Question

我尝试从我的ArrayList按顺序排序字符串。我有点困惑如何做到这一点。例如，我有我的订单清单：

def my_order = ['Version', 'Author', 'Somethink']
def files_sort = pack.files // here have string's to files

files_sort中包含db/author/test.sql，db/Somethink/test1.sql等随机字符串以及其他字符串。我认为最简单的方法是使用循环遍历我的文件并使用基于我的order_list的排序？例如：

files.sort.each {
   it.sort(here give my_order ??)
}

感谢您的提示！

编辑：

输入列表：

def files_to_sort = ['db/author/test1.sql', 'db/author/foo.sql', 'db/version/test1.sql', 'db/Somethink/foo.sql']

按my_order列表排序的输出列表：

def files_after_sort = ['db/version/test1.sql', 'db/author/test1.sql', 'db/author/foo.sql', 'db/Somethink/foo.sql']

Answer 1

def files = ['db/author/test1.sql', 'db/author/foo.sql', 'db/version/test1.sql', 'db/Somethink/foo.sql']
//convert array to uppercase values to support ignore-case
def my_order = ['Version', 'Author', 'Somethink'].collect{it.toUpperCase()} 

//closure to calculate sorting index of string `x` 
def my_index = {x->
    x = x.toUpperCase()
    return my_order.findIndexOf{ x.contains(it) }
}
//the compare in sort: first - by index, if index is same then by value itself
files.sort{a,b->  my_index(a)<=>my_index(b) ?: a<=>b }
println files

Answer 2

因此，要获得排序数组，我循环遍历my_order并再次循环遍历文件。

my_order.each{ type ->
        files.sort.each{ file ->
            if(file =~ /.*$type.*/){
                println file
            }
        }
}

由ArrayList对Groovy排序字符串

2 个答案: