为了学习的目的,我喜欢解析如下的查找表:我想检查分支名称并获取相应的名称值,但是我想要获取每个分支的名称,例如:=>:"test“=> name "a”than分支:"test-1“=> name "d”,等等,
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'],
]
例如,结果可能如下所示:
result = {
"test": [{ "name":"a" }],
"test-1": [{ "name":"d" }],
"test-2": [{ "name":"f" }],
"test-3": [{ "name":"i" }],
"test-4": [{ "name":"k" }],
"test-5": [{ "name":"l" }],
}
另外,稍后我可能会向键添加额外的值,例如test-1:"test-1":{ "name":"a","new_name":"new" }
发布于 2022-09-22 15:28:24
我不确定你的花括号,但这会创建一个地图:
def original = [
[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'],
]
println original
.groupBy { branch -> branch.branch }
.collectEntries { key,value -> [(key) : [name : value.get(0).name]] }
发布于 2022-09-19 14:36:52
不是积极的,如何按您的要求格式化结果,但找到每个分支的第一个实例相当容易。
result = [:]
for (map in LUT)
{
if (!result.containsKey(map['branch']))
{
println map['name'] // prints the unique name
result.put(map['branch'], map['name'])
}
}
https://stackoverflow.com/questions/73773498
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