问题
In jq, I can select an item in a list fairly easily:
$ echo '["a","b","c","d","e"]' | jq '.[] | select(. == ("a","c"))'
Or if you prefer to get it as an array:
$ echo '["a","b","c","d","e"]' | jq 'map(select(. == ("a","c")))'
But how do I select all of the items that are not in the list? Certainly . != ("a","c")
does not work:
$ echo '["a","b","c","d","e"]' | jq 'map(select(. != ("a","c")))'
[
"a",
"b",
"b",
"c",
"d",
"d",
"e",
"e"
]
The above gives every item twice, except for "a"
and "c
"
Same for:
$ echo '["a","b","c","d","e"]' | jq '.[] | select(. != ("a","c"))'
"a"
"b"
"b"
"c"
"d"
"d"
"e"
"e"
How do I filter out the matching items?
回答1:
The simplest and most robust (w.r.t. jq versions) approach would be to use the builtin -
:
$ echo '["a","b","c","d","e"]' | jq -c '. - ["a","c"]'
["b","d","e"]
If the blacklist is very long and riddled with duplicates, then it might be appropriate to remove them (e.g. with unique
).
Variations
The problem can also be solved (in jq 1.4 and up) using index
and not
, e.g.
["a","c"] as $blacklist
| .[] | select( . as $in | $blacklist | index($in) | not)
Or, with a variable passed in from the command-line (jq --argjson blacklist ...):
.[] | select( . as $in | $blacklist | index($in) | not)
To preserve the list structure, one can use map( select( ...) )
.
With jq 1.5 or later, you could also use any
or all
, e.g.
def except(blacklist):
map( select( . as $in | blacklist | all(. != $in) ) );
Special case: strings
See e.g. Select entries based on multiple values in jq
回答2:
I'm sure it is not the most simple solution, but it works :)
$ echo '["a","b","c","d","e"]' | jq '.[] | select(test("[^ac]"))'
Edit: one more solution - this is even worse :)
$ echo '["a","b","c","d","e"]' | jq '.[] | select(. != ("a") and . != ("b"))'
来源:https://stackoverflow.com/questions/44563115/how-to-use-jq-to-filter-select-items-not-in-list