Mule 4 - 如何将具有相同 id 字段的嵌套数组中的数组合并为一个

问题描述

假设我有以下带有嵌套数组的有效负载，我如何将嵌套数组内的数组组合为相同的 externalId 以及某些字段上的某些逻辑，例如

shipQty - 此字段将在fillingOrder 下具有相同externalId 的记录求和或相加

serialNumbers - 如果 externalId 相同，serialNumbers 下的所有记录将一起显示

请参考下面的输入和预期输出

Json 有效载荷输入

     {
        "Identifier": "9i098p-898j-67586k","transactionDate": "2019-09-08T10:01:00-04:00","order": [
            {
                "orderNumber": "123456789","CourierOrderId": "1300-88-2525","fillingOrder": [
                    {
                        "numberOfBoxes": 0,"tracking": [
                            {
                                "carrier": "Orange","trackNum": "3333444","trackUrl": "https://www.orange.com/track/status","shipDate": "2019-09-08T10:01:00-04:00","SerialNumber": "00000123"
                            }
                        ],"row": [
                            {
                                "externalId": "1","unitNo": "OP04-123456-789","shipQty": 2,"serialNumbers": [
                                    {
                                        "serialNumber": "USD333555","quantity": 1
                                    },{
                                        "serialNumber": "USD235678","quantity": 1
                                    }
                                ]
                            }
                        ]
                    },{
                        "tracking": [
                            {
                                "carrier": "Apple","trackNum": "555666","trackUrl": "https://www.apple.com/track/status","SerialNumber": "00000645"
                            }
                        ],"shipQty": 3,"serialNumbers": [
                                    {
                                        "serialNumber": "USD123456",{
                                        "serialNumber": "USD98765",{
                                        "serialNumber": "USD45689","quantity": 1
                                    }
                                    
                                ]
                            }
                        ]
                    },{
                        "tracking": [
                            {
                                "carrier": "banana","trackNum": "587390","trackUrl": "https://www.banana.com/track/status","SerialNumber": "00000365"
                            }
                        ],"row": [
                            {
                                "externalId": "2","unitNo": "OP05-123456-111","serialNumbers": [
                                    {
                                        "serialNumber": "USD00045",{
                                        "serialNumber": "USD00046","quantity": 1
                                    }
                                    
                                ]
                            }
                        ]
                    }
                ]
            }
        ]
    }

预期的 Json 输出

{
    "row": [
        {
            "externalId": "1","shipQty": 5,//the shipQty should be add up when the externalId is same
            "serialNumbers": [ //the serialNumbers should display all the data inside the serialNumbers when the externalId is same
                {
                    "serialNumber": "USD333555","quantity": 1
                },{
                    "serialNumber": "USD235678",{
                    "serialNumber": "USD123456",{
                    "serialNumber": "USD98765",{
                    "serialNumber": "USD45689","quantity": 1
                }
            ]
        },{
            "externalId": "2","serialNumbers": [
                {
                    "serialNumber": "USD00045",{
                    "serialNumber": "USD00046","quantity": 1
                }
        }
    ]
}

解决方法

看起来您只需要payload的fillingOrder字段中的“row”数据。因此，简化问题的第一件事是将所有行作为单个数组。一旦你有了它们，你只需要按外部 ID 将它们分组，问题就会开始变得更小。

%dw 2.0
output application/json

//First get all rows since it looks like you only need them. 
//If you find this confusing try to use flatten with some simpler payloads. 
var allRows = flatten(flatten(payload.order.fillingOrder).row)

//Group them according to external id.
var groupedExtId = allRows groupBy $.externalId
---
{
    row: groupedExtId pluck ((value,extId,index) -> do {
        var sumShipQuant = sum(value.shipQty default [])
        ---
        {
            externalId: (extId),//the key after grouping is external id
            unitNo: value.unitNo[0],//assuming it is same across diff external id
            shipQty: sumShipQuant,serialNumbers: flatten(value.serialNumbers) //Flatten because value is an array and it has multiple serielNumbers array
        }
    })
}

,

这应该会有所帮助。我从 Harshank Bansal 的帖子中获得了一些灵感

%dw 2.0
output application/json
var groupFlat = flatten(flatten (payload.order.fillingOrder).row) groupBy ($.externalId)
---

row: [groupFlat mapObject ((value,key,index) -> {
    externalId: value.externalId[0],unitNO: value.unitNo[0],shipQty: sum(value.shipQty),serialNumbers: flatten(value.serialNumbers)
})]

,

试试这个：

%dw 2.0
output application/json
---
row:[ if (payload..order..row..externalId[0] == payload..order..row..externalId[1]) {
externalId :  payload..order..row..externalId[0],unitNo: payload..order..row..unitNo[0],shipQty: payload..order..row..shipQty[0] + payload..order..row..shipQty[1],serialNumbers:  flatten (payload..order..row..serialNumbers)  
}
else null]