我试图转换一个字符串列表:
def description = """user:some.mail@gmail.com;groups:Team-1, Team-2
user:some.othermail@gmail.com;groups:Team-2, Team-3
user:another.mail@gmail.com;groups:Team-1, Team-3
some other text"""
description = description.split('\\r\\n|\\n|\\r').findAll { it.startsWith('user') }变成了一张地图,看起来是这样的:
[some.mail@gmail.com: "Team-2, Team-3", some.othermail@gmail.com: "Team-2, Team-3", another.mail@gmail.com: "Team-1, Team-3"]所以我可以迭代得到一个电子邮件地址和相应的团队。
遗憾的是,使用下面的代码,我只能部分地实现它,而且只能实现列表中的一项。我被困在一个循环中,得到了完整的结果。
def userData = [:]
userData = description[0].split(';').inject([:]) { map, token ->
token.split(':').with {
map[it[0].trim()] = it[1].trim()
}
map
}你能给我一个提示吗?我怎样才能从清单上得到一张包含所有物品的地图?
发布于 2022-04-05 05:32:24
您可以在列表上使用collectEntries方法:
def description = """user:some.mail@gmail.com;groups:Team-1, Team-2
user:some.othermail@gmail.com;groups:Team-2, Team-3
user:another.mail@gmail.com;groups:Team-1, Team-3
some other text"""
description = description.split('\\r\\n|\\n|\\r').findAll { it.startsWith('user') }
def map = description.collectEntries {
// split "user:some.mail@gmail.com;groups:Team-1, Team-2"
def split = it.split(';')
// remove "user:" prefix
def email = split[0].split(':')[1]
// remove "groups:" prefix
def groups = split[1].split(':')[1]
// create a map entry
[(email), groups]
}然后运行map.forEach {k, v -> println "key: '${k}', value: '${v}'"}打印如下:(在本例中,标准映射到字符串可能有点混乱)
key: 'some.mail@gmail.com', value: 'Team-1, Team-2'
key: 'some.othermail@gmail.com', value: 'Team-2, Team-3'
key: 'another.mail@gmail.com', value: 'Team-1, Team-3'https://stackoverflow.com/questions/71744137
复制相似问题