List 将值重新指定给列表长生不老药
我有一个结构(数据),其中实际值取决于客户提出请求的情况,但我无法重新分配它们。这是我的代码: 基础结构:List 将值重新指定给列表长生不老药,list,enums,elixir,List,Enums,Elixir,我有一个结构(数据),其中实际值取决于客户提出请求的情况,但我无法重新分配它们。这是我的代码: 基础结构: Party: [ Sender: [ ] ] 还有一个关于客户真实价值的例子: Party: [ Sender: [ AgentUserSender: [ Name: "The_name", Type: "The_type", OtherIDs: [ {:OtherID, %{Description: "value
Party: [
Sender: [
]
]
还有一个关于客户真实价值的例子:
Party: [
Sender: [
AgentUserSender: [
Name: "The_name",
Type: "The_type",
OtherIDs: [
{:OtherID, %{Description: "value"}, "_value"},
{:OtherID, %{Description: "value_1"}, "_value_1"}
],
OtherKey: "other_value"
]
]
]
Party: [
Sender: [
TravelAgencySender: [
Name: "NAME",
IATA_Number: "xxxxxxxx",
AgencyID: "agency"
]
]
以及我(糟糕)如何努力做到:
data[:Party][:Sender] = customer[:Party][:Sender]
另外还有一个问题:并非所有客户在结构上都有相同的字段。这是其他客户的另一个示例:
Party: [
Sender: [
AgentUserSender: [
Name: "The_name",
Type: "The_type",
OtherIDs: [
{:OtherID, %{Description: "value"}, "_value"},
{:OtherID, %{Description: "value_1"}, "_value_1"}
],
OtherKey: "other_value"
]
]
]
Party: [
Sender: [
TravelAgencySender: [
Name: "NAME",
IATA_Number: "xxxxxxxx",
AgencyID: "agency"
]
]
非常感谢。我相信您要做的是:
> data = [Party: [Sender: []]]
[Party: [Sender: []]]
> data = put_in(data[:Party][:Sender], ["something"])
[Party: [Sender: ["something"]]]
或者这个:
> data = [Party: [Sender: ["something"]]]
[Party: [Sender: ["something"]]]
> data = put_in(data[:Party][:Sender], data[:Party][:Sender] ++ ["something_else"])
[Party: [Sender: ["something", "something_else"]]]
#Alternatively update_in instead of put_in
> data = [Party: [Sender: ["something"]]]
[Party: [Sender: ["something"]]]
> data = update_in(data[:Party][:Sender], &(&1 ++ ["something_else"]))
[Party: [Sender: ["something", "something_else"]]]
最直接的方法是使用: 如果您想要合并值,或者以某种复杂的方式修改它,只需修改
{ sender, cust[:Party][:Sender] }
行返回作为第二个元组项所需的内容(第一个是获得的值,因此保持不变)。仅将
客户
用作数据
有什么不对?为什么需要更新一个基本为空的对象?