如何:
session.ids = session.ids.unique(false)
更新
unique()
和unique(false)
之間的差異:第二個不修改原始列表。希望有所幫助。
def originalList = [1, 2, 4, 9, 7, 10, 8, 6, 6, 5]
//Mutate the original list
def newUniqueList = originalList.unique()
assert newUniqueList == [1, 2, 4, 9, 7, 10, 8, 6, 5]
assert originalList == [1, 2, 4, 9, 7, 10, 8, 6, 5]
//Add duplicate items to the original list again
originalList << 2 << 4 << 10
// We added 2 to originalList, and they are in newUniqueList too! This is because
// they are the SAME list (we mutated the originalList, and set newUniqueList to
// represent the same object.
assert originalList == newUniqueList
//Do not mutate the original list
def secondUniqueList = originalList.unique(false)
assert secondUniqueList == [1, 2, 4, 9, 7, 10, 8, 6, 5]
assert originalList == [1, 2, 4, 9, 7, 10, 8, 6, 5, 2, 4, 10]
它說空session.ids是空的,當我使用它像session.ids = session.ids.unique(),然後我把它作爲session.ids?.unique()和在一天會議結束。 ids是null.first case。 – danielad
,clouser的工作人員可以增加更多的意義,我的進一步名單業務謝謝。 – danielad
也許只是使用Set而不是List – Daniele