如何使用jq过滤不在列表中的选择项？

Question

在JQ，我可以很容易在列表中选择一个项目：

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

或者如果您希望将其作为数组：

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

但是，如何选择列表中没有的所有项目？ 当然. != ("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"
]

上面给出了每个项目两次，除了"a"和"c ”

同样的：

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

如何过滤掉匹配的项目？

Answer 1

最简单和最强大（wrt jq版本）的方法是使用内置- ：

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

如果黑名单很长并且有重复，那么删除它们可能是合适的（例如，使用unique ）。

变化

问题也可以使用index解决（在jq 1.4及以上），而not ，例如

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

或者，从命令行传入变量（jq --argjson blacklist ...）：

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

要保留列表结构，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

请参阅例如， 基于jq中的多个值选择条目

Answer 2

我敢肯定这不是最简单的解决方案，但它确实有效:)

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

编辑：还有一个解决方案 - 这更糟糕:)

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