Question

我需要将列表>的格式更改为列表>的另一种格式。我有这个收藏：

[{
    "OPERATION_NAME": "COPY",
    "OPERATION_STATUS": "READY",
    "COUNT(1)": 1
}, {
    "OPERATION_NAME": "ARCHIVE",
    "OPERATION_STATUS": "READY",
    "COUNT(1)": 1
}, {
    "OPERATION_NAME": "EXPORT_CA",
    "OPERATION_STATUS": "READY",
    "COUNT(1)": 1
}, {
    "OPERATION_NAME": "MERGE",
    "OPERATION_STATUS": "READY",
    "COUNT(1)": 1
}, {
    "OPERATION_NAME": "MERGE",
    "OPERATION_STATUS": "DONE",
    "COUNT(1)": 1
}]

我将其更改为：

[{
    "READY": 0,
    "NEW": 0,
    "DONE": 0,
    "OPERATION_NAME": "MERGE"
}, {
    "READY": 0,
    "NEW": 0,
    "DONE": 0,
    "OPERATION_NAME": "ARCHIVE"
}, {
    "READY": 0,
    "NEW": 0,
    "DONE": 0,
    "OPERATION_NAME": "COPY"
}, {
    "READY": 0,
    "NEW": 0,
    "DONE": 0,
    "OPERATION_NAME": "EXPORT_CA"
}]

但是我无法获取COUNT（1）的值并将其设置为我正在创建的新对象。

最后我应该有这个：

[{
    "READY": 1,
    "NEW": 0,
    "DONE": 1,
    "OPERATION_NAME": "MERGE"
}, {
    "READY": 1,
    "NEW": 0,
    "DONE": 0,
    "OPERATION_NAME": "ARCHIVE"
}, {
    "READY": 1,
    "NEW": 0,
    "DONE": 0,
    "OPERATION_NAME": "COPY"
}, {
    "READY": 1,
    "NEW": 0,
    "DONE": 0,
    "OPERATION_NAME": "EXPORT_CA"
}]

这是我的功能，可以将一个对象转换为另一个对象：

private List<Map<String, Object>> transformObject(List<Map<String, Object>> resDB) {
        Gson gson = new Gson();
        LOGGER.info("Json response : {}", gson.toJson(resDB));

        List<Map<String, Object>> result = new ArrayList<>();

        List<String> listOfOperationName = new ArrayList<>();
        List<String> listOfOperationStatus = new ArrayList<>();

        // Retrieve OP Name and OP Status
        for (Map<String, Object> map : resDB) {
            for (Map.Entry<String, Object> entry : map.entrySet()) {
                String key = entry.getKey();
                Object value = entry.getValue();
                if(key.equals("OPERATION_NAME")) {
                    listOfOperationName.add(value.toString());
                } else if(key.equals("OPERATION_STATUS")) {
                    listOfOperationStatus.add(value.toString());
                }
            }
        }

        // Remove duplicates
        Set<String> setName = new HashSet<>(listOfOperationName);
        listOfOperationName.clear();
        listOfOperationName.addAll(setName);
        // Remove duplicates
        Set<String> setStatus = new HashSet<>(listOfOperationStatus);
        listOfOperationStatus.clear();
        listOfOperationStatus.addAll(setStatus);

        // Create field OP Name with value
        for (String tempOpName : listOfOperationName) {
            Map<String, Object> tempObjName = new HashMap<>();
            tempObjName.put("OPERATION_NAME", tempOpName);
            for (String tempOpStatus : listOfOperationStatus) {
                tempObjName.put(tempOpStatus, 0);
            }
            result.add(tempObjName);
        }

        Gson gsonResult = new Gson();
        LOGGER.info("Json response : {}", gsonResult.toJson(result));

        return result;
    }

如果您有想法如何做，那就太好了！非常感谢！

Answer 1

使用这两种帮助方法

Received and deleted message: 250e37a8-d779-48a1-9941-84219a82513e

您可以轻松进行转换

private static Map<String, Object> transform(Map<String, Object> initial) {
    Map<String, Object> transformed = new HashMap<>();
    transformed.put("OPERATION_NAME", (String) initial.get("OPERATION_NAME"));
    transformed.put("READY", 0);
    transformed.put("NEW", 0);
    transformed.put("DONE", 0);
    transformed.put((String) initial.get("OPERATION_STATUS"),
                    Integer.valueOf((String) initial.get("COUNT(1)")));
    return transformed;
}

private static Map<String, Object> mergeMaps(Map<String, Object> left, Map<String, Object> right) {
    for (Entry<String, Object> re : right.entrySet()) {
        if (!"OPERATION_NAME".equals(re.getKey())) {
            left.merge(re.getKey(), re.getValue(), (l, r) -> {
                Integer res = 0;
                res += (Integer) l;
                res += (Integer) r;
                return res;
            });
        }
    }
    return left;
}

希望有帮助！

从List <map>检索值并将其插入另一个List <map>

1 个答案: