如何删除除jq以外的所有键? [英] How do I remove all keys except one with jq?

问题描述

给出一个对象列表，其中包含许多我不想要的键:

Given a list of objects, with many keys I don't want:

[{
    "name": "Alice",
    "group": "Admins",
    "created": "2014"
}, {
    "name": "Bob",
    "group": "Users",
    "created": "2014"
}]

如何过滤这些对象以仅包含我想要的键?

How do I filter these objects to only include keys I want?

[{
    "name": "Alice"
}, {
    "name": "Bob"
}]

我已经尝试过jq '.[].name'，但这会提取值，而不是保留对象.

I've tried jq '.[].name' but that extracts the values, rather than preserving the objects.

推荐答案

您可以使用map()函数来过滤任何键:

You can use the map() function to filter any key:

jq 'map({name: .name})'

更新

由 @WilfredHughes 建议:上面的过滤器可以缩写为:

Update

Suggested by @WilfredHughes: The above filter can be abbreviated as follows:

jq 'map({name})'

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