下面的Groovy代码打印一个空列表：
List<String> list = ["test-${1+2}","test-${2+3}","test-${3+4}"]
List<String> subl = ["test-1","test-2","test-3"]
println subl.findAll { it in list }
输出：
[]
List<String> list = ["test-${1+2}" as String,"test-${2+3}" as String,"test-${3+4}" as String]
List<String> subl = ["test-1","test-3"]
println subl.findAll { it in list }
输出：
[test-3]
解决方法
你可以使用*.扩展运算符以轻松获取字符串(请参阅下面的list2示例).但是你的检查可以通过交叉更容易完成.
List<String> list = ["test-${1+2}","test-3"]
assert subl.findAll{ it in list }==[] // wrong
// use intersect for a shorter version,which uses equals
assert subl.intersect(list)==['test-3']
// or with sets...
assert subl.toSet().intersect(list.toSet())==['test-3'].toSet()
// spread to `toString()` on your search
List<String> list2 = ["test-${1+2}","test-${3+4}"]*.toString()
assert subl.findAll{ it in list2 }==['test-3']
