当文件中有其他对象时，合并具有公共键值的两个JSON文件

Question

我有两个要合并的JSON文件，其中一个在这里：

文件1：

https://raw.githubusercontent.com/zcreativelabs/react-simple-maps/master/topojson-maps/world-110m.json

文件2：

https://gist.github.com/ArchieHicklin/764812b232a6cdc157d50da40bc033a5

文件1开始有一个拓扑对象和数组，以及许多可以忽略的东西：

{
"type": "Topology",
"arcs": [
    [
        [16814, 15074],
        [71, -45],
        [53, 16],
        [15, 54],
        [55, 18],
        [39, 37],
        [14, 95],
        [59, 24],
        [11, 42],
        [32, -32],
        [21, -3]
    ],

..。

然后进入一个国家数组(这是我想要增加价值的部分)：

"objects": {
    "ne_110m_admin_0_countries": {
        "type": "GeometryCollection",
        "geometries": [{
            "arcs": [
                [0, 1, 2, 3, 4, 5]
            ],
            "type": "Polygon",
            "properties": {
                "NAME": "Afghanistan",
                "NAME_LONG": "Afghanistan",
                "ABBREV": "Afg.",
                "FORMAL_EN": "Islamic State of Afghanistan",
                "POP_EST": 34124811,
                "POP_RANK": 15,
                "GDP_MD_EST": 64080,
                "POP_YEAR": 2017,
                "GDP_YEAR": 2016,
                "ISO_A2": "AF",
                "ISO_A3": "AFG",
                "CONTINENT": "Asia",
                "REGION_UN": "Asia",
                "SUBREGION": "Southern Asia"
            }
        },

文件2具有相同国家/地区的数组，但具有附加值：

{ "NAME": "Afghanistan", "BRANDONE": "Xiaomi", "BRANDTWO": "Apple", "BRANDTHREE": "Huawei", "BRANDFOUR": "Unknown", "BRANDFIVE": "Samsung", "BRANDSIX": "Oppo" },

正如你所看到的，文件1和文件2之间的公共值键是"NAME“。

理想情况下，我希望合并这些文件，以便原始文件包含来自另一个文件的额外值，例如：

"objects": {
"ne_110m_admin_0_countries": {
    "type": "GeometryCollection",
    "geometries": [{
        "arcs": [
            [0, 1, 2, 3, 4, 5]
        ],
        "type": "Polygon",
        "properties": {
            "NAME": "Afghanistan",
            "NAME_LONG": "Afghanistan",
            "ABBREV": "Afg.",
            "FORMAL_EN": "Islamic State of Afghanistan",
            "POP_EST": 34124811,
            "POP_RANK": 15,
            "GDP_MD_EST": 64080,
            "POP_YEAR": 2017,
            "GDP_YEAR": 2016,
            "ISO_A2": "AF",
            "ISO_A3": "AFG",
            "CONTINENT": "Asia",
            "REGION_UN": "Asia",
            "SUBREGION": "Southern Asia",
            "BRANDONE": "Xiaomi", 
            "BRANDTWO": "Apple", 
            "BRANDTHREE": "Huawei", 
            "BRANDFOUR": "Unknown", 
            "BRANDFIVE": "Samsung", 
            "BRANDSIX": "Oppo"
        }
    },

我在这里尝试了很多合并答案，但当我合并时

(如jq -s '[ .[0] + .[1] | group_by(.NAME)[] | select(length > 1) | add ]' clean.json world.json)

但我总是以一个错误结束：

jq: error (at file2.json:13575): array ([{"NAME":"A...) and object ({"type":"To...) cannot be added

有没有人知道我想做的事情在jq中是不是可以实现的？为什么我总是得到一个错误？

谢谢。

Answer 1

首先构造一个字典，然后使用它执行更新：

jq --argfile new 2.json '
 ($new | INDEX(.NAME)) as $dict
 | .objects.ne_110m_admin_0_countries.geometries[].properties
     |= (if $dict[.NAME] then . + $dict[.NAME] else . end)
' 1.json

顺便说一下..。

这里最困难的部分是确定相关的路径。jq出手相救：

jq -c 'paths(. == "Afghanistan")' 1.json

另一种可能性是盲目地更新与“properties”相关的每个值：

($new | INDEX(.NAME)) as $dict
| walk(if type == "object" and .properties
       then .properties |= . + $dict[.NAME]
       else . end)

这实际上会产生相同的结果。