Java/Groovy中的列表解析
Java/Groovy中的列表解析
提问于 2022-09-20 07:19:06
我喜欢解析以下列表。
LUT = [
[branch: "test", name: 'a', image_name: 'abc'],
[branch: "test", name: 'b', image_name: 'abc'],
[branch: "test", name: 'c', image_name: 'abc'],
[branch: "test-1", name: 'd', image_name: 'abc'],
[branch: "test-1", name: 'e', image_name: 'abc'],
[branch: "test-2", name: 'f', image_name: 'abc'],
[branch: "test-2", name: 'g', image_name: 'abc'],
[branch: "test-2", name: 'h', image_name: 'abc'],
[branch: "test-3", name: 'i', image_name: 'abc'],
[branch: "test-3", name: 'j', image_name: 'abc'],
[branch: "test-4", name: 'k', image_name: 'abc'],
[branch: "test-5", name: 'l', image_name: 'abc'],
]现在我要做的是：
result = [:]
for (map in LUT)
{
if (!result.containsKey(map['branch']))
{
println map['name'] // prints the unique name
result.put(map['branch'], map['name'])
}
}这将导致列表中的1到1张地图：
[test:a, test-1:d, test-2:f, test-3:i, test-4:k, test-5:l]但我想要的是一份清单，如：
[test:[a:abc], test-1:[d:abc], test-2:[f:abc], test-3:[i:abc], test-4:[k:abc], test-5:[l:abc]]有人能帮我吗，谢谢
回答 1
Stack Overflow用户
回答已采纳
发布于 2022-09-20 07:35:09
我不知道很帅，但我想我能帮上忙。
而不是这样：
result = [:]
for (map in LUT)
{
if (!result.containsKey(map['branch']))
{
println map['name'] // prints the unique name
result.put(map['branch'], map['name'])
}
}你需要这样的东西：
result = [:]
for (map in LUT)
{
if (!result.containsKey(map['branch']))
{
println map['name'] // prints the unique name
newMap = [:]
newMap.put(map['name'], map['image_name'])
result.put(map['branch'], newMap)
}
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/73782922
复制相关文章
相似问题
腾讯云开发者
