使用 JOLT 转换将示例 JSON 转换为嵌套 JSON 数组

Question

我遇到了一个问题，使用 jolt 转换将平面 JSON 转换为嵌套 JSON。而且我对颠簸转型很陌生。输入和输出详细信息如下。

我的输入：

[
{
"policyNo": 1,
"lProdCode": 500,
"name": "Prasad",
"id": "10",
"Age": "56"
},
{
"policyNo": 1,
"lProdCode": 500,
"name": "Mahapatra",
"id": "101",
"Age": "56"
},
{
"policyNo": 2,
"lProdCode": 500,
"name": "Pra",
"id": "109",
"Age": "56"
},
{
"policyNo": 3,
"lProdCode": 400,
"name": "Pra",
"id": "108",
"Age": "56"
},
{
"policyNo": 1,
"lProdCode": 500,
"name": "Pra",
"id": "108",
"Age": "56"
}
]

预期产出

[
{
"policyNo": 1,
"lProdCode": 500,
"beneficiaries": [
{
"name": "Prasad",
"id": "10900629001",
"Age": "56"
},
{
"name": "Mahapatra",
"id": "10900629001",
"Age": "56"
},
{
"name": "Pra",
"id": "108",
"Age": "56"
}
]
},
{
"policyNo": 2,
"lProdCode": 500,
"beneficiaries": [
{
"name": "Pra",
"id": "10900629001",
"Age": "56"
}
]
},
{
"policyNo": 3,
"lProdCode": 400,
"beneficiaries": [
{
"name": "Pra",
"id": "108",
"Age": "56"
}
]
}
]

Answer 1

原则上你需要按属性分组policyNo并为policyNo和lProdCode以外的属性生成一个新的列表(beneficiaries)。这可以在班次转换中处理。然后再添加三个步骤来修剪第一次转换产生的粗糙度，例如

[
{
"operation": "shift",
"spec": {
"*": {
"policyNo": "@(1,policyNo).&",
"lProdCode": "@(1,policyNo).&",
"*": "@(1,policyNo).beneficiaries[&1].&"
}
}
},
{
"operation": "modify-overwrite-beta",
"spec": {
"*": "=recursivelySquashNulls"
}
},
{
"operation": "cardinality",
"spec": {
"*": {
"policyNo": "ONE",
"lProdCode": "ONE"
}
}
},
{
"operation": "shift",
"spec": {
"*": ""
}
}
]