java 如何对时间戳列表进行排序？
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How to sort timestamp list?
提问by K Roy
I have a list of timestamp(long) in which I have to sort list base on time or list added.
我有一个时间戳（长）列表，我必须根据添加的时间或列表对列表进行排序。
I have tried and searched but its now working.
我已经尝试和搜索，但它现在工作。
Collections.sort(vehicles, new Comparator<VehicleListModel.Vehicle>() {
@Override
public int compare(VehicleListModel.Vehicle o1, VehicleListModel.Vehicle o2) {
try {
DateFormat format = new SimpleDateFormat("MM-dd-yyyy hh:mm:ss");
return format.parse(o1.getCreated()).compareTo(format.parse(o2.getCreated()));
} catch (Exception e) {
e.printStackTrace();
return 0;
}
}
});
customerListAdapter.notifyDataSetChanged();
its not working then I tried
thisbut this Date(timeStamp)is deprecated
它不工作然后我尝试了
这个，但这Date(timeStamp)已被弃用
please help
请帮忙
回答by Darshan Mehta
If getCreatedreturns a Datethen you can compare two dates using compareToand sort the list based on this comparison, e.g.:
如果getCreated返回 aDate那么您可以使用比较两个日期compareTo并根据此比较对列表进行排序，例如：
List<VehicleListModel.Vehicle> list = new ArrayList<>();
list.sort((e1, e2) -> e1.getCreated().compareTo(e2.getCreated()));
update
更新
If getCreatedreturns longvalue then you can box it and use compareTomethod of Longclass, e.g.:
如果getCreated返回long值，那么您可以将其装箱并使用类的compareTo方法Long，例如：
List<ChainCode> list = new ArrayList<ChainCode>();
list.sort((e1, e2) -> new Long(e1.getCreated()).compareTo(new Long(e2.getCreated())));
回答by Jd Prajapati
Use long timestamp value for soring
使用长时间戳值进行排序
Collections.sort(vehicles, new Comparator<VehicleListModel.Vehicle>() {
@Override
public int compare(VehicleListModel.Vehicle o1, VehicleListModel.Vehicle o2) {
try {
DateFormat format = new SimpleDateFormat("MM-dd-yyyy hh:mm:ss");
return Long.valueOf(format.parse(o1.getCreated()).getTime()).compareTo(format.parse(o2.getCreated()).getTime());
} catch (Exception e) {
e.printStackTrace();
return 0;
}
}
});
customerListAdapter.notifyDataSetChanged();
