用jq解析json树中的多个键/值
使用jq,我想从以下json中挑选键/值对:用jq解析json树中的多个键/值,json,parsing,key,jq,extraction,Json,Parsing,Key,Jq,Extraction,使用jq,我想从以下json中挑选键/值对: { "project": "Project X", "description": "This is a description of Project X", "nodes": [ { "name": "server001", "detail001": "foo", "detail002": "bar", "networks": [ { "net_t
{
"project": "Project X",
"description": "This is a description of Project X",
"nodes": [
{
"name": "server001",
"detail001": "foo",
"detail002": "bar",
"networks": [
{
"net_tier": "network_tier_001",
"ip_address": "10.1.1.10",
"gateway": "10.1.1.1",
"subnet_mask": "255.255.255.0",
"mac_address": "00:11:22:aa:bb:cc"
}
],
"hardware": {
"vcpu": 1,
"mem": 1024,
"disks": [
{
"disk001": 40,
"detail001": "foo"
},
{
"disk002": 20,
"detail001": "bar"
}
]
},
"os": "debian8",
"geo": {
"region": "001",
"country": "Sweden",
"datacentre": "Malmo"
},
"detail003": "baz"
}
],
"detail001": "foo"
}
为了举个例子,我想解析以下键及其值:“Project”、“name”、“net_tier”、“vcpu”、“mem”、“disk001”、“disk002”
我能够解析单个元素而没有太多问题,但是由于完整解析的层次性,我没有太多机会解析不同的分支(即网络和硬件>磁盘)
谢谢你的帮助
编辑:
为清楚起见,我要使用的输出是逗号分隔的CSV。就解析所有组合而言,现在覆盖示例中的示例数据即可。我希望能够详细介绍任何建议。这里有一种方法可以让您实现所需的输出 program.jq:
["project","name","net_tier","vcpu","mem","disk001","disk002"],
[.project]
+ (.nodes[] | .networks[] as $n |
[
.name,
$n.net_tier,
(.hardware |
.vcpu,
.mem,
(.disks | add["disk001","disk002"])
)
]
)
| @csv
基本上,您需要将所需的字段投影到数组中,以便将这些数组转换为csv行。您的输入使一个给定节点看起来可能有多个网络。因此,如果您想输出所有组合,则必须将其平展。这里有一种方法可以实现所需的输出 program.jq:
["project","name","net_tier","vcpu","mem","disk001","disk002"],
[.project]
+ (.nodes[] | .networks[] as $n |
[
.name,
$n.net_tier,
(.hardware |
.vcpu,
.mem,
(.disks | add["disk001","disk002"])
)
]
)
| @csv
基本上,您需要将所需的字段投影到数组中,以便将这些数组转换为csv行。您的输入使一个给定节点看起来可能有多个网络。因此,如果要输出所有组合,必须将其平铺。这里有另一种方法,它足够简短,足以说明问题:
def s(f): first(.. | f? // empty) // null;
[s(.project), s(.name), s(.net_tier), s(.vcpu), s(.mem), s(.disk001), s(.disk002)]
| @csv
调用:
$ jq -r -f value-pairs.jq input.json
结果:
"Project X","server001","network_tier_001",1,1024,40,20
带标题
使用与上述相同的s/1
:
. as $d
| ["project", "name", "net_tier", "vcpu", "mem", "disk001","disk002"]
| (., map( . as $v | $d | s(.[$v])))
| @csv
.project as $p
| ["project", "name", "net_tier", "vcpu", "mem", "disk001","disk002"] as $h
| ($h,
(.nodes[] as $d
| $h
| map( . as $v | $d | s(.[$v]) )
| .[0] = $p)
) | @csv
具有多个节点
再次使用s/1
如上所述:
. as $d
| ["project", "name", "net_tier", "vcpu", "mem", "disk001","disk002"]
| (., map( . as $v | $d | s(.[$v])))
| @csv
.project as $p
| ["project", "name", "net_tier", "vcpu", "mem", "disk001","disk002"] as $h
| ($h,
(.nodes[] as $d
| $h
| map( . as $v | $d | s(.[$v]) )
| .[0] = $p)
) | @csv
输出说明性多节点数据:
"project","name","net_tier","vcpu","mem","disk001","disk002"
"Project X","server001","network_tier_001",1,1024,40,20
"Project X","server002","network_tier_002",1,1024,,40
下面是另一种方法,它足够简短,足以说明问题:
def s(f): first(.. | f? // empty) // null;
[s(.project), s(.name), s(.net_tier), s(.vcpu), s(.mem), s(.disk001), s(.disk002)]
| @csv
调用:
$ jq -r -f value-pairs.jq input.json
结果:
"Project X","server001","network_tier_001",1,1024,40,20
带标题
使用与上述相同的s/1
:
. as $d
| ["project", "name", "net_tier", "vcpu", "mem", "disk001","disk002"]
| (., map( . as $v | $d | s(.[$v])))
| @csv
.project as $p
| ["project", "name", "net_tier", "vcpu", "mem", "disk001","disk002"] as $h
| ($h,
(.nodes[] as $d
| $h
| map( . as $v | $d | s(.[$v]) )
| .[0] = $p)
) | @csv
具有多个节点
再次使用s/1
如上所述:
. as $d
| ["project", "name", "net_tier", "vcpu", "mem", "disk001","disk002"]
| (., map( . as $v | $d | s(.[$v])))
| @csv
.project as $p
| ["project", "name", "net_tier", "vcpu", "mem", "disk001","disk002"] as $h
| ($h,
(.nodes[] as $d
| $h
| map( . as $v | $d | s(.[$v]) )
| .[0] = $p)
) | @csv
输出说明性多节点数据:
"project","name","net_tier","vcpu","mem","disk001","disk002"
"Project X","server001","network_tier_001",1,1024,40,20
"Project X","server002","network_tier_002",1,1024,,40
这里有一个不同的过滤器,它计算网络层和磁盘名称的唯一集合,然后生成一个结果,其中包含适合数据的列
{
tiers: [ .nodes[].networks[].net_tier ] | unique
, disks: [ .nodes[].hardware.disks[] | keys[] | select(startswith("disk")) ] | unique
} as $n
| def column_names($n): [ "project", "name" ] + $n.tiers + ["vcpu", "mem"] + $n.disks ;
def tiers($n): [ $n.tiers[] as $t | .networks[] | if .net_tier==$t then $t else null end ] ;
def disks($n): [ $n.disks[] as $d | map(select(.[$d]!=null)|.[$d])[0] ] ;
def rows($n):
.project as $project
| .nodes[]
| .name as $name
| tiers($n) as $tier_values
| .hardware
| .vcpu as $vcpu
| .mem as $mem
| .disks
| disks($n) as $disk_values
| [$project, $name] + $tier_values + [$vcpu, $mem] + $disk_values
;
column_names($n), rows($n)
| @csv
如果我们将另一个节点添加到样本数据中,这种方法的优点就会变得明显:
{
"name": "server002",
"networks": [
{
"net_tier": "network_tier_002"
}
],
"hardware": {
"vcpu": 1,
"mem": 1024,
"disks": [
{
"disk002": 40,
"detail001": "foo"
}
]
}
}
示例运行(假设在filter.jq
中使用filter,在data.json
中使用修改后的数据)
这里有一个不同的过滤器,它计算网络层和磁盘名称的唯一集合,然后生成一个结果,其中包含适合数据的列
{
tiers: [ .nodes[].networks[].net_tier ] | unique
, disks: [ .nodes[].hardware.disks[] | keys[] | select(startswith("disk")) ] | unique
} as $n
| def column_names($n): [ "project", "name" ] + $n.tiers + ["vcpu", "mem"] + $n.disks ;
def tiers($n): [ $n.tiers[] as $t | .networks[] | if .net_tier==$t then $t else null end ] ;
def disks($n): [ $n.disks[] as $d | map(select(.[$d]!=null)|.[$d])[0] ] ;
def rows($n):
.project as $project
| .nodes[]
| .name as $name
| tiers($n) as $tier_values
| .hardware
| .vcpu as $vcpu
| .mem as $mem
| .disks
| disks($n) as $disk_values
| [$project, $name] + $tier_values + [$vcpu, $mem] + $disk_values
;
column_names($n), rows($n)
| @csv
如果我们将另一个节点添加到样本数据中,这种方法的优点就会变得明显:
{
"name": "server002",
"networks": [
{
"net_tier": "network_tier_002"
}
],
"hardware": {
"vcpu": 1,
"mem": 1024,
"disks": [
{
"disk002": 40,
"detail001": "foo"
}
]
}
}
示例运行(假设在filter.jq
中使用filter,在data.json
中使用修改后的数据)
那么,您的预期产出是多少?您提到了字段,但您希望它是如何构造的?数据中还有一堆数组。如果有比示例中更多的内容,它将如何影响您的预期输出?您希望输出所有的组合吗?请在您的问题中更清楚地说明这一点。谢谢Jeff,为了清晰起见,我进行了编辑。那么,您的预期输出是什么呢?您提到了字段,但您希望它是如何构造的?数据中还有一堆数组。如果有比示例中更多的内容,它将如何影响您的预期输出?您希望输出所有的组合吗?请在你的问题中更清楚地说明这一点。谢谢杰夫,为了清楚起见,我已经编辑了。