将所有动态 Json 响应字符串存储到放心的 java 列表中的正确方法是什么？

Question

public void getresponseString() {
    for (int i = 0; i < getInfoResp.jsonPath().getInt("Map.List.size()"); i++) {
        ArrayList<String> country = getInfoResp.jsonPath().get("Map.List[" + i + "]" + ".country");
    }
}

响应：

{
"plan": {
    "program": "gtr",
    "syter": "yes"
},

    "Map": {
        "List": [
            {
                "id": "tyt6577",
                "proxy": "ENABLED",
                "type": "BENEFIT",
                "country": "us",
                "triag": null
            },
            {
                "id": "yyaqtf6327",
                "proxy": "ENABLED",
                "type": "BENEFIT",
                "country": "aus",
                "triag": null
            },
            {
                "id": "676hwjsgvhgv",
                "proxy": "ENABLED",
                "type": "BENEFIT",
                "country": "rus",
                "triag": null
            },
           {
                "id": "676hsdhgv",
                "proxy": "ENABLED",
                "type": "BENEFIT",
                "country": "spa",
                "triag": null
            },
           {
                "id": "623ujhhgv",
                "proxy": "ENABLED",
                "type": "BENEFIT",
                "country": "cha",
                "triag": null
            }
           
        ]
    }

}

在上述方法中，我尝试将 Jsonresponse 的所有字符串存储在 List 中，但最终出现此错误“ java.lang.String cannot be cast to java.util.ArrayList ”
将所有动态字符串存储到列表的正确方法是什么？

Answer 1

为什么不拆分json并将响应存储在ArrayList<String>

ArrayList<String> list = new ArrayList<>(Arrays.stream(response.split(",")).collect(Collectors.toList()));

您可以用逗号或任何其他字符串/字符拆分它

Answer 2

在这里提取country的最简单方法是：

ArrayList<String> country = getInfoResp.jsonPath().get("Map.List.country");
//[us, aus, rus, spa, cha]