Question

在jq中，我可以非常轻松地在列表中选择项：

$ 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不的所有项目？当然. != ("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" ]

上述内容为每个项目提供两次，"a"和"c＆＃34除外;

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 out 匹配的项目？

Answer 1

最简单，最强大（w.r.t.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带有重复项，那么删除它们可能是合适的（例如使用unique）。

变体形式

使用index和not也可以解决问题（在jq 1.4及更高版本中），例如

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

或者，从命令行传递一个变量（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，您还可以使用any或all，例如

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

特例：字符串

参见例如Select entries based on multiple values in jq

Answer 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"))'

如何使用jq过滤不在列表中的选择项？

2 个答案:

变体形式

特例：字符串