Question

我实际上是在寻找一种基于几个＆＃34;索引＆＃34;合并几个（3-4）数组的解决方案。键。

示例阵列1：

{"Fan":[{ "Last Name":Mueller,"Firstname":Martin,"Adress":Madisson Square,"City":"New York","DegreesIndex":3,"SchoolIndex2":1,}]

DegreesIndex和Schoolindex引用另外两个阵列中的两个不同的键：

示例数组2：

{"Degrees":[
{"DegreesIndex":3,
"Key":"12759303,
"Degrees":1.6}]}

示例数组3：

{＆＃34;学校＆＃34;：[ {＆＃34; SchoolIndex＆＃34;：1，＆＃34;预告＆＃34;：＆＃34; 12759303.8，＆＃34;师＆＃34;：米勒}]}

如何基于＆＃34;索引＆＃34;合并该数组？在Windows10下使用JQ 1.5的键？

此致蒂莫

Answer 1

我相信这可能接近你正在寻找的东西：

# input: an object
def merge_by_index(obj; ix):
  ix as $index
  | . + (obj | map( select( ix == $index)) [0] )
  | del(ix) ;

清理完样本输入后，假定a1，a2和a3为三个顶级对象：

a1
| .Fan[0] |= merge_by_index(a2|.Degrees; .DegreesIndex)
| .Fan[0] |= merge_by_index(a3|.School; .SchoolIndex)

产生

{
  "Fan": [
    {
      "Last Name": "Mueller",
      "Firstname": "Martin",
      "Adress": "Madisson Square",
      "City": "New York",
      "Key": 12759303,
      "Degrees": 1.6,
      "Teaser": 12759303.8,
      "Trainer": "Miller"
    }
  ]
}

Answer 2

以下是我将如何处理这个问题。这比peak的方法更加冗长。

首先用数据定义一些数组。请注意，我冒昧地添加了另一个条目，用于测试当给定粉丝的 DegreesIndex 或 SchoolIndex 匹配时会发生什么。我喜欢在这里使用函数，因为您可以使用获取实际数据所需的任何逻辑轻松替换它们。

def array1: {
  "Fan": [
    {
      "Last Name":"Mueller", "First Name":"Martin",
      "Address": "Madisson Square", "City": "New York",
      "DegreesIndex": 3, "SchoolIndex": 1
    },
    {
      "Last Name":"Roberts", "First Name":"Bob",
      "DegreesIndex": 2, "SchoolIndex": 4
    },
    {
      "Last Name":"Skywalker", "First Name":"Luke",
      "DegreesIndex": 5, "SchoolIndex": 1
    }
  ]};

def array2: {
 "Degrees": [
     { "DegreesIndex":3, "Key": "12759303", "Degrees":1.6 },
     { "DegreesIndex":2, "Key": "2",        "Degrees":2 }
  ]};

def array3: {
  "School": [
    { "SchoolIndex":1, "Teaser":"12759303.8", "Trainer":"Miller" },
    { "SchoolIndex":2, "Teaser":"2",          "Trainer":"Miller" }
  ]};

现在定义一些简单的查找函数，它们将返回与指定键匹配的记录。请注意，如果找不到该项，则使用[...] [0]构造返回null，而不是完全省略 Fan 。

def LookupDegrees($i):
  [
      array2
    | .Degrees[]
    | select(.DegreesIndex == $i)
  ][0]
;

def LookupSchool($i):
  [
      array3
    | .School[]
    | select(.SchoolIndex == $i)
  ][0]
;

所有这一切使主要例程变得简单：

  array1
| .Fan[] 
| .Degrees = LookupDegrees(.DegreesIndex)
| .School = LookupSchool(.SchoolIndex)

这是我用jq -n -f file.jq

运行时得到的结果

{
  "Last Name": "Mueller",
  "First Name": "Martin",
  "Address": "Madisson Square",
  "City": "New York",
  "DegreesIndex": 3,
  "SchoolIndex": 1,
  "Degrees": {
    "DegreesIndex": 3,
    "Key": "12759303",
    "Degrees": 1.6
  },
  "School": {
    "SchoolIndex": 1,
    "Teaser": "12759303.8",
    "Trainer": "Miller"
  }
}
{
  "Last Name": "Roberts",
  "First Name": "Bob",
  "DegreesIndex": 2,
  "SchoolIndex": 4,
  "Degrees": {
    "DegreesIndex": 2,
    "Key": "2",
    "Degrees": 2
  },
  "School": null
}
{
  "Last Name": "Skywalker",
  "First Name": "Luke",
  "DegreesIndex": 5,
  "SchoolIndex": 1,
  "Degrees": null,
  "School": {
    "SchoolIndex": 1,
    "Teaser": "12759303.8",
    "Trainer": "Miller"
  }
}

如果您需要不同的嵌套或输出，这应该很容易调整。

JQ：在索引键上合并几个json数组

2 个答案: