我有一个JSON文件:
{
"ClassIdentifier": "consumer-leads",
"StateMode": "txt-2300",
"StateGroups": []
}
{
"ClassIdentifier": "main",
"StateMode": null,
"StateGroups": [
{
"Status": "active",
"StateGroupName": "default"
},
{
"Status": "active",
"StateGroupName": "brown-space"
},
{
"Status": "active",
"StateGroupName": "txt-hosts"
}
]
}
{
"ClassIdentifier": "paid-media",
"StateMode": "txt-2300",
"StateGroups": []
}
{
"ClassIdentifier": "reports",
"StateMode": null,
"StateGroups": [
{
"Status": "active",
"StateGroupName": "txt-hosts"
},
{
"Status": "active",
"StateGroupName": "grey-space"
},
{
"Status": "active",
"StateGroupName": "default"
}
]
}
我需要的输出:
consumer-leads,txt-2300,null
main,null,brown-space|default|txt-hosts
paid-media,txt-2300,null
reports,null,default|grey-space|txt-hosts
请注意,StateGroups(如果它们一直存在)按StateGroupName排序为(或之前)它们被转换为由垂直条分隔的值字符串。
我所尝试的已经给了我部分结果,但没有真正的工作:
cat json_file |
jq -r '[ .ClassIdentifier,
.StateMode,
.StateGroups[]
]'
cat json_file |
jq -r '{ ClassIdentifier,
StateMode
} +
( .StateGroups[] | { StateGroupName, Status } )
'
cat json_file |
jq -r ' [ .ClassIdentifier,
.StateMode,
.StateGroups |= sort_by( .StateGroupName )
]'
更新:我们必须使用JQ 1.3,所以请记住这一点以便做出回应。
答案 0 :(得分:7)
这应该有效:
[
.ClassIdentifier,
.StateMode // "null",
(.StateGroups
| map(select(.Status=="active").StateGroupName)
| sort
| join("|")
| if .=="" then "null" else . end
)
] | @csv
产生:
"consumer-leads","txt-2300","null"
"main","null","brown-space|default|txt-hosts"
"paid-media","txt-2300","null"
"reports","null","default|grey-space|txt-hosts"
请注意,由于您使用的是1.3,因此join/1
无法使用def join(sep): sep as $sep
| reduce .[1:][] as $item (.[0]|tostring; . + $sep + $item)
;
。但是自己实施起来并不困难。
parameterized