使用 jq 删除与另一个 JSON 文件中的列表匹配的键

Question

I have a json file A:

{
  "remove" : ["foo", "bar"]
}

and a json file B:

{
    "someDynamicKey" : {
        "foo" : 1,
        "xyz" : 2,
        "bar" : "x"
     }

}

I want to remove all keys in file B that match in "remove" section in file A. The problem is that I don't know which keys would be in file A.

Expect:

{
    "someDynamicKey" : {
        "xyz" : 2
     }

}

I was trying

jq --slurpfile a A.json '. as $b | reduce  $a[] as $key ($b; . = del($b.$key))'  B.json

and got error :

jq: error: syntax error, unexpected '$', expecting FORMAT or QQSTRING_START (Unix shell quoting issues?) at <top-level>, line 1:
. as $b | reduce  $a[] as $key ($b; . = del($b.$key)) 

I am not sure how to do next or is it possible to achieve using jq? I appreciate any help!!

Answer 1

Keeping it simple:

jq --argfile A A.json '
  $A.remove as $keys 
  | .someDynamicKey
    |= with_entries( .key as $k
                     | if $keys | index($k)
                       then empty 
                       else . end)' B.json

Or if you want a one-liner and don't like deprecated features and don't mind not :

jq --slurpfile A A.json '$A[0].remove as $keys | .someDynamicKey |= with_entries(select( .key as $k | $keys | index($k) | not))' B.json