question: "question 1",
        answers: [
            { value: 1, text: "rafael" },
            { value: 2, text: "dontaelo" },
            { value: 3, text: "leonardo" },
            { value: 4, text: "michelangelo" }
        ]
    },
    {
        id: 1,
        question: "question 2",
        answers: [
            { value: 1, text: "rafael" },
            { value: 2, text: "dontaelo" },
            { value: 3, text: "leonardo" },
            { value: 4, text: "michelangelo" }
        ]
    }
];

export default qData;

我试图像这样对答案值求和：

handleShowScore = () => {
  var i, newScore;
  var a = qData.answers;

  for (i = 0; i < a.length; i++) {
    newScore = newScore + a[i].value;
  }
}

Answer 1

如果我正确理解了您的问题，以下示例将满足您的要求

// Loop through all questions. Each question is stored in "q".
qData.forEach((q) => {
  console.log('qData:', q);

  // Make an array of all answer values   
  const answerValues = q.answers.map((a) => {
     return a.value;
  });
  console.log('answerValues:', answerValues);

  // Sum all answer values together
  const totalValues = answerValues.reduce((a, b) => a + b, 0)
  console.log('totalValues: ', totalValues);
});

如果您想获得特定ID的总和

// Find specific ID.
const question = qData.find((q) => q.id === 0);

// Make an array of all answer values   
const answerValues = question.answers.map((a) => {
   return a.value;
});
console.log('answerValues:', answerValues);

// Sum all answer values together
const totalValues = answerValues.reduce((a, b) => a + b, 0)
console.log('totalValues: ', totalValues);

Answer 2

letter_opener是一个数组，没有属性qData，所以answers将不起作用并返回qData.answers。

您首先必须遍历问题，然后在每个问题中必须遍历答案：

undefined

如果使用flatMap，则可以减少循环次数。将答案汇总成一个大数组。

const qData = [{id:0,question:"question 1",answers:[{value:1,text:"rafael"},{value:2,text:"dontaelo"},{value:3,text:"leonardo"},{value:4,text:"michelangelo"}]},{id:1,question:"question 2",answers:[{value:1,text:"rafael"},{value:2,text:"dontaelo"},{value:3,text:"leonardo"},{value:4,text:"michelangelo"}]}];

let sum = 0;
for (const question of qData) {
  for (const answer of question.answers) {
    sum += answer.value;
  }
}

console.log(sum);

除了使用for循环求和之外，还可以使用reduce，它会在数组上进行迭代以将其减少为单个值。

const qData = [{id:0,question:"question 1",answers:[{value:1,text:"rafael"},{value:2,text:"dontaelo"},{value:3,text:"leonardo"},{value:4,text:"michelangelo"}]},{id:1,question:"question 2",answers:[{value:1,text:"rafael"},{value:2,text:"dontaelo"},{value:3,text:"leonardo"},{value:4,text:"michelangelo"}]}];

const answers = qData.flatMap(question => question.answers);

let sum = 0;
for (const answer of answers) {
  sum += answer.value;
}

console.log(sum);

Answer 3

有点测验，对吧？假设我的答案是：

const myAnswers = [{question: 'question 1', myAnswer:'rafael'},{question: 'question 2', myAnswer:'dontaelo'}]

我从第一个问题得1分，从第二个问题得2分。我们需要经历2个循环：1找到相应的问题，1找到相应的点数：

const myPoints = myAnswers.map(answer => qData.find(question => question.question === answer.question).answers.find(possibleAnswer => possibleAnswer.text === answer.myAnswer).value)

那给了我[1，2]。现在我们需要用reduce求和：

const reducer = (accumulator, currentValue) => accumulator + currentValue;
const myScore = myPoints.reduce(reducer, 0);

我有3分；-）

努力遍历对象数组

3 个答案: