Sort array in Swift using a variable property?

Question

I have an NSTableView with an array datasource that I'm sorting when the user clicks on a column header. At the moment I have some IF statements to sort ascending/descending for every column name. Is it possible to perform a sort using a variable property name? That is, instead of hard coding .name in the following example, I could use a variable containing the name of the property instead of .name?

tableContents.sort {
    $0.name.localizedCaseInsensitiveCompare($1.name) == NSComparisonResult.OrderedAscending
}

This does work, but I have to duplicate this for every column in the table

consider to use an NSArrayController to get the sorting stuff for free — vadian
– vadian, Commented Aug 12, 2015 at 10:40

hennes · Accepted Answer · 2015-08-12 11:28:00Z

1

If all the properties that are used for sorting have the same type, you could use the built-in key-value access. Consider the following class A with two properties a and b both of which are strings.

class A: NSObject, Printable {
    var a: String
    var b: String

    init(a: String, b: String) {
        self.a = a
        self.b = b
    }

    override var description: String {
        return "{ a: \(self.a), b: \(self.b) }"
    }
}

Let's say your table contents are build from instances of this class.

var tableContents = [A(a: "1", b: "2"), A(a: "2", b: "1")]

To sort this array based on the values of either a or b you could use the following.

tableContents.sort {
    let key = "a"
    let oneValue = $0.valueForKey(key) as! String
    let otherValue = $1.valueForKey(key) as! String
    return oneValue.localizedCaseInsensitiveCompare(otherValue) == NSComparisonResult.OrderedAscending
}

println(tableContents) // [{ a: 1, b: 2 }, { a: 2, b: 1 }]

tableContents.sort {
    let key = "b"
    let oneValue = $0.valueForKey(key) as! String
    let otherValue = $1.valueForKey(key) as! String
    return oneValue.localizedCaseInsensitiveCompare(otherValue) == NSComparisonResult.OrderedAscending
}

println(tableContents) // [{ a: 2, b: 1 }, { a: 1, b: 2 }]

The only difference in the comparison block is the usage of either let key = "a" or let key = "b".

answered Aug 12, 2015 at 11:28

hennes

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1 Comment

Marc Fearby Over a year ago

My properties are not all of the same type, so I take your point. I'm probably asking the impossible. Your solution will probably still come in handy, though. Thanks.

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