LINQ для XML.как мне получить доступ к этим данным? - PullRequest
0 голосов
/ 30 мая 2011

Элемент XML, к которому я обращаюсь: здесь, если нижеприведенный образец не читается: http://api.petfinder.com/pet.find?&animal=dog&count=150&location=90605&key=a94884c2349b5b6975ec84fb51ac3ec4

<pets> 
    <pet> 
        <id>19575609</id> 
        <shelterId>CA699</shelterId> <shelterPetId/> 
        <name>Louie</name> 
        <animal>Dog</animal> 
        <breeds> 
            <breed>Basset Hound</breed> 
        </breeds> 
        <mix>no</mix> 
        <age>Adult</age> 
        <sex>M</sex> 
        <size>M</size> 
        <options> ... </options>
        <description>
            <![CDATA[<div>Hi, my name is Louie. I&#39;m a great looking houndthat      really really wants his next home to be his final loving one. I&#39;m looking for an adult home that will have patience with me as I learn to trust humans again. I like other dogs and tolerate cats. I&#39;ve got lovely big front feet - just like a Basset should have. I walk well on a leash, and I&#39;m house-trained. I&#39;m a shy boy but I warm up once I get to know you and with lots of love and patience I&#39;m sure in no time I&#39;ll turn into a hound that is a happy-go-lucky boy. I&#39;m up to date on all vaccines and my adoption fee is $200.00. Please consider meeting me - I think once you do you will want to take me home and love me for the rest of my life. </div> <br/><a href="http://www.petfinder.com/petdetail/19575609">View this pet on Petfinder.com</a> ]]>
        </description> 
        <lastUpdate>2011-05-16T19:22:23Z</lastUpdate> 
        <status>A</status> 
        <media> 
            <photos>
                <photo size="x"  id="1">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-1-x.jpg</photo>
                 <photo size="t" id="1">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-1-t.jpg</photo> 
                 <photo size="pn" id="1">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-1-pn.jpg</photo> 
                 <photo size="pnt" id="1">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-1-pnt.jpg</photo> 
                 <photo size="fpm" id="1">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-1-fpm.jpg</photo> 
                 <photo size="x" id="2">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-2-x.jpg</photo> 
                 <photo size="t" id="2">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-2-t.jpg</photo> 
                 <photo size="pn" id="2">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-2-pn.jpg</photo> 
                 <photo size="pnt" id="2">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-2-pnt.jpg</photo> 
                 <photo size="fpm" id="2">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-2-fpm.jpg</photo> 
                 <photo size="x" id="3">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-3-x.jpg</photo> 
                 <photo size="t" id="3">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-3-t.jpg</photo> 
                 <photo size="pn" id="3">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-3-pn.jpg</photo> 
                 <photo size="pnt" id="3">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-3-pnt.jpg</photo> 
                 <photo size="fpm" id="3">http://photocache.petfinder.com/fotos/CA699/CA699.19575609-3-fpm.jpg</photo> 
             </photos> 
        </media> 
    </pet>
</pets>

Код, который я пытаюсь использовать, приведен ниже.атрибут, который я пытаюсь вернуть - это URL.

void processXML(XElement _xml)
    {

        XNamespace ns = "";
        var pets = from item in
                        _xml.Elements(ns + "pets").Elements(ns + "pet")
                    select item;

        foreach (var pet in pets)
        {
            Pet _pet = new Pet();

            _pet.id = (string)pet.Element(ns + "id");
            _pet.breeds = (string)pet.Element(ns + "breeds").Element(ns + "breed");
            _pet.name = (string)pet.Element(ns + "name");
            _pet.animal = (string)pet.Element(ns + "animal");

            **_pet.media = (string)pet.Element(ns + "media")
                                                        .Element(ns + "photos").Element(ns + "photo")
                                                        .Attribute(ns + "fpm");**


            Results.Items.Add(_pet);

        }
    }

Заранее благодарим за любые рекомендации.

rich

1 Ответ

0 голосов
/ 30 мая 2011

Если я понимаю, что вы должны сделать что-то вроде этого:

XmlElement photoList=pet.Element(ns + "media").Element(ns + "photos");

var url = from i in photoList.Descendants("photo")
    where i.attribute("size").Equals("fpm")
    select i.value;

В данный момент я не могу скомпилировать его, но я думаю, что это хорошая отправная точка!

обновлено

 void processXML(XElement _xml)
    {


        var pets = from item in
                       _xml.Elements("pet")
                   select item;

        foreach (var pet in pets)
        {
            //Pet _pet = new Pet();

             string id = (string)pet.Element("id");
             string breeds = (string)pet.Element("breeds").Element("breed");
             string name = (string)pet.Element("name");
             string animal = (string)pet.Element("animal");

             /*string media = (string)pet.Element(ns + "media")
                                                        .Element(ns + "photos").Element(ns + "photo")
                                                        .Attribute(ns + "fpm");*/



             var url = from i in pet.Descendants().Elements("photo")
                       where i.Attribute("size").Value.Equals("fpm")
                       select i.Value;
             //Results.Items.Add(_pet);

        }
    }
...