Re: [Simple-support] Need Help deserializing

[Niall G. <gal...@ya...>]

It should be entry="Open" name should not exist as its inline. So change the openList annotation to use entry="Open" and not name="Open"

--- On Thu, 19/1/12, Sony Ericsson <101...@gm...> wrote:


From: Sony Ericsson <101...@gm...>
Subject: [Simple-support] Need Help deserializing
To: sim...@li...
Received: Thursday, 19 January, 2012, 5:23 PM


Hello,

Source XML:

<List>
        <ItemType>Fruits</ItemType>
        <Order>Yes</Order>
        <Open>
                  <id>111</id>
                  <name>Apple</name>
         </Open>
         <Open>
                   <id>222</id>
                   <name>Mango</name>
         </Open>
</List>

My Code looks like below:

     @Root(name = "List", strict = false)
      public class List {

           @ElementList(name = "Open", required = false, inline = true)
           private List<Openl> openList;

          @Element(name = "ItemType", required = false)
           private String itemType;

          @Element(name = "Order", required = false)
          private String shouldWeOrder;

          public String getOpenItemList() {
                return openList;
          }

          public String getItemType() {
                return itemType;
          }

          public String getOrder() {
                return shouldWeOrder;
          }
     }

    @Root(name="Open")
    public class Open {

          @Element(name = "id", required = false)
           private String id;

          @Element(name = "name", required = false)
          private String name;

          public String getId() {
                 return id;
          }

          public String getName() {
                 return name;
          }
    }


I am only able to deserialize ItemType and Order. My List - openList is always null, that may be because I am not constructing one. So, how to do that ? How to deserialize such XML ?


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