Question

这是我的JSON文件，

  [{
    "Title": "AY",
    "Items": [{
        "Name": "Alex",
        "RoadNameShort": "AY"
    }, {
        "Name": "Kep",
        "RoadNameShort": "AY"
    }, {
        "Name": "Lower",
        "RoadNameShort": "AY"
    }]
  }, {
    "Title": "BK",
    "Items": [{
        "Name": "Chantek",
        "RoadNameShort": "BK"
    }, {
        "Name": "Wood",
        "RoadNameShort": "BK"
    }]
  }]

这是我的Java代码，

listDataHeader = new ArrayList<String>();
listDataHeade = new ArrayList<String>();

listDataChild = new HashMap<String, List<String>>();
HashMap<String, String> contact = new HashMap<>();

// Adding child data for lease offer
List<String> lease_offer = new ArrayList<String>();


JSONArray array = new JSONArray(jsonstr);
for (int i = 0; i < array.length(); i++) {
    // tmp hash map for single contact
    JSONObject c = array.getJSONObject(i);
    listDataHeader.add(c.getString("Title"));
    String title = c.getString("Title");

    JSONArray items = c.getJSONArray("Items");
    for (j = 0; j < items.length(); j++) {
        JSONObject item = items.getJSONObject(j);
        String name = item.getString("Name");
        Log.d("email", name);

        listDataHeade.add(item.getString("Name"));

        lease_offer.add(item.getString("Name"));
        // Header into Child data
        listDataChild.put(listDataHeader.get(i), lease_offer);
    }
}

对于父母，我使用了Title。孩子是Name。我正确地获得了标题为AY，BK的组。但我的问题是给孩子的。名称Chantek和木材应属于BK组，而不是AY组。但在AY组，我也让Chantek和木头像孩子一样。

Answer 1

尝试将listDataChild移到内部循环之外，因为您需要在迭代所有子项之后将所有子项添加到父列表，并在内部循环开始之前初始化lease_offer因为每个父项需要创建新的子列表，如下所示：

JSONArray array = new JSONArray(jsonstr);
for (int i = 0; i < array.length(); i++) {
    // tmp hash map for single contact
    JSONObject c = array.getJSONObject(i);
    String title = c.getString("Title");
    listDataHeader.add(title);

    JSONArray items = c.getJSONArray("Items");
    / Adding child data for lease offer
    List<String> lease_offer = new ArrayList<String>();
    for (j = 0; j < items.length(); j++) {
        JSONObject item = items.getJSONObject(j);
        String name = item.getString("Name");
        Log.d("email", name);
        listDataHeade.add(name);
        lease_offer.add(name);
    }
    // Header into Child data
    listDataChild.put(listDataHeader.get(i), lease_offer);
}

列出json消耗性儿童不准确

1 个答案: