访问多个 JSON 数组中的元素

Question

我使用 JSON-Simple 创建了以下 JSON。

问题：

我如何从这个 JSON 数组中访问“情绪”、“分数”和“评论”等元素。

我已经提到了这个链接here，但仍然很困惑。请帮忙 JSON：

{
    "Restaurant3": [
        {
            "sentiment": "positive",
            "score": "0.665205",
            "review": "Good Indian food in campbell that's better than average at a fair price. Will come back."
        }
    ],
    "Restaurant1": [
        {
            "sentiment": "positive",
            "score": "0.457413",
            "review": "People can say what they will about this being McIndians but I have ALWAYS had a solid meal at Tandoori Oven and was NEVER disappointed with the food.The saag lamb may be my favorite dish in the world as long as I can enjoy a cucumber salad and plenty of mint cilantro chutney to cool down my mouth. The vegetarian specials are also excellent every time. My only gripe is the menu can be a bit pricey and they have screwed around with student specials that used to get you two meals for under $10. ... Perhaps they realized it was too good of a deal?"
        }
    ],
    "Restaurant2": [
        {
            "sentiment": "positive",
            "score": "0.684585",
            "review": "We live near by and this is one of our favorite quick Indian food stop.I really like their butter chicken which is very creamy and fairly thick. The chicken inside is well spiced and grilled beforehand to give a good texture. Their baagen bharta is good as well, not as mashed up as other places and had a good mix of spice. Another thing to try is the tandoori salmon. It's juicy on the inside and crunchy on the outside. Do expect long lines during lunch, but dinner time it's fairly open."
        }
    ]
}

我使用 JSON-Simple 来生成这个 JSON-Array：

 List  l1 = new LinkedList(); //LinkedList.
 List  l2 = new LinkedList();
 List  l3 = new LinkedList();
 JSONObject obj2 = new JSONObject(); //JSON-Simple Library
 obj2.put("Restaurant1", l1); //I'm making a linked list and passing them to make nested json
 obj2.put("Restaurant2", l2);
 obj2.put("Restaurant3", l3);
 String answer = obj2.toJSONString();
 System.out.println(answer); //prints JSON
 return answer; //Returning entire JSON (for now)

Answer 1

var yourJSON = { ... }; // your json goes here
var restaurant3sentiment = yourJSON.Restaurant3[0].sentiment;

您的 JSON 包含数组，因此您必须指定子元素的索引