需要关于迭代 Java List<Map<String, String>> 的建议

Question

我需要关于如何最好地遍历 List<Map<String, String>> 对象以获得以下结果的建议：

该对象包含从 sql 数据库中获取的数据。每个 Map 条目描述了返回数据的单列：

0 =
    0 =
      key = "ColumnA"
      value = "1"
    1 =
      key = "ColumnB"
      value = "2"
    2 =
      key = "ColumnC"
      value = "3"
1 =
    0 =
      key = "ColumnA"
      value = "1"
    1 =
      key = "ColumnB"
      value = "2"
    2 =
      key = "ColumnC"
      value = "3"

一个实际的数据示例如下所示：

0 =
    0 =
      key = "Itemtype"
      value = "1"
    1 =
      key = "Itemdate"
      value = "01.01.2018"
    2 =
      key = "Subitem"
      value = "A"
    3 =
      key = "Subitemdetail"
      value = "A"
    4 =
      key = "Subitemdetail2"
      value = "A"
1 =
    0 =
      key = "Itemtype"
      value = "1"
    1 =
      key = "Itemdate"
      value = "01.01.2018"
    2 =
      key = "Subitem"
      value = "B"
    3 =
      key = "Subitemdetail"
      value = "B"
    4 =
      key = "Subitemdetail2"
      value = "B"
2 =
    0 =
      key = "Itemtype"
      value = "2"
    1 =
      key = "Itemdate"
      value = "01.01.2018"
    2 =
      key = "Subitem"
      value = "A"
    3 =
      key = "Subitemdetail"
      value = "A"
    4 =
      key = "Subitemdetail2"
      value = "A"
3 =
    0 =
      key = "Itemtype"
      value = "2"
    1 =
      key = "Itemdate"
      value = "01.01.2018"
    2 =
      key = "Subitem"
      value = "B"
    3 =
      key = "Subitemdetail"
      value = "B"
    4 =
      key = "Subitemdetail2"
      value = "B"

目前，数据通过 sn-p 以与 JSON 类似的格式返回：

  JSONArray resultSet = new JSONArray();   

  for (Map<String, String> res : data) {
      JSONObject resObject = new JSONObject();

      for (Entry<String, String> subres : res.entrySet()) {
         resObject.put(subres.getKey(), subres.getValue());
      }

     resultSet.put(resObject);
  }

上面的sn-p返回如下JSON：

{
"res":[
    {
        "Itemtype": "1",
        "Itemdate": "01.01.2018",
        "Subitem": "A",
        "Subitemdetail": "A",
        "Subitemdetail2": "A"
    },
    {
        "Itemtype": "1",
        "Itemdate": "01.01.2018",
        "Subitem": "B",
        "Subitemdetail": "B",
        "Subitemdetail2": "B"
    },
    {
        "Itemtype": "2",
        "Itemdate": "01.01.2018",
        "Subitem": "A",
        "Subitemdetail": "A",
        "Subitemdetail2": "A"
    },
    {
        "Itemtype": "2",
        "Itemdate": "01.01.2018",
        "Subitem": "B",
        "Subitemdetail": "B",
        "Subitemdetail2": "B"
    }
]}

期望的结果：

但是，我现在想根据 Itemtype 值对 JSON 进行分组。期望的结果如下：

{
"result":[
{
    "Itemtype": "1",
    "Itemdate": "01.01.2018",
    "Subitem": [
        {
            "Subitem": "A",
            "Subitemdetail": "A",
            "Subitemdetail2": "A"
        },
        {
            "Subitem": "B",
            "Subitemdetail": "B",
            "Subitemdetail2": "B"
        }
    ]
},
{
    "Itemtype": "2",
    "Itemdate": "01.01.2018",
    "Subitem": [
        {
            "Subitem": "A",
            "Subitemdetail": "A",
            "Subitemdetail2": "A"
        },
        {
            "Subitem": "B",
            "Subitemdetail": "B",
            "Subitemdetail2": "B"
        }
    ]
}
]}

我正在想办法如何最好地遍历List<Map<String, String>> 对象。你能给我一个建议吗？

因为我目前只能想到非常丑陋的解决方案，例如，首先将对象作为一个整体进行检查，然后存储一个 itemtypes 列表及其位置，对于上面的示例意味着： 第 1 项：0、1 和第 2 项：1、2。 然后，我将浏览列表并为自己构建 JSON。但也许你可以给我一个更好的方法的建议？或者甚至有一个 Java 8 流可以更好地解决这个问题？

提前致谢！

Answer 1

您可以使用 groupingBy 流收集器来获取一个 Map，然后通过 reduce 将每个条目转换为您想要的最终结构

有点像

// Group into Itemtype -> objects Map
Map<String, List<JSONObject>> grouped = results.stream().collect(groupingBy(obj -> obj.getString("Itemtype"))

// Reduce entries into single JSON array where the Itemtype is a top level property and the entries are under Subitem
grouped.entries().stream().reduce(result, entry-> {
    JSONObject obj = new JSONObject();
    obj.putString("Itemtype", entry.getKey());
    obj.putObject("Subitem", entry.getValue());

    result.put(obj);
    return result;
}, new JSONArray())

这并不能完美地完成您想要的属性，但我相信您可以弄清楚其余的。