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AnswerRe: Coding help Pin
Wendelius21-Jan-15 7:19
mentorWendelius21-Jan-15 7:19 
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Member 1139217221-Jan-15 7:43
Member 1139217221-Jan-15 7:43 
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Pete O'Hanlon21-Jan-15 9:06
mvePete O'Hanlon21-Jan-15 9:06 
QuestionDeserialize Parts of an XML Pin
Antonio Cambule20-Jan-15 22:32
Antonio Cambule20-Jan-15 22:32 
AnswerRe: Deserialize Parts of an XML Pin
Daniel Pfeffer20-Jan-15 22:55
professionalDaniel Pfeffer20-Jan-15 22:55 
GeneralRe: Deserialize Parts of an XML Pin
Antonio Cambule20-Jan-15 23:13
Antonio Cambule20-Jan-15 23:13 
GeneralRe: Deserialize Parts of an XML Pin
Daniel Pfeffer21-Jan-15 0:40
professionalDaniel Pfeffer21-Jan-15 0:40 
GeneralRe: Deserialize Parts of an XML Pin
Antonio Cambule21-Jan-15 0:56
Antonio Cambule21-Jan-15 0:56 
Oh, I think I've been unclear.

It's not the same implementation for both examples.
It was just a showcase for two different xml files, that both will be handled in their
own implementation.

At least i don't know how to handle exmple 2 as example 1 works fine.

Here again a different xml example with the same problematic
With my kind of implementation i'm able to deserialize an xml like this:

XML
<?xml version="1.0"?>
<root>
  <project project_id="16576">
    <project_id><![CDATA[16576]]></project_id>
    <name><![CDATA[xxxx]]></name>
    <active><![CDATA[1]]></active>
  </project>
<root>
<root>
  <project project_id="16577">
    <project_id><![CDATA[16577]]></project_id>
    <name><![CDATA[xxxx]]></name>
    <active><![CDATA[0]]></active>
  </project>
<root>


I have an class List<item> with an class Item
and in List<t> i have the method to deserialize and with the given root information,
and the given <xmltype> and so on I'm able to deserialize every xml file, that has a list of tags
like above...

Now I have an xml file (complete different file and implementaion, means i would write other two classes for it)
that doesn't start directly with the needed list of data after the <root> node but later in the hierarchy

example again:

XML
<rootX>
  <data1>some info here</data1>
  <data2>some info there</data2>
  <coolList>
    <item kw="online email marketing" position="14" competition="0" traffic="5" url="http://www.marmato.de/"></result>
    <item kw="listbroking" position="20" competition="50" traffic="20" url="http://www.marmato.de/leistungen/e-mail-marketing/listbroking.html">
  </coolList>


I'm interested in the <coollist> and it's items. So how do i have to change my implementation to get out this list?

I hope i could clarify.
GeneralRe: Deserialize Parts of an XML Pin
Daniel Pfeffer21-Jan-15 2:51
professionalDaniel Pfeffer21-Jan-15 2:51 
AnswerRe: Deserialize Parts of an XML Pin
Gerry Schmitz21-Jan-15 13:00
mveGerry Schmitz21-Jan-15 13:00 
QuestionIE 11 Registry Entries Pin
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QuestionHex to UUEncode in C# Pin
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