I have the following service class:
public class SheetSorter {
// sorts the sheet
public static Sheet sortSheet(Sheet sheet) {
List<String[]> list = sheet.getSheet().stream().collect(Collectors.toList());
Collections.sort(list, new Comparator<String[]>() {
@Override
public int compare(String[] arr1, String[] arr2) {
if (Long.parseLong(arr1[3]) == Long.parseLong(arr2[3])) {
return (int) (Long.parseLong(arr1[4]) - Long.parseLong(arr2[4]));
}
else {
return (int) (Long.parseLong(arr1[3]) - Long.parseLong(arr2[3]));
}
}
});
sheet.setSheet(list);
return sheet;
}
}
As you can see, the sortSheet
methods implements sort
on column 4
and thenby
on column 5
. Is there any way I can convert this to take an Integer... columnOrder
as parameter so that this will run sort on columnOrder[0]
and 'thenBies' (or is it 'thenBys'?) on the remaining arguments supplied in that order?
EDIT:
@Holger's Suggestion
// sorts the sheet
public static Sheet sortSheet(Sheet sheet) {
List<String[]> list = new ArrayList<>(sheet.getSheet());
Collections.sort(list, new Comparator<String[]>() {
@Override
public int compare(String[] arr1, String[] arr2) {
if (Long.parseLong(arr1[3]) == Long.parseLong(arr2[3])) {
return Long.compare(Long.parseLong(arr1[4]), Long.parseLong(arr2[4]));
}
else {
return Long.compare(Long.parseLong(arr1[3]), Long.parseLong(arr2[3]));
}
}
});
sheet.setSheet(list);
return sheet;
}
CodePudding user response:
An easy way is to just compose your Comparator
based on the int array:
public static Sheet sortSheet(Sheet sheet, int...indices) {
if(indices.length == 0) throw new IllegalArgumentException();
Comparator<String[]> comp = Comparator.comparingLong(s -> Long.parseLong(s[indices[0]]));
for(int i = 1; i < indices.length; i ) {
int e = i;
comp = comp.thenComparing(Comparator.comparingLong(s -> Long.parseLong(s[indices[e]])));
}
List<String[]> list = sheet.getSheet().stream().collect(Collectors.toList());
Collections.sort(list, comp);
sheet.setSheet(list);
return sheet;
}
comp
above is created by composing other Comparator
s based on indices after indices[0]
.
CodePudding user response:
My Solution based on the accepted answer:
public static Sheet sortSheet(Sheet sheet, Integer... indices) {
List<String[]> list = new ArrayList<>(sheet.getSheet());
Collections.sort(list, new Comparator<String[]>() {
@Override
public int compare(String[] arr1, String[] arr2) {
for (int i = 0; i < indices.length;) {
if (Long.parseLong(arr1[i]) == Long.parseLong(arr2[i])) {
i ;
}
else {
return Long.compare(Long.parseLong(arr1[i]), Long.parseLong(arr2[i]));
}
}
return 0;
}
});
sheet.setSheet(list);
return sheet;
}