Java/Groovy中的列表解析

dwthyt8l 于 2022-09-20 发布在 Java

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我喜欢分析以下列表。

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'])
        }
    }

这将在列表中产生1对1的Map：

[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，比如：

[test:[a:abc], test-1:[d:abc], test-2:[f:abc], test-3:[i:abc], test-4:[k:abc], test-5:[l:abc]]

有人能帮我这个忙吗？谢谢

Jenkins

来源：https://stackoverflow.com/questions/73782922/list-parsing-in-java-groovy

1条答案

按热度按时间

3wabscal1#

我不知道，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)
    }
}

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Java/Groovy中的列表解析

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