获取 jq json 解析中的第一个（或第 n 个）元素
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get the first (or n'th) element in a jq json parsing
提问by Demian Glait
I can get the 1st element in a json inside []
我可以在 [] 中获取 json 中的第一个元素
$ echo '[{"a":"x", "b":true}, {"a":"XML", "b":false}]' | jq '.[1]'
{
"a": "XML",
"b": false
}
But if the json is already disassembled (for instance, after filtering entries using 'select'), how can I choose a single entry and avoid the error seen here?
但是，如果 json 已经被反汇编（例如，在使用“select”过滤条目之后），我该如何选择单个条目并避免出现此处的错误？
$ echo '[{"a":"x", "b":true}, {"a":"x", "b":false},{"a":"XML", "b":false}]' | jq '.[] | select( .a == "x")'
{
"a": "x",
"b": true
}
{
"a": "x",
"b": false
}
$ echo '[{"a":"x", "b":true}, {"a":"x", "b":false},{"a":"XML", "b":false}]' | jq '.[] | select( .a == "x") | .[1]'
jq: error (at <stdin>:1): Cannot index object with number
回答by hek2mgl
You can wrap the results from selectin an array:
您可以将结果包装select在一个数组中：
jq '[.[]|select(.a=="x")][0]' your.json
Output:
输出：
{
"a": "x",
"b": false
}
回答by jq170727
jq also provides first/0, last/0, nth/1so in this case the filter
jq 还提供first/0, last/0，nth/1所以在这种情况下过滤器
( map(select(.a == "x")) | first )
, ( map(select(.a == "x")) | last )
, ( map(select(.a == "x")) | nth(1) )
produces
产生
{
"a": "x",
"b": true
}
{
"a": "x",
"b": false
}
{
"a": "x",
"b": false
}
Additional streaming forms 'first/1', 'last/1'and 'nth/2'are also available so with this data
其他流形式'first/1'、'last/1'和'nth/2'也可用，因此使用此数据
( first(.[] | select(.a == "x")) )
, ( last(.[] | select(.a == "x")) )
, ( nth(1; .[] | select(.a == "x")) )
produces
产生
{
"a": "x",
"b": true
}
{
"a": "x",
"b": false
}
{
"a": "x",
"b": false
}
回答by tink
use map
用 map
cat raw.json|jq -r -c 'map(select(.a=="x"))|.[1]'
maprecivce a filterto filter an array.
map接收一个过滤器来过滤一个数组。
this command
这个命令
cat raw.json|jq -r -c 'map(select(.a=="x"))'
give the middle result
给出中间结果
[{"a":"x","b":true},{"a":"x","b":false}]
[{"a":"x","b":true},{"a":"x","b":false}]
.[1]take the first element
.[1]取第一个元素
