Question

我有一个带有三个值（字符串）的ArrayList，但第二个是数字，我希望将数字从0加到例如100.在这个时间它排序1,10,2,21,5等。在输出中我希望有1,2,5,10,21。

JSONObject jsonObj = new JSONObject(jsonStr);

                JSONArray devices = jsonObj.getJSONArray("list");

                for (int i = 0; i < devices.length(); i++) {
                    JSONObject c = devices.getJSONObject(i);
                    String id = c.getString("id");
                    String name = c.getString("name");
                    String known = c.getString("known");
                    String description = c.getString("description");
                    String controllerId = c.getString("controllerId");


                    JSONObject frequencySurvey = c.getJSONObject("frequencySurvey");
                    if (frequencySurvey.has("5000")) {
                        JSONObject fivezero = frequencySurvey.getJSONObject("5000");

                        String timestamp = fivezero.getString("timestamp");
                        String clients = fivezero.getString("clients");
                        String enabled = fivezero.getString("enabled");

                        HashMap<String, String> device = new HashMap<>();

                        device.put("id", id);
                        device.put("name", name);
                        device.put("enabled", enabled);
                        device.put("clients", clients);

                        DevicesList.add(device);

                    }


ListAdapter adapter = new SimpleAdapter(Devices_5_0.this, DevicesList,
                R.layout.list_item, new String[]{ "name","clients","enabled"},
                new int[]{R.id.name,R.id.clients, enabled});

        lv.setAdapter(adapter);

我使用这种类型，但它按字符串排序：

class SortValues implements Comparator {
    public int compare(Object obj1, Object obj2) {
        HashMap<String, String> test1 = (HashMap<String, String>) obj1;
        HashMap<String, String> test2 = (HashMap<String, String>) obj2;
        return test1.get("clients").compareTo(test2.get("clients"));
    }
}

Answer 1

如果客户价值是您的数字值，这将起作用：

class SortValues implements Comparator {
    public int compare(Object obj1, Object obj2) {
        HashMap<String, String> test1 = (HashMap<String, String>) obj1;
        HashMap<String, String> test2 = (HashMap<String, String>) obj2;
        Long firstClients = Long.valueOf(test1.get("clients"));
        Long secondClients = Long.valueOf(test2.get("clients"));
        return firstClients.compareTo(secondClients);
    }
}

Answer 2

您可以将数字字符串转换为整数

class SortValues implements Comparator {
    public int compare(Object obj1, Object obj2) {
        HashMap<String, String> test1 = (HashMap<String, String>) obj1;
        HashMap<String, String> test2 = (HashMap<String, String>) obj2;
        return new Integer(test1.get("clients"))
            .compareTo(new Integer(test2.get("clients")));
    }
}

Answer 3

试试这个：

Collections.sort(arrayList,new ArrayListHashMapComparator());

 public class ArrayListHashMapComparator implements Comparator<HashMap<String,String>> {

        @Override
        public int compare(HashMap<String, String> lhs, HashMap<String, String> rhs) {

            try {
                return lhs.get(lhs.keySet()).compareToIgnoreCase(rhs.get(rhs.keySet()));
            } catch (Exception e) {
                e.printStackTrace();
                return -1;
            }
        }
    }

上面的代码是按升序排序所有hashmap值

如何排序值arraylist <hashmap <string，string>＆gt;

3 个答案: