Mulesoft dataweave 外连接

如何解决Mulesoft dataweave 外连接

在使用 DataWeave 2.0 转换如下消息时需要您的帮助。

如果我有以下用户和订单的 JSON 有效负载。需要的输出是有订​​单（内连接）的用户和没有订单的用户列表。如果用户有订单列表中不存在的订单，则列表不应包含用户数据。

我尝试从 dataweave 使用外连接，但这会拉出有订单但不在订单列表中的用户。

请在下面找到示例。

var users = [
{
  "id": "1","name": "Jerney","order_id": "101",},{
  "id": "2","name": "Marty"
  "order_id": "102",{
  "id": "3","name": "Marty"
}
]


var orders = [
{
   "order_id": "101","product": "bread"
},{
   "order_id": "104","product": "broccoli"
}

]

需要的输出是

[
{
  "id": "1","product": "bread"
}
{
  "id": "3","name": "Marty"
}
]

我希望示例清除预期的输出。我尝试过滤（如外连接），但它也会在输入中包含订单 ID 102。

感谢您的帮助！

解决方法

我找到了一种方法来实现这一点，但我在测试中遇到了一些奇怪的问题，即连接中缺少 order_id，即使在过滤之后也是如此。我使用了额外的地图来解决。它不应该被需要。此脚本准确返回您想要的输出。

%dw 2.0
output application/json
import * from dw::core::Arrays
var users = [
    {
        "id": "1","name": "Jerney","order_id": "101"
    },{
        "id": "2","name": "Marty","order_id": "102"
    },{
        "id": "3","name": "Marty"
    } 
]

var orders = [
    {
        "order_id": "101","product": "bread"
    },{
        "order_id": "104","product": "broccoli"
    }
]

fun getUserWithOrders(users) = (users filter $.order_id?) 
        map ($ mapObject ($$):$) // for some reason had to add this to avoid a null error in the join

fun getUserWithoutOrders(users) = (users filter not $.order_id?) 
---
join(getUserWithOrders(users),orders,(u) -> u.order_id,(o) -> o.order_id)
    map ($.l ++ {product: $.r.product})
    ++ getUserWithoutOrders(users)

,

我建议如下（代码中嵌入了解释）：

%dw 2.0
output application/json

var users = [
{
  "id": "1","order_id": "101",},{
  "id": "2","order_id": "102",{
  "id": "3","name": "Marty"
}
]

var orders = [
{
   "order_id": "101","product": "bread"
},{
   "order_id": "104","product": "broccoli"
}
]

import leftJoin from dw::core::Arrays

---
// Do a left outer join to get all data from the users in the output
leftJoin(
    users,(u) -> u.order_id default "",(o) -> o.order_id default ""
)
// Iterate over the data and via pattern matching detect:
//   1. whether you have data in the l and the r nodes where 
//      you concatenate them and add them to the results
//   2. whether a sole l node has an order_id fields where
//      you add them to the result
//   3. throw aways all else
reduce (e,acc=[]) -> e match {
    case node if (node.l? and node.r?) -> acc + (node.l ++ node.r - "orderBy")
    case node if (node.l.order_id?) -> acc + node.l
    else node -> acc
}