英文:
Jolt spec to apply all keys in a parent object to all objects in an array
问题
Here is the translation of the code part you provided:
给定一个包含各种键的输入数组和一个 active_candidates_group 键，该键是候选对象的数组，创建一个新的输出数组，其中每个候选对象与父对象的键组合在一起。
需要注意以下几点：
- 父对象中有数百个键，最好不要硬编码它们。
- 父对象的键的顺序不重要。
- 父对象和候选对象之间不应有重复的键。
输入
[
{
"id": 123,
"name": "foo",
// { 数百个其他键 }
"active_candidates_group": [
{
"person_id": 123,
"person_name": "john"
// { 与‘候选’相关的许多字段 ——不希望与父对象重复 }
},
{
"person_id": 321,
"person_name": "bob"
// { 与‘候选’相关的许多字段 ——不希望与父对象重复 }
}
]
}
]
期望的输出
[
{
"id": 123,
"name": "foo",
// { 来自父对象的数百个其他键 }
"person_id": 123,
"person_name": "john"
// { 其他候选字段 }
},
{
"id": 123,
"name": "foo",
// { 来自父对象的数百个其他键 }
"person_id": 321,
"person_name": "bob"
// { 其他候选字段 }
}
]
我已翻译了代码部分，如果需要进一步解释或帮助，请随时告诉我。
英文:
Given an input array with objects containing various keys and an active_candidates_group key, with this key being an array of candidate objects, create a new output array where each candidate object is combined with the keys from the parent object.
It is important to note the following points:
- There are hundreds of keys in the parent object and it's not ideal to hard-code them.
- The parent object's keys' order does not matter.
- There should be no duplicate keys between the parent object and the candidate objects.
Input
[
{
"id": 123,
"name": "foo",
// { hundreds of other keys }
"active_candidates_group": [
{
"person_id": 123,
"person_name": "john"
// { many fields specific to ‘candidate’ —no duplicates to parent expected }
},
{
"person_id": 321,
"person_name": "bob"
// { many fields specific to ‘candidate’ —no duplicates to parent expected }
}
]
}
]
Desired Output
[
{
"id": 123,
"name": "foo",
// { hundreds of other keys from parent }
"person_id": 123,
"person_name": "john"
// { other candidate fields }
},
{
"id": 123,
"name": "foo",
// { hundreds of other keys from parent }
"person_id": 321,
"person_name": "bob"
// { other candidate fields }
}
]
What I tried so far:
[
{
"operation": "shift",
"spec": {
"*": { // Iterate for each object in the input array
"active_candidates_group": { // Match 'active_candidates_group' key
"*": { // Iterate for each object inside 'active_candidates_group'
"@2": "&2[&1]", // Copy all keys from the parent object to the same index in the output array
"*": "&2[&1].&" // Copy the current object to its corresponding index in the output array
}
}
}
}
},
{
"operation": "remove",
"spec": {
"active_candidates_group": {
"*": {
"active_candidates_group": ""
}
}
}
}
]
答案1
得分: 2
你离解决方案很近了，只需在两个不同级别使用"*": "[&1].&"稍微扭曲一下，而不需要硬编码单独的属性，如下所示：
[
{
"operation": "shift",
"spec": {
"*": {
"active_candidates_group": {
"*": {
"@2": { "*": "[&1].&" }, // 在更深的级别匹配
"*": "[&1].&"
}
}
}
}
},
{ // 仅对数组进行清除处理
"operation": "remove",
"spec": {
"*": {
"active_candidates_group": ""
}
}
}
]
在网站http://jolt-demo.appspot.com/上的演示如下：
英文:
You are so close to the solution, just distort a bit through use of "*": "[&1].&" at two different level without hardcoding the attributes individually such as
[
{
"operation": "shift",
"spec": {
"*": {
"active_candidates_group": {
"*": {
"@2": { "*": "[&1].&" }, // match at the deeper level
"*": "[&1].&"
}
}
}
}
},
{ // only purify for the array everyehere
"operation": "remove",
"spec": {
"*": {
"active_candidates_group": ""
}
}
}
]
the demo on the site http://jolt-demo.appspot.com/ is
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