JQ-根据其中一个值查找JSON对象并从中获取另一个值 [英] jq - Find a JSON object based on one of its values and get another value from it
本文介绍了JQ-根据其中一个值查找JSON对象并从中获取另一个值的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我最近才开始使用JQ,我想知道这样的事情是否可能。
示例:
{
"name": "device",
"version": "1.0.0",
"address": [
{
"address": "10.1.2.3",
"interface": "wlan1_wifi"
},
{
"address": "10.1.2.5",
"interface": "wlan2_link"
},
{
"address": "10.1.2.4",
"interface": "ether1"
}
],
"wireless": [
{
"name": "wlan1_wifi",
"type": "5Ghz",
"ssid": "wifi"
},
{
"name": "wlan2_link",
"type": "2Ghz",
"ssid": "link"
}
]
}
首先,让我们将该示例转换为以下json对象:
cat json | jq '. | {"name": ."name", "version": ."version", "wireless": [."wireless"[] | {"name": ."name", "type": ."type", "ssid": ."ssid"}]}'
{
"name": "device",
"version": "1.0.0",
"wireless": [
{
"name": "wlan1_wifi",
"type": "5Ghz",
"ssid": "wifi"
},
{
"name": "wlan2_link",
"type": "2Ghz",
"ssid": "link"
}
]
}
现在有一个问题。我需要为"wireless"
数组分配一个地址。地址存储在"address"
数组中。
所以问题是:有没有办法根据"name"
(在无线阵列中)和interface"
(在地址数组中)为"wireless"
数组中的每个json对象找到正确的json对象,然后将"address"
赋给它?
最终结果应如下所示:
{
"name": "device",
"version": "1.0.0",
"wireless": [
{
"name": "wlan1_wifi",
"type": "5Ghz",
"ssid": "wifi",
"address": "10.1.2.3"
},
{
"name": "wlan2_link",
"type": "2Ghz",
"ssid": "link",
"address": "10.1.2.5"
}
]
}
答案:
以下是我基于@Peak的answer的答案。我没有复制.wireless
的内容,然后使用map
,而是精心挑选了我只想包括的关键字。这也让我可以随心所欲地定位"地址"。
(INDEX(.address[]; .interface)) as $dict
| {name: .name, version: .version,
wireless: [.wireless[] | {name, address: ($dict[.name]|.address), type, ssid}]}
推荐答案
以下代码按照最初的请求生成输出:
(.wireless[].name) as $name
| .address[]
| select(.interface == $name)
| { wireless: {name: $name, address}}
但是,上述筛选器可能会产生多个结果,因此您可能需要进行相应的修改。
修订后的要求
如果您的JQ有INDEX/2
(仅在JQ 1.5发布后才可用),则只需使用它来创建查找表:
(INDEX(.address[]; .interface)) as $dict
| {name,
version,
wireless: (.wireless
| map(. + {address: ($dict[.name]|.address) }) ) }
或(可能取决于具体要求):
(INDEX(.address[]; .interface)) as $dict
| del(.address)
| .wireless |= map(. + {address: ($dict[.name]|.address) })
如果您的JQ没有INDEX/2
,那么您可以很容易地调整上面的(使用reduce
),或者更容易从https://github.com/stedolan/jq/blob/master/src/builtin.jq
INDEX/2
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