Question

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：

[]

但是，这种修改会产生正确的输出：

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＆＃34;解决方案＆＃34;感觉非常笨重什么是正确的Groovy方法来实现这一目标？

Answer 1

您可以使用*.点差运算符轻松获取字符串（请参阅下面的list2示例）。但是使用intersect可以更轻松地完成检查。

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']

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

1 个答案: