blocks|key|2361578|text|您应该改进您的API。但是，如果您不能这样做，那么试试下面的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2361579|class+User:+Mappable+{

++++var+name:+String?
++++var+val:+Int?

++++required+init?(map:+Map)+{+}

++++func+mapping(map:+Map)+{
++++++++name+<-+map["name"]

++++++++//+1:+Get+the+value+of+JSON+and+store+it+in+a+variable+of+type+Any?
++++++++var+userIdData:+Any?
++++++++userIdData+<-+map["userId"]

++++++++//+2:+Check+the+value+type+of+the+value+and+convert+to+Int+type
++++++++if+userIdData+is+Int+{
++++++++++++val+=+(userIdData+as!+Int)+
++++++++}+else+if+userIdData+is+String+{
++++++++++++val+=+Int(userIdData+as!+String)+
++++++++}
++++}
}|code-block|syntax|javascript|2361580|您可以参考以下文档：型铸造|offset|length|2361581|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/TypeCasting.html^0|0|0|A|3|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

You should improve your API. However, if you can't do that, then try this code:

<pre><code>class User: Mappable {

 var name: String?
 var val: Int?

 required init?(map: Map) { }

 func mapping(map: Map) {
 name &lt;- map["name"]

 // 1: Get the value of JSON and store it in a variable of type Any?
 var userIdData: Any?
 userIdData &lt;- map["userId"]

 // 2: Check the value type of the value and convert to Int type
 if userIdData is Int {
 val = (userIdData as! Int) 
 } else if userIdData is String {
 val = Int(userIdData as! String) 
 }
 }
}
</code></pre>

You can refer to this document: <a href="https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/TypeCasting.html" rel="nofollow noreferrer">Type Casting</a>

blocks|key|1031926|text|如果您的API是这样的-它是非常糟糕的API。您可以做的是有两个变量而不是一个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1031927|class+User:+Mappable+{

++++var+name:+String?
++++var+valInt:+Int?
++++var+valString:+String?

++++required+init?(map:+Map)+{+}

++++func+mapping(map:+Map)+{
++++++++name+<-+map["name"]
++++++++valInt++<-+map["val"]
++++++++valString+<-+map["val"]
++++}
}|code-block|syntax|javascript|1031928|您甚至可以添加将使您获得以下值的函数：|1031929|//+bellow+func+mapping(map:+Map){
func+getUserId()+->+String+{
+++++if(self.valInt+!=+nil)+{
+++++++++return+"\(valInt!)"
+++++}
+++++else+{
+++++++++return+valString!
+++++}
}|1031930|或者，如果让：|1031931|func+getUserId()+->+String+{
+++++if+let+userId++=+self.valInt+{
+++++++++return+"\(userId)"
+++++}
+++++if+let+userId+=+valString+{
+++++++++return+userId
+++++}
+++++return+""
}|1031932|或者使用选项，以便以后可以使用if+let+userId+=+object.getUserId()|offset|length|style|CODE|1031933|func+getUserId()+->+String?+{
+++++if(self.valInt+!=+nil)+{
+++++++++return+String(valInt)
+++++}
+++++else+{
+++++++++return+valString
+++++}
}|1031934|entityMap^0|0|0|0|0|0|0|F|Y|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@$Q|15|R|16|S|T]]|9|@]|A|$]]|$1|U|3|V|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

If your API is like this - it is very bad API. What you could do is have two variables instead of one:

<pre><code>class User: Mappable {

 var name: String?
 var valInt: Int?
 var valString: String?

 required init?(map: Map) { }

 func mapping(map: Map) {
 name &lt;- map["name"]
 valInt &lt;- map["val"]
 valString &lt;- map["val"]
 }
}
</code></pre>

You can even add function that will get you value like:

<pre><code>// bellow func mapping(map: Map){
func getUserId() -&gt; String {
 if(self.valInt != nil) {
 return "\(valInt!)"
 }
 else {
 return valString!
 }
}
</code></pre>

Or, using if let:

<pre><code>func getUserId() -&gt; String {
 if let userId = self.valInt {
 return "\(userId)"
 }
 if let userId = valString {
 return userId
 }
 return ""
}
</code></pre>

Or using optionals so later on you can use <code>if let userId = object.getUserId()</code>

<pre><code>func getUserId() -&gt; String? {
 if(self.valInt != nil) {
 return String(valInt)
 }
 else {
 return valString
 }
}
</code></pre>

blocks|key|2361651|text|在阅读了ObjectMapper的代码之后，我找到了一种更容易解决问题的方法，就是定制转换。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2361652|public+class+IntTransform:+TransformType+{

++++public+typealias+Object+=+Int
++++public+typealias+JSON+=+Any?

++++public+init()+{}

++++public+func+transformFromJSON(_+value:+Any?)+->+Int?+{

++++++++var+result:+Int?

++++++++guard+let+json+=+value+else+{
++++++++++++return+result
++++++++}

++++++++if+json+is+Int+{
++++++++++++result+=+(json+as!+Int)
++++++++}
++++++++if+json+is+String+{
++++++++++++result+=+Int(json+as!+String)
++++++++}

++++++++return+result
++++}

++++public+func+transformToJSON(_+value:+Int?)+->+Any??+{

++++++++guard+let+object+=+value+else+{
++++++++++++return+nil
++++++++}

++++++++return+String(object)
++++}
}|code-block|syntax|javascript|2361653|然后，使用对mapping函数的自定义转换。|style|CODE|2361654|class+User:+Mappable+{

++++var+name:+String?
++++var+userId:+Int?

++++required+init?(map:+Map)+{+}

++++func+mapping(map:+Map)+{
++++++++name+++<-+map["name"]
++++++++userId+<-+(map["userId"],+IntTransform())+//+here+use+the+custom+transform.
++++}
}|2361655|希望它能帮助那些有同样问题的人。:)|2361656|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Hearst-DD/ObjectMapper^0|4|C|0|0|0|6|7|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|13|8|@$A|14|B|15|K|L]]|9|@]|C|$]]|$1|M|3|N|5|F|7|16|8|@]|9|@]|C|$G|H]]|$1|O|3|P|5|6|7|17|8|@]|9|@]|C|$]]|$1|Q|3|-4|5|6|7|18|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]]]

After reading the code of <a href="https://github.com/Hearst-DD/ObjectMapper" rel="noreferrer">ObjectMapper</a>, I found an easier way to solve the problem, it's to custom the transform.

<pre><code>public class IntTransform: TransformType {

 public typealias Object = Int
 public typealias JSON = Any?

 public init() {}

 public func transformFromJSON(_ value: Any?) -&gt; Int? {

 var result: Int?

 guard let json = value else {
 return result
 }

 if json is Int {
 result = (json as! Int)
 }
 if json is String {
 result = Int(json as! String)
 }

 return result
 }

 public func transformToJSON(_ value: Int?) -&gt; Any?? {

 guard let object = value else {
 return nil
 }

 return String(object)
 }
}
</code></pre>

then, use the custom transform to the <code>mapping</code> function.

<pre><code>class User: Mappable {

 var name: String?
 var userId: Int?

 required init?(map: Map) { }

 func mapping(map: Map) {
 name &lt;- map["name"]
 userId &lt;- (map["userId"], IntTransform()) // here use the custom transform.
 }
}
</code></pre>

Hope it can help others who have the same problem. :)

I'm using <a href="https://github.com/Hearst-DD/ObjectMapper" rel="nofollow noreferrer">ObjectMapper</a> to map my JSON to Swift object.

I have the following Swift object:

<pre><code>class User: Mappable {

 var name: String?
 var val: Int?

 required init?(map: Map) { }

 func mapping(map: Map) {
 name &lt;- map["name"]
 val &lt;- map["userId"]
 }
}
</code></pre>

I have this JSON structure:

<pre><code>{
 "name": "first",
 "userId": "1" // here is `String` type.
},
{
 "name": "second",
 "userId": 1 // here is `Int` type.
}
</code></pre>

After mapping the JSON, the <code>userId</code> of <code>User</code> which <code>name</code> is <code>"first"</code> is null. 

How can I map <code>Int/String</code> to <code>Int</code>?

How to map different type using ObjectMapper?

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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我使用将JSON映射到Swift对象。我有以下Swift对象：class User: Mappable {    var name: String?    var val: Int?    required init?(map: Map) { }    func mapping(map: Map) {        n...

问如何使用ObjectMapper绘制不同类型的地图？
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问如何使用ObjectMapper绘制不同类型的地图？EN