ios - 使用 Swift 4 及更高版本对十六进制数进行排序

Question

我有像

这样的字符串

[
 "71ADDC23D2048",
 "9BF4F631EBA69",
 "B123454F4F631EBA69",
 "123454F4F6345FGT"
]

我当前的排序代码是:

// Sorting the Tans
func sortByTagID(arrayTags:[TansIDS],orderBy:String) -> [TansIDS] {

    if orderBy.uppercased() == OrderBy.Ascending {
        let sortedArray = arrayTags.sorted {String($0.trans-id!).localizedStandardCompare(String($1.trans-id!)) == .orderedAscending}
        return sortedArray
    }
    else {
        let sortedArray = arrayTags.sorted {String($0.trans-id!).localizedStandardCompare(String($1.trans-id!)) == .orderedDescending}
        return sortedArray
    }

}

请告诉我是否有比这更好的排序方法？因为当我与其他平台(Android)进行比较时，它不一样。

Answer 1

试试这个

func sortByTagID(arrayTags:[TansIDS],orderBy:String) -> [TansIDS] {

    if orderBy.uppercased() == OrderBy.Ascending {
        let sortedArray = arrayTags.sorted {String($0.trans-id!) < String($1.trans-id!)}
        return sortedArray
    }
    else {
        let sortedArray = arrayTags.sorted {String($0.trans-id!) > String($1.trans-id!)}
        return sortedArray
    }
}

输出升序:

[
 "123454F4F6345FGT",
 "71ADDC23D2048",
 "9BF4F631EBA69",
 "B123454F4F631EBA69"
] 

输出降序:

[
 "B123454F4F631EBA69",
 "9BF4F631EBA69",
 "71ADDC23D2048",
 "123454F4F6345FGT"
]