entityMap|blocks|key|c2me0|text|要扩展JustMichael的答案，您可以首先使用String.to_existing_atom/1将键转换为原子，然后使用Kernel.struct/2来构建结构：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|eq697|user_with_atom_keys+=+for+{key,+val}+<-+user,+into:+%25{}+do
++{String.to_existing_atom(key),+val}
end

user_struct+=+struct(UserInfo,+user_with_atom_keys)
#+%25UserInfo{basic_auth:+"Basic+Ym1hOmphYnJhMTc=",+firstname:+"foo",
+lastname:+"boo"}|code-block|syntax|javascript|79u6d|请注意，这使用String.to_existing_atom/1来防止VM达到全局Atom限制。|ctvm7^0|P|P|1Q|F|0|0|7|P|0^^$0|$]|1|@$2|3|4|5|6|7|8|O|9|@$A|P|B|Q|C|D]|$A|R|B|S|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|T|9|@]|E|@]|F|$J|K]]|$2|L|4|M|6|7|8|U|9|@$A|V|B|W|C|D]]|E|@]|F|$]]|$2|N|4|-4|6|7|8|X|9|@]|E|@]|F|$]]]]

To expand on JustMichael's answer, you can first convert the keys to atoms, using <code>String.to_existing_atom/1</code>, then <code>Kernel.struct/2</code> to build the struct:

<pre><code>user_with_atom_keys = for {key, val} &lt;- user, into: %{} do
 {String.to_existing_atom(key), val}
end

user_struct = struct(UserInfo, user_with_atom_keys)
# %UserInfo{basic_auth: "Basic Ym1hOmphYnJhMTc=", firstname: "foo",
 lastname: "boo"}
</code></pre>

Note that this uses <code>String.to_existing_atom/1</code> to prevent the VM from reaching the global Atom limit.

entityMap|blocks|key|agbb4|text|我的猜测是它不起作用，因为你在map中有键作为字符串，在struct中有原子。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|b5v2h^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|D|9|@]|A|@]|B|$]]|$2|C|4|-4|6|7|8|E|9|@]|A|@]|B|$]]]]

My wild guess would that it doesn't work because you have keys as strings in the map and as atoms in the struct.

entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://github.com/appcues/exconstructor|1|http://hexdocs.pm/exconstructor/ExConstructor.html|blocks|key|drgnl|text|你可以试试exconstructor。它做的正是你需要的。|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|csha5|下面是一个小示例：|bjhqq|defmodule+TestStruct+do
++defstruct+field_one:+nil,
++++++++++++field_two:+nil,
++++++++++++field_three:+nil,
++++++++++++field_four:+nil
++use+ExConstructor
end

TestStruct.new(%25{"field_one"+=>+"a",+"fieldTwo"+=>+"b",+:field_three+=>+"c",+:FieldFour+=>+"d"})
#+=>+%25TestStruct{field_one:+"a",+field_two:+"b",+field_three:+"c",+field_four:+"d"}|code-block|syntax|javascript|hr5m|文档链接：http://hexdocs.pm/exconstructor/ExConstructor.html|24dle^0|5|D|0|0|0|0|5|1E|1|0^^$0|$1|$2|3|4|5|6|$7|8]]|9|$2|3|4|5|6|$7|A]]]|B|@$C|D|E|F|2|G|H|W|I|@]|J|@$K|X|L|Y|C|Z]]|6|$]]|$C|M|E|N|2|G|H|10|I|@]|J|@]|6|$]]|$C|O|E|P|2|Q|H|11|I|@]|J|@]|6|$R|S]]|$C|T|E|U|2|G|H|12|I|@]|J|@$K|13|L|14|C|15]]|6|$]]|$C|V|E|-4|2|G|H|16|I|@]|J|@]|6|$]]]]

You can try <a href="https://github.com/appcues/exconstructor" rel="nofollow noreferrer">exconstructor</a>. It does exactly what you need.

Here's a small example:

<pre><code>defmodule TestStruct do
 defstruct field_one: nil,
 field_two: nil,
 field_three: nil,
 field_four: nil
 use ExConstructor
end

TestStruct.new(%{"field_one" =&gt; "a", "fieldTwo" =&gt; "b", :field_three =&gt; "c", :FieldFour =&gt; "d"})
# =&gt; %TestStruct{field_one: "a", field_two: "b", field_three: "c", field_four: "d"}
</code></pre>

Link to the documentation: <a href="http://hexdocs.pm/exconstructor/ExConstructor.html" rel="nofollow noreferrer">http://hexdocs.pm/exconstructor/ExConstructor.html</a>

entityMap|blocks|key|5gtm7|text|我会使用Ecto的变更集来实现这一点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|28q0f|defmodule+UserInfo+do
++use+Ecto.Schema

