JavaScript: group items by a key, sort items within(!) the groups by another key - how?
grouping, javascript, jquery, sorting
Solution
The sort function needed to be expanded to cover when the rooms are equal. When the rooms are equal we then compare the age.
students.sort(function(a, b) {
if(a.room_id < b.room_id){
return -1;
}else if(a.room_id > b.room_id){
return 1;
}else{
if(a.age < b.age){
return -1
}else if(a.age > b.age){
return 1;
}else{
return 0;
}
}
});
Working Example http://jsfiddle.net/n9KNx/1/
Problem
in my results Students should be grouped(!) by Room-Id and then, within the group, sorted by age. Results grouped/sorted by Room: ``` Name | R | Age ------------------- Student 2 | 1 | 22 Student 4 | 1 | 25 Student 3 | 3 | 21 Student 6 | 3 | 27 Student 1 | 5 | 29 <- Wrong, should be: Student 5, Room 5, Age 23 Student 5 | 5 | 23 <- Wrong, should be: Student 1, Room 5, Age 29 ``` My current code (JSFiddle: http://jsfiddle.net/n9KNx/): ``` <script type="text/javascript"> /* add students */ var students = new Array(); students.push({ name: "Student 1", room_id: 5, age: 29 }); students.push({ name: "Student 2", room_id: 1, age: 22 }); students.push({ name: "Student 3", room_id: 3, age: 21 }); students.push({ name: "Student 4", room_id: 1, age: 25 }); students.push({ name: "Student 5", room_id: 5, age: 23 }); students.push({ name: "Student 6", room_id: 3, age: 27 }); /* sort by room_id */ students.sort(function(a, b) { return ((a.room_id < b.room_id) ? -1 : ((a.room_id > b.room_id) ? 1 : 0)); }); /* FIXME: sort by age without changing the room order */ /* ... code ... */ /* build HTML output */ var html = '<table>'; html += '<tr><th>Name</th><th>Room</th><th>Age</th></tr>'; for (var i = 0; i < students.length; i++) { html += '<tr><td>'+ students[i].name +'</td><td>'+ students[i].room_id +'</td> <td>'+ students[i].age +'</td></tr>'; } html += '</table>'; $("#students").html(html); </script> <div id="students">...</div> ``` How would you sort the students by age within each group? I added the code on JSFiddle, feel free to play with it!