Unable to Sort ArrayList<HashMap<String, Object>> by Single HashMap object field in Java Android

Question

i am adding data to ArrayList < HashMap < String, Object>> contactList;

i want to sort the contactList as per their single field .

i am adding data like this :

HashMap<String, Object> docs = new HashMap<>();
                        docs.put("name", key);
                        docs.put("speciality", speci);
                        docs.put("status", status);  // sort by this field , if contains online it should appear first .
                        docs.put("picture", bp);
                        docs.put("education", educate);
                        docs.put("experience", experi);
                        docs.put("rating", ft);
                        contactList.add(docs);

data is added in a loop and later i assign contactList to simpleAdapter for listview.

Now i want my contact list to compare the 'status' field , if status is 'online' show first or if all status are offline do nothing just show all ;

how can i sort my data , i need to use this data later in listView to show online status contacts first in listview. Any help Would be appreciated. Thanks

i tried this solution but no success stackoverflow.com/questions/31342658/… — Adnan Quraishi
– Adnan Quraishi, Commented Mar 30, 2017 at 22:58
Couldn't you just use another set or map for online status? Instead of time complexity, growing heap is better solution i guess. — Abdullah Tellioglu
– Abdullah Tellioglu, Commented Mar 30, 2017 at 23:00
actually the structure used in code is according to this pattern . so i am bound to trigger some workaround to bring those maps object first whose status value is online in listview by simpleadapter — Adnan Quraishi
– Adnan Quraishi, Commented Mar 30, 2017 at 23:06

Darshan Mehta · Accepted Answer · 2017-03-30 23:56:58Z

1

You can define a comparator by anonymous inner class and sort based on value of status field assuming it's always going to be a String, e.g.:

contactList.sort(new Comparator<Map<String, Object>>() {
    @Override
    public int compare(Map<String, Object> o1, Map<String, Object> o2) {
        if(null != o1.get("status") && null != o1.get("status")){
            return o2.get("status").toString().compareTo(o1.get("status").toString());
        }else if(null != o1.get("status")){
            return 1;
        }else{
            return -1;
        }
    }
});

Update

You can use Collections.sort if list.sort is not compatible with current API version, e.g.:

Collections.sort(contactList, new Comparator<Map<String, Object>>() {
        @Override
        public int compare(Map<String, Object> o1, Map<String, Object> o2) {
            if(null != o1.get("status") && null != o1.get("status")){
                return o2.get("status").toString().compareTo(o1.get("status").toString());
            }else if(null != o1.get("status")){
                return 1;
            }else{
                return -1;
            }
        }
    });

edited Mar 30, 2017 at 23:56

answered Mar 30, 2017 at 23:01

Darshan Mehta

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8 Comments

Adnan Quraishi Over a year ago

where are you checking the status contains value online||offline ..?

Darshan Mehta Over a year ago

In the comparison, I am comparing o2 to o1 which means online status will always get sorted before offline due to reverse alphabetical ordering

Adnan Quraishi Over a year ago

How can i do this below api 24 ... ? contactList.sort( requires api 24 current min api 18

Darshan Mehta Over a year ago

You can change the API version as per this answer : stackoverflow.com/questions/19465049/…

Adnan Quraishi Over a year ago

changing api will not allow the app to run on lower platforms ? is it ?

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