++schema+"user_info"+do
++++field+:basic_auth,+:string
++++field+:firstname,+:string
++++field+:lastname,+:string
++end
end|code-block|syntax|javascript|4idoa|然后在你的代码中的某个地方|atnfj|import+Ecto.Changeset

user_struct+=+cast(%25UserInfo{},+
+++++++++++++++++++your_user_map,+
+++++++++++++++++++UserInfo.__schema__(:fields))+
++++++++++++++%7C>+apply_changes|bppfd|使用ecto的好处是，您还添加了验证等好处，并且不必担心map中的字段是字符串或原子。|29t9e^0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|O|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|P|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|Q|9|@]|A|@]|B|$]]|$2|J|4|K|6|E|8|R|9|@]|A|@]|B|$F|G]]|$2|L|4|M|6|7|8|S|9|@]|A|@]|B|$]]|$2|N|4|-4|6|7|8|T|9|@]|A|@]|B|$]]]]

I'd use Ecto's changesets for this:

<pre><code>defmodule UserInfo do
 use Ecto.Schema

 schema "user_info" do
 field :basic_auth, :string
 field :firstname, :string
 field :lastname, :string
 end
end
</code></pre>

then somewhere in your code

<pre><code>import Ecto.Changeset

user_struct = cast(%UserInfo{}, 
 your_user_map, 
 UserInfo.__schema__(:fields)) 
 |&gt; apply_changes
</code></pre>

the nice thing about this use of ecto is that you also have added benefits like validation and you do not have to worry fields in your map are strings or atoms.

entityMap|blocks|key|f1rd2|text|++%25{"basic_auth"+=>+"Basic+Ym1hOmphYnJhMTc=",+"firstname"+=>+"foo",+"lastname"+=>+"boo"}
++%7C>+Poison.encode
++%7C>+(fn+{:ok,+json}+->+json+end).()
++%7C>+Poison.decode(as:+%25SapOdataService.Auth.UserInfo{})|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1hhhp|或者|unstyled|3naop|++~S({"basic_auth":"Basic+Ym1hOmphYnJhMTc=","firstname":"foo","lastname":"boo"})
++%7C>+Poison.decode(as:+%25SapOdataService.Auth.UserInfo{})|7ghic|请注意，这不会在UserInfo上使用@enforce_keys进行编译。|offset|length|style|CODE|63r4l^0|0|0|0|8|8|J|D|0^^$0|$]|1|@$2|3|4|5|6|7|8|Q|9|@]|A|@]|B|$C|D]]|$2|E|4|F|6|G|8|R|9|@]|A|@]|B|$]]|$2|H|4|I|6|7|8|S|9|@]|A|@]|B|$C|D]]|$2|J|4|K|6|G|8|T|9|@$L|U|M|V|N|O]|$L|W|M|X|N|O]]|A|@]|B|$]]|$2|P|4|-4|6|G|8|Y|9|@]|A|@]|B|$]]]]

<pre><code> %{"basic_auth" =&gt; "Basic Ym1hOmphYnJhMTc=", "firstname" =&gt; "foo", "lastname" =&gt; "boo"}
 |&gt; Poison.encode
 |&gt; (fn {:ok, json} -&gt; json end).()
 |&gt; Poison.decode(as: %SapOdataService.Auth.UserInfo{})
</code></pre>

or

<pre><code> ~S({"basic_auth":"Basic Ym1hOmphYnJhMTc=","firstname":"foo","lastname":"boo"})
 |&gt; Poison.decode(as: %SapOdataService.Auth.UserInfo{})
</code></pre>

note that this will not compile with <code>@enforce_keys</code> on <code>UserInfo</code>.

I am trying to convert maps to struct as follow:

I have a map:

<pre><code>iex(6)&gt; user 
%{"basic_auth" =&gt; "Basic Ym1hOmphYnJhMTc=", "firstname" =&gt; "foo",
 "lastname" =&gt; "boo"}
</code></pre>

The value should be applied to struct:

<pre><code>iex(7)&gt; a = struct(UserInfo, user)
%SapOdataService.Auth.UserInfo{basic_auth: nil, firstname: nil, lastname: nil}
</code></pre>

As you can see, the values of struct is nil, why?

Convert maps to struct

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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 我尝试将map转换为struct，如下所示： 我有一张地图： iex(6)> user                      %{"basic_auth" => "Basic Ym1hOmphYnJhMTc=", "firstname" => "foo",  "lastname" => "boo"} 该值应应用于struct： iex(7)> a = struct(UserInfo, use

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