Question

如何将每个属性作为XElement？

我基本上试图将IEnumerable对象转换为Web服务中的XML。

这是我尝试转换为XML的代码。

CollectionChanged

修改

类别类定义（使用EntityFramework 6 Code First）

observableCollection.CollectionChanged += (ss, ee) =>
{
    if(ee.Action == System.Collections.Specialized.NotifyCollectionChangedAction.Add)
    {
        SomeType newItem = ee.NewItems[0] as SomeType;
        someObject.AddItem(newItem);
    }
};

修改2

折扣类别定义

DataGrid

Product_Category_Mapping类定义

        using (var db = new nopMass())
        {
            var cats = db.Categories
                        .Where(x => x.Deleted == false
                                    && x.Published == true)
                        .OrderBy(x => x.DisplayOrder)
                        .AsEnumerable()
                        .Select(cat => new Category
                        {
                            Id = cat.Id,
                            Name = cat.Name,
                            Description = cat.Description,
                            MetaKeywords = cat.MetaKeywords,
                            MetaDescription = cat.MetaDescription,
                            MetaTitle = cat.MetaTitle,

                            PictureId = cat.PictureId,

                            DisplayOrder = cat.DisplayOrder,
                            CreatedOnUtc = cat.CreatedOnUtc,
                            Product_Category_Mapping = cat.Product_Category_Mapping,
                            ParentCategoryId = cat.ParentCategoryId,
                        })
                        .ToArray();


            XElement Configuration = new XElement("Collection",
                  cats
                  .ToList()
                  .Select(c => new XElement("Element", c)));

            return Configuration.ToString();
        }

Answer 1

使用节点的OuterXml属性。它会准确地返回你想要的东西。

 return Configuration.OuterXml;

Answer 2

因此你实际上需要xml字符串，最简单的方法就是简单的xml序列化。例如。如果您有以下类别：

var cats = new List<Category> {
    new Category { Id = 1, Name = "Auto", Description = "blah-blah-blah" },
    new Category { Id = 2, Name = "Moto", Description = "bikes!" }
};

序列化将类似于

var serializer = new XmlSerializer(typeof(List<Category>), 
                                   new XmlRootAttribute("Categories"));
using (var writer = new StringWriter())
{
    serializer.Serialize(writer, cats);
    return writer.ToString();
}

输出：

<?xml version="1.0" encoding="utf-16"?>
<Categories xmlns:xsd="http://www.w3.org/2001/XMLSchema"
            xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
  <Category>
    <Id>1</Id>
    <Name>Auto</Name>
    <Description>blah-blah-blah</Description>
  </Category>
  <Category>
    <Id>2</Id>
    <Name>Moto</Name>
    <Description>bikes!</Description>
  </Category>
</Categories>

Answer 3

这是我能想到的最佳解决方案。它有效，所以接受它作为我的答案。

$doc = new DOMDocument('1.0', 'UTF-8');

$internalErrors = libxml_use_internal_errors(true);

if ($doc->loadHTMLfile($url)) {

    libxml_use_internal_errors($internalErrors);
    $xpath = new DOMXpath($doc);

    $namepath = '//*[@id="mod-product-detail-1"]/div/div[1]/div/div[2]/h1';
    $pricepath = '//*[@id="mod-product-detail-1"]/div/div[1]/div/div[2]/div[3]/div[1]/span/span[2]';

    //get name
    $elements = $xpath->query($namepath);
    if (!is_null($elements) && isset($elements[0]->nodeValue)) {
        $data['name'] = $elements[0]->nodeValue;
    } else {
        $data['name'] = '';
    }

    //get price
    $price = $xpath->query($pricepath);
    if (!is_null($price) && isset($price[0]->nodeValue)) {
        $data['price'] = filter_var($price[0]->nodeValue, FILTER_SANITIZE_NUMBER_INT);
    } else {
        $data['price'] = '';
    }
}

将每个Property输出为XElement

3 个答案: