Groovy'；s"；在；具有字符串和GString元素的列表的运算符

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 }

但是，此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 }

但是这个“解决方案”感觉非常笨拙。

---

实现这一点的正确Groovy方法是什么？

您可以使用

*。

扩展运算符轻松获取字符串（请参见下面的

列表2

示例）。但是通过

intersect

，您可以更轻松地在那里进行检查

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']

List List=[“test-${1+2}”、“test-${2+3}”、“test-${3+4}]
列表subl=[“测试-1”、“测试-2”、“测试-3”]
assert subl.findAll{it in list}=[]错误
//对于较短的版本使用intersect，该版本使用equals
assert subl.intersect（列表）=['test-3']
//或者用布景。。。
assert subl.toSet（）.intersect（list.toSet（））==['test-3'].toSet（）
//搜索时传播到'toString（）'
列表列表2=[“test-${1+2}”、“test-${2+3}”、“test-${3+4}”]*.toString（）
assert subl.findAll{it in list2}=['test-3']

[test-3]

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']