Groovy的"in“运算符用于字符串和GString元素的列表

Question

下面的Groovy代码输出一个空列表：

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，这种修改将导致正确的输出：

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。

实现这一点的正确Groovy方法是什么？

Answer 1

您可以使用*.扩展运算符轻松获取字符串(参见下面的list2示例)。但是您在那里的检查可以更轻松地使用intersect。

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']