使用jq获取特定密钥的所有父密钥

Question

我想做的是在json上"walk"，并为每个名为"base"的键添加一个键，该键是该键的完整"路径"。

这是我的JSON:

{
"entity": {
"product": {
"att": {
"number_of_values": "Number of values"
},
"title": {
"base": ["product", "products", "pdt", "pdts"]
}
},
"rabbit": {
"att": {
"number_of_values": "Number of values"
},
"title": {
"base": ["rabbit", "rabbits"]
}
}
}
}

我想要这样的结果：

{
"entity": {
"product": {
"att": {
"number_of_values": "Number of values"
},
"title": {
"base": ["product", "products", "pdt", "pdts"],
"path": "entity.product.title"
}
},
"rabbit": {
"att": {
"number_of_values": "Number of values"
},
"title": {
"base": ["rabbit", "rabbits"]
"path": "entity.rabbit.title"
}
}
}
}

事实上，我做到了，但我只得到一个子密钥：

walk(if type == "object" and .base then  keys[] as $k | .base |= {path: $k} else . end)

结果：

{
"entity": {
"product": {
"att": {
"number_of_values": "Number of values"
},
"title": {
"base": {
"path": "base"
}
}
},
"rabbit": {
"att": {
"number_of_values": "Number of values"
},
"title": {
"base": {
"path": "base"
}
}
}
}
}

我也试过这个：

walk(if type == "object" and .base then  paths(..) as $v | .base |= {path: $v} else . end)

结果：

{
"entity": {
"product": {
"att": {
"number_of_values": "Number of values"
},
"title": {
"base": {
"path": [
"base",
3
]
}
}
},
"rabbit": {
"att": {
"number_of_values": "Number of values"
},
"title": {
"base": {
"path": [
"base",
1
]
}
}
}
}
}

非常感谢您的建议和专业知识！

Answer 1

这里有一种方法：

reduce paths(objects | has("base")) as $p (.;
setpath($p + ["path"]; $p | join("."))
)

^在线演示

Answer 2

您可以使用path获得路径的数组表示，使用join(假设简单的、类似标识符的字段名(连接它，并使用setpath:设置它

jq '
reduce path((.. | objects | select(.base))) as $p (.;
setpath($p + ["path"]; $p | join("."))
)
'

{
"entity": {
"product": {
"att": {
"number_of_values": "Number of values"
},
"title": {
"base": [
"product",
"products",
"pdt",
"pdts"
],
"path": "entity.product.title"
}
},
"rabbit": {
"att": {
"number_of_values": "Number of values"
},
"title": {
"base": [
"rabbit",
"rabbits"
],
"path": "entity.rabbit.title"
}
}
}
}

演示

Answer 3

这是一次关于path、paths、setpath和objects的真实学习体验。在阅读其他答案之前，我显然重新实现了setpath:

def add_path(p; val):
if p|length == 0
then .path = val
else .[p[0]] |= add_path(p[1:]; val)
end;
reduce ([paths(..) | select(last == "base")] | unique | map(.[:-1]))[] as $path (.;
add_path($path; ($path | join(".")))
)