Groovy 的 "in"运算符，用于包含 String 和 GString 元素的列表

Question

以下 Groovy 代码打印一个空列表:

List<String> list = ["test-${1+2}", "test-${2+3}", "test-${3+4}"]
List<String> subl = ["test-1", "test-2", "test-3"]
println subl.findAll { it in list }

输出:

[]

但是，此修改会导致正确的输出:

List<String> list = ["test-${1+2}" as String, "test-${2+3}" as String, "test-${3+4}" as String]
List<String> subl = ["test-1", "test-2", "test-3"]
println subl.findAll { it in list }

输出:

[test-3]

但是这个“解决方案”感觉很笨重。
实现这一目标的正确 Groovy 方法是什么？

Answer 1

您可以使用 *.展开运算符以轻松获取字符串(请参见下面的 list2 示例)。但是您可以使用 intersect 更轻松地进行检查。 .

List<String> list = ["test-${1+2}", "test-${2+3}", "test-${3+4}"]
List<String> subl = ["test-1", "test-2", "test-3"]
assert subl.findAll{ it in list }==[] // wrong

// use intersect for a shorter version, which uses equals
assert subl.intersect(list)==['test-3']

// or with sets...
assert subl.toSet().intersect(list.toSet())==['test-3'].toSet()

// spread to `toString()` on your search
List<String> list2 = ["test-${1+2}", "test-${2+3}", "test-${3+4}"]*.toString()
assert subl.findAll{ it in list2 }==['test-3']