如何使用jq過濾不在列表中的選擇項？

Question

在JQ，我可以很容易在列表中選擇一個項目：

$ echo '["a","b","c","d","e"]' | jq '.[] | select(. == ("a","c"))'

或者如果您希望將其作為數組：

$ echo '["a","b","c","d","e"]' | jq 'map(select(. == ("a","c")))'

但是，如何選擇列表中沒有的所有項目？ 當然. != ("a","c") . != ("a","c")不起作用：

$ echo '["a","b","c","d","e"]' | jq 'map(select(. != ("a","c")))'
[
  "a",
  "b",
  "b",
  "c",
  "d",
  "d",
  "e",
  "e"
]

上面給出了每個項目兩次，除了"a"和"c ”

同樣的：

$ echo '["a","b","c","d","e"]' | jq '.[] | select(. != ("a","c"))'
"a"
"b"
"b"
"c"
"d"
"d"
"e"
"e"

如何過濾掉匹配的項目？

Answer 1

最簡單和最強大（wrt jq版本）的方法是使用內置- ：

$ echo '["a","b","c","d","e"]' | jq -c '. - ["a","c"]'
["b","d","e"]

如果黑名單很長並且有重復，那么刪除它們可能是合適的（例如，使用unique ）。

變化

問題也可以使用index解決（在jq 1.4及以上），而not ，例如

["a","c"] as $blacklist
| .[] | select( . as $in | $blacklist | index($in) | not) 

或者，從命令行傳入變量（jq --argjson blacklist ...）：

.[] | select( . as $in | $blacklist | index($in) | not) 

要保留列表結構，可以使用map( select( ...) ) 。

使用jq 1.5或更高版本，您也可以使用any或all ，例如

def except(blacklist):
  map( select( . as $in | blacklist | all(. != $in) ) );

特例：字符串

請參閱例如， 基於jq中的多個值選擇條目

Answer 2

我敢肯定這不是最簡單的解決方案，但它確實有效:)

$ echo '["a","b","c","d","e"]' | jq '.[] | select(test("[^ac]"))'

編輯：還有一個解決方案 - 這更糟糕:)

$ echo '["a","b","c","d","e"]' | jq '.[] | select(. != ("a") and . != ("b"))'