根据 Mule 3 中的条件从数组中删除对象

Question

我想从列表中删除配置文件为 Aftersales_Manager 且 raflag 为假的对象，但我Mule 3 中的 DataWeave 1.0 面临和问题。请建议 Mule 3 中正确的 DataWeave 代码。

%dw 1.0
%output application/json
---
payload filter ($.raflag != false and $.mule_temp_profile ='Aftersales_Manager') 
        map {
        "ldap": $.ldap,
        "status": $.status,
        "m_value": $.m_value,
        "raflag": $.raflag,
        "profile": $.profile
        }

输入：

[{
    "ldap": "V00075",
    "raflag": true,
    "profile": "Aftersales_Manager"
}, {
    "ldap": "V00076",
    "raflag": true,
    "profile": "Aftersales_Manager"
}, {
    "ldap": "V00077",
    "raflag": false,
    "profile": "Aftersales_Manager"
}, {
    "ldap": "V00078",
    "raflag": true,
    "profile": "Worker"
}, {
    "ldap": "V00079",
    "raflag": true,
    "profile": "Manager"
}]

预期输出：

[{
    "ldap": "V00075",
    "raflag": true,
    "profile": "Aftersales_Manager"
},
{
    "ldap": "V00076",
    "raflag": true,
    "profile": "Aftersales_Manager"
},
{
    "ldap": "V00078",
    "raflag": true,
    "profile": "Worker"
},
{
    "ldap": "V00079",
    "raflag": true,
    "profile": "Manager"
}
]

Answer 1

您输入的内容没有 mule_temp_profile。我怀疑你的意思是 profile。并且不需要将 raflag 与 false 进行比较，因为它已经是一个布尔值。该地图也不需要，因为您毕竟输出相同的 key-values。

%dw 1.0
%output application/json
---
payload filter ($.raflag or $.profile !='Aftersales_Manager')

输出：

[
  {
    "ldap": "V00075",
    "raflag": true,
    "profile": "Aftersales_Manager"
  },
  {
    "ldap": "V00076",
    "raflag": true,
    "profile": "Aftersales_Manager"
  },
  {
    "ldap": "V00078",
    "raflag": true,
    "profile": "Worker"
  },
  {
    "ldap": "V00079",
    "raflag": true,
    "profile": "Manager"
  }
]

根据 Mule 3 中的条件从数组中删除对象

Remove objects from array based on conditions in Mule 3

mule

mule-studio

dataweave

mulesoft

mule-component