Java 如何解析嵌套JSON结果中的动态JSON键？

  • Z时代
  • 2024-01-10
  • 分类：问答

有以下格式的JSON结果，JSON Lint将其显示为“有效响应”。

我的问题是：由于“ 141”，“ 8911”等都是动态值，我如何访问“ question_mark”的内容？

我的示例代码用于访问“产品”的价值。

//Consider I have the first <code>JSONObject</code> of the "search_result" array and 
//I access it's "product" value as below.
String product = jsonObject.optString("product"); //where jsonObject is of type JSONObject.
//<code>product<code> now contains "abc".
JSON：
{
 "status": "OK",
 "search_result": [
            {
                "product": "abc",
                "id": "1132",
                "question_mark": {
                    "141": {
                        "count": "141",
                        "more_description": "this is abc",
                        "seq": "2"
                    },
                    "8911": {
                        "count": "8911",
                        "more_desc": "this is cup",
                        "seq": "1"
                    }
                },
                "name": "some name",
                "description": "This is some product"
            },
            {
                "product": "XYZ",
                "id": "1129",
                "question_mark": {
                    "379": {
                        "count": "379",
                        "more_desc": "this is xyz",
                        "seq": "5"
                    },
                    "845": {
                        "count": "845",
                        "more_desc": "this is table",
                        "seq": "6"
                    },
                    "12383": {
                        "count": "12383",
                        "more_desc": "Jumbo",
                        "seq": "4"
                    },
                    "257258": {
                        "count": "257258",
                        "more_desc": "large",
                        "seq": "1"
                    }
                },
                "name": "some other name",
                "description": "this is some other product"
            }
       ]
}

我的问题标题“动态键”可能是错误的，但目前我不知道该问题的最佳名称是什么。

任何帮助将不胜感激！

回答：

使用JSONObject keys（）获取密钥，然后迭代每个密钥以获取动态值。

大致的代码如下所示：

    // searchResult refers to the current element in the array "search_result"
    JSONObject questionMark = searchResult.getJSONObject("question_mark");
    Iterator keys = questionMark.keys();
    while(keys.hasNext()) {
        // loop to get the dynamic key
        String currentDynamicKey = (String)keys.next();
        // get the value of the dynamic key
        JSONObject currentDynamicValue = questionMark.getJSONObject(currentDynamicKey);
        // do something here with the value...
    }

以上是 Java 如何解析嵌套JSON结果中的动态JSON键？ 的全部内容， 来源链接： utcz.com/qa/400427.html

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