Groovy:如何按最后一位对文本+数字的字符串数组进行排序 [英] Groovy: How to sort String array of text+numbers by last digit

问题描述

我有这样的数组:

    def arr = [ 
      "v3.1.20161004.0",
      "v3.1.20161004.1",
      "v3.1.20161004.10",
      "v3.1.20161004.11",
      "v3.1.20161004.2",
      "v3.1.20161004.3",
      "v3.1.20161004.30",
    ]

我需要得到这个:

    def arr = [ 
      "v3.1.20161004.0",
      "v3.1.20161004.1",
      "v3.1.20161004.2",
      "v3.1.20161004.3",
      "v3.1.20161004.10",
      "v3.1.20161004.11",
      "v3.1.20161004.30",
    ]

如何按最后一个数字'.x'排序?

How to sort it by last number '.x' ?

推荐答案

您可以将.上的每个字符串标记化，然后将最后一个元素作为Integer进行排序，并对此进行排序(传递false以返回新列表) )

You can tokenize each string on . and then grab the last element as an Integer, and sort on this (passing false to return a new list)

def newArray = arr.sort(false) { it.tokenize('.')[-1] as Integer }

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