Sort indexPathsForSelectedRows in Swift - objective-c

Question

When you invoke tableView.indexPathsForSelectedRows(), it returns an array of AnyObject type with the indexPaths of the rows selected in the order that user has selected them.
What I want is to sort the indexPaths in order to get ascending order. Before Swift, it could be achieved with this:
NSArray *sortedIndexPaths = [[tableView indexPathsforSelectedRows]
sortedArrayUsingSelector:#selector(compare:)];
But I have tried to implement it with Swift and it does not work :S
Anyone knows how to deal with this?
Thanks in advance!

Answer 1

it's simple code to sort array of NSIndexPath objects stored in paths variable .
The trick is in casting to [NSIndexPath]. Now you can have your array sorted.
let paths = tableView.indexPathsForSelectedRows() as [NSIndexPath]
let sortedArray = paths.sorted {$0.row < $1.row}
OR
if you wish to have separate function for that like:
func compare (obj0: NSIndexPath, obj1: NSIndexPath) -> Bool {
return obj0.row < obj1.row
}
then
let sortedArray = paths.sorted { compare($0, $1) }

Answer 2

I'm on Windows 7 at the moment so I cannot test but this is what I would expect to work. It may need a type annotation.
let paths = tableView.indexPathsforSelectedRows()
let sorted = paths.sortedArrayUsingSelector("compare:")

Answer 3

You can overload the < and > operators and then just call sort on it.
Define this globally
func <(left: NSIndexPath, right: NSIndexPath) -> Bool {
return left.section < right.section || left.row < right.row
}
Then you can just do this for ascending
let sortedIndexPaths = tableView.indexPathsForSelectedRows?.sort(<)
Obviously because of it returns an optional you would guard against it somehow, for example
guard let sortedIndexPaths = tableView.indexPathsForSelectedRows?.sort(<) else {
return
}

Answer 4

Here is Sort indexPathsForSelectedRows swift 4 code.
if var selection = tableView.indexPathsForSelectedRows
{
selection = selection.sorted{ $1.compare($0) == .orderedAscending }
}
If you have any issue ping me.