每個變量返回$ .each函數

Question

我對JSON中的多維數組有點新，而且我現在一直在用一些東西來敲打我的頭。讓我後下

JSON我的代碼：

{ 
    "easy_questions": [ 
     { 
      "q": "Question 1", 
      "a": [ 
       {"option":"Answer 1", "correct": false}, 
       {"option":"Answer 2", "correct":true}, 
       {"option":"Answer 3", "correct":false}, 
       {"option":"Answer 4", "correct": false} 
      ], 
      "image": false, 
      "source": "dsa" 
     }, 
     { 
      "q": "Question 2", 
      "a": [ 
       {"option":"Answer 1", "correct": false}, 
       {"option":"Answer 2", "correct":true}, 
       {"option":"Answer 3", "correct":false}, 
       {"option":"Answer 4", "correct": false} 
      ], 
      "image": false, 
      "source": false 
     }, // this goes on for about 7 more questions... 
    ], 
    "step2_easy": [ 
     { 
      "q": "Question 1", 
      "a": [ 
       {"option":"Answer 1", "correct": false}, 
       {"option":"Answer 2", "correct":true}, 
       {"option":"Answer 3", "correct":false}, 
       {"option":"Answer 4", "correct": false} 
      ], 
      "image": false, 
      "source": false 
     }, 
     { 
      "q": "Question 2", 
      "a": [ 
       {"option":"Answer 1", "correct": false}, 
       {"option":"Answer 2", "correct":true}, 
       {"option":"Answer 3", "correct":false}, 
       {"option":"Answer 4", "correct": false} 
      ], 
      "image": false, 
      "source": false 
     }, //goes on again 

而且我的JavaScript

function quizFun() { 
    var questions = []; 
    var answers = []; 
    var quiz = "url.to/questions.json"; 

    $.getJSON(quiz, function(data) { 
     lvl_1(data); 
     function lvl_1(data) { 
      var easy = data.easy_questions; 

      function shuffle(array) { 
       var counter = array.length, temp, index; 
        // While there are elements in the array 
        while (counter--) { 
         // Pick a random index 
         index = (Math.random() * counter) | 0; 
         // And swap the last element with it 
         temp = array[counter]; 
         array[counter] = array[index]; 
         array[index] = temp; 
        } 

        return array; 
      } 
      var randoms = shuffle(easy.slice(0)); 
      randoms.length = 3; 

      jQuery.each(randoms, function(i, val) { 
       var question = val.q, 
        sourse = val.source; 
        img = val.image; 
        answ = val.a; 
        option = null; 
        correct = null; 

       $.each(answ,function(option, correct) { 
        option = this.option; 
        correct = this.correct; 
        answers = '<span data-true="'+correct+'">'+option+'</span>'; 

        return answers; 

       }); 
       if(!img) { 
        html_strucutre = question + answers; 
       } else { 
        html_strucutre = question + '<img src="'+img+'" width="300">' + answers; 
       } 

       $('body').append(html_strucutre+'<br>'); 
      }); 
     } 
    }); 

我知道這是不是最好的，但我想這一個與梳理小步驟。

首先，我需要得到3個隨機問題'easy_questions'，以及相應的答案（true或false）。一旦我經歷了其中的3個，我會跳到下一個等等。

我確實設法得到問題甚至答案，但每當我試圖退出$.each(answ,function(option, correct) {..}時，它只會爲每個問題返回一個答案，並且總是最後一個答案。

這裏有什麼嚴重錯誤？

Answer 1

jQuery.each與原生Array.prototype.forEach方法大致相同，這意味着它對集合中的每個事物運行一次給定函數。這個想法是，你將它用於副作用，一次用於每件事情。正如其他地方所陳述的那樣，您可以使用它在循環中每次將某些東西推入數組中。

然而，更清潔的方式，如果你想返回東西從功能，將是用jQuery.map()來代替。

var answers = $.map(answ, function(option, correct) { 
     option = this.option; 
     correct = this.correct; 
     answer = '<span data-true="'+correct+'">'+option+'</span>'; 
     return answer; 
    }); 

這產生相同長度的陣列作爲answ但含有構建爲answer對於每一個字符串。然後你可以調用.join()將它變成一個字符串。

var myHtml = answers.join(""); 

你可以完全避免的jQuery通過使用內置的$.map陣列instyead的map功能：

var answers = answ.map(function(option, correct) { 
     option = this.option; 
     correct = this.correct; 
     answer = '<span data-true="'+correct+'">'+option+'</span>'; 
     return answer; 
    });