CODEWITHSUNDEEP
c#xmlserialization
SexyMF
SexyMF

Reputation: 11165

XML Deserializing list of members to class

I have the following xml and I want to deserialize it blindly to this class:

public class FieldTranslations
{
    public FieldTranslation FirstName { get; set; }
    public FieldTranslation LastName { get; set; }
    public FieldTranslation Email { get; set; }
    public FieldTranslation PhoneNumber { get; set; }
    public FieldTranslation Country { get; set; }
}

public class FieldTranslation
{
    public string Title { get; set; }
    public string Helper { get; set; }
    public string ErrorMessage { get; set; }
}

XML:

<FieldsTranslations> 
    <FirstName>
        <Title>First Name</Title>
        <Helper>please insert your first name</Helper>
        <ErrorMessage>First name is suppose to be more that 2 characters</ErrorMessage>
    </FirstName>

    <LastName>
        <Title>Last Name</Title>
        <Helper>please insert your last name</Helper>
        <ErrorMessage>Last name is suppose to be more that 2 characters</ErrorMessage>
    </LastName>

    <Email>
        <Title>Email</Title>
        <Helper>please insert your email address (ex:[email protected])</Helper>
        <ErrorMessage>email address is incorrect. the format should be: ex: [email protected]</ErrorMessage>
    </Email>

    <PhoneNumber>
        <Title>Phone Number</Title>
        <Helper>Please insert your phone number ex:(972) 14262315</Helper>
        <ErrorMessage>Phone number is incorrect see example: (972) 14262315 </ErrorMessage>
    </PhoneNumber>

    <Country>
        <Title>Country</Title>
        <Helper>Please choose your country of residence</Helper>
        <ErrorMessage>Country must be chosen</ErrorMessage>
    </Country>
</FieldsTranslations>

When I say blindly it means that each time Im adding a member to FieldTranslations , if there is name match, it will be deserialized without explicitly doing anything.

Upvotes: 1

Views: 78

Answers (1)

Marc Gravell
Marc Gravell

Reputation: 1062965

var serializer = new XmlSerializer(typeof(FieldTranslations));
var obj = (FieldTranslations)serializer.Deserialize(source);

with one tweak to the classes (where the class name and root element name are not an exact match):

[XmlRoot("FieldsTranslations")]
public class FieldTranslations
{
    // ... your code as before
}

Upvotes: 3

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