如果你想从fetchedStudyName中删除所有存在于excludeStudyName中的元素，你可以这样做：

def toRemove = excludeStudyName.intersect(fetchedStudyName)
fetchedStudyName.removeAll(toRemove)

更多信息：

• DefaultGroovyMethods.intersect（）
• Groovy Goodness: Intersect Collections
• Apache Groovy: What Is Groovy Used For?

你很接近了：

def valid = fetchedStudyName.findAll  { a -> 
    !excludeStudyName.contains(a)
}

应该工作。
如果excludeStudyName是一个很大的列表，你应该考虑将它改为一个set来加速contains(a)：

Set excludeStudyName = new JsonSlurper().parseText(tempExcludeStudyName)

def valid = fetchedStudyName.findAll  { a -> 
    !excludeStudyName.contains(a)
}

这个方法如何
对你的代码做一个小改动

def valid = fetchedStudyName.findAll  { a -> 
    excludeStudyName.every { val ->
        !a.contains(val)
    }
}

您当前的代码获取所有与excludeStudyName匹配的值，并以valid返回它们。如果您的目标是获取与excludeStudyName不匹配的值，那么这里有一些选项。

def included = ["this", "is", "not", "here"]

def excluded = ["is"]

def valid = included.findAll  { a -> 
    excluded.any {a.contains(it)}
}

def valid1 = included.findAll  { a -> 
    excluded.every { val ->
        !a.contains(val)
    }
}

def valid2 = included.findAll  { a -> 
    !excluded.contains(a)
}

println("valid " + valid)
println("valid1 " + valid1)
println("valid2 " + valid2)

Output:
valid [this, is]
valid1 [not, here]
valid2 [this, not, here]