Json jq:将对象数组转换为对象
我收到curl的回复,格式如下:Json jq:将对象数组转换为对象,json,bash,translation,jq,Json,Bash,Translation,Jq,我收到curl的回复,格式如下: [ { "list": [ { "value": 1, "id": 12 }, { "value": 15, "id": 13 }, { "value": -4, "id": 14 } ] }, ... ] { "12": "newId1", "13": "n
[
{
"list": [
{
"value": 1,
"id": 12
},
{
"value": 15,
"id": 13
},
{
"value": -4,
"id": 14
}
]
},
...
]
{
"12": "newId1",
"13": "newId2",
"14": "newId3"
}
.[].list |= reduce .[] as $i ({};
($i.id|tostring) as $k
| (select($mapping | has($k))[$mapping[$k]] = $i.value) // .
)
给定ID之间的映射,如下所示:
[
{
"list": [
{
"value": 1,
"id": 12
},
{
"value": 15,
"id": 13
},
{
"value": -4,
"id": 14
}
]
},
...
]
{
"12": "newId1",
"13": "newId2",
"14": "newId3"
}
.[].list |= reduce .[] as $i ({};
($i.id|tostring) as $k
| (select($mapping | has($k))[$mapping[$k]] = $i.value) // .
)
我想说:
[
{
"list": {
"newId1": 1,
"newId2": 15,
"newId3": -4,
}
},
...
]
这样我就得到了从ID到值的映射(在这一过程中,我希望重新映射ID)
我已经在这方面工作了一段时间了,每次我都会遇到死路一条
注:如有必要,我可以使用Shell或类似产品来预成型环
编辑:以下是我目前开发的一个版本:
jq '[].list.id = ($mapping.[] | select(.id == key)) | del(.id)' -M --argjson "mapping" "$mapping"
我不认为这是最好的,但我想看看是否能找到一个更接近我需要的旧版本。[编辑:下面的回答是对问题的回答,它描述了(a)如下所示的映射,以及(b)具有表单的输入数据:
[
{
"list": [
{
"value": 1,
"id1": 12
},
{
"value": 15,
"id2": 13
},
{
"value": -4,
"id3": 14
}
]
}
]
编辑结束]
在下文中,我将假设映射通过以下函数可用,但这是一个无关紧要的假设:
def mapping: {
"id1": "newId1",
"id2": "newId2",
"id3": "newId3"
} ;
def mapping:
{
"12": "newId1",
"13": "newId2",
"14": "newId3"
} ;
map( .list
|= (map( mapping[.id|tostring] as $mapped
| select($mapped)
| {($mapped): .value} )
| add) )
然后,以下jq滤波器将产生所需的输出:
map( .list
|= (map( to_entries[]
| (mapping[.key]) as $mapped
| select($mapped)
| {($mapped|tostring): .value} )
| add) )
剥猫皮的方法很多。我会这样做:
[
{
"list": [
{
"value": 1,
"id": 12
},
{
"value": 15,
"id": 13
},
{
"value": -4,
"id": 14
}
]
},
...
]
{
"12": "newId1",
"13": "newId2",
"14": "newId3"
}
.[].list |= reduce .[] as $i ({};
($i.id|tostring) as $k
| (select($mapping | has($k))[$mapping[$k]] = $i.value) // .
)
您只需通过单独的文件或参数提供映射
$ cat program.jq
.[].list |= reduce .[] as $i ({};
($i.id|tostring) as $k
| (select($mapping | has($k))[$mapping[$k]] = $i.value) // .
)
$ cat mapping.json
{
"12": "newId1",
"13": "newId2",
"14": "newId3"
}
$ jq --argfile mapping mapping.json -f program.jq input.json
[
{
"list": {
"newId1": 1,
"newId2": 15,
"newId3": -4
}
}
]
这是一个修正后的问题的免还原解决方案 在下文中,我将假设映射通过以下函数可用,但这是一个无关紧要的假设:
def mapping: {
"id1": "newId1",
"id2": "newId2",
"id3": "newId3"
} ;
def mapping:
{
"12": "newId1",
"13": "newId2",
"14": "newId3"
} ;
map( .list
|= (map( mapping[.id|tostring] as $mapped
| select($mapped)
| {($mapped): .value} )
| add) )
“选择”是为了安全(即,它检查所考虑的.id是否确实映射)。通过在第3行写入
{($mapped | tostring):.value}
来确保$mapped
是一个字符串也是合适的,.key
=0、1或2不是吗?映射变量中的索引不总是未定义的吗。我成功了!非常感谢,我已经为此挣扎了几个小时。@user2752635-有人改变了问题,所以我添加了一个单独的回答。