我有这个XML很棒:
<Products>
<Product ProductCode="C1010" CategoryName="Coins" />
<Product ProductCode="C1012" CategoryName="Coins" />
<Product ProductCode="C1013" CategoryName="Coins" />
</Products>
但它输出到这个JSON:
{"Products":{"Product":[{"@ProductCode":"C1010","@CategoryName":"Coins"},
{"@ProductCode":"C1012","@CategoryName":"Coins"},
{"@ProductCode":"C1013","@CategoryName":"Coins"}]}}
我想在我的json中没有“产品”分级,因为所有三行都是产品。这是我的C#代码:
//x is an XDocument.
JsonConvert.SerializeXNode(x, Formatting.None, false)
//JsonConvert.SerializeXNode(x); //I also tried without the formatting and the boolean.
当我将一个XDocument转换为XmlDocument并使用:
var xmlDocument = new System.Xml.XmlDocument();
using (var xmlReader = x.CreateReader())
{
xmlDocument.Load(xmlReader);
}
JsonConvert.SerializeXmNode(xmlDocument);
它给了我完全相同的输出。那么如何修改我的JSON解析,以便我有一个简单的产品列表。我更喜欢最干净的解决方案。
或许更清楚一点,我会像输出一样:
[{"@ProductCode":"C1010","@CategoryName":"Coins"},
{"@ProductCode":"C1012","@CategoryName":"Coins"},
{"@ProductCode":"C1013","@CategoryName":"Coins"}]
答案 0 :(得分:5)
使用方法调用
JsonConvert.SerializeXNode(x, Formatting.None, true);
这将省略根节点,并应创建您期望的内容。
答案 1 :(得分:0)
而不是使用xml writer或con version只需从原始xml文件中读取新文件中的写入文件流编写器。
基本上你会:
List xml_retrevedData = new List();
FileStramWriter fr = new FileStramWriter();
fr.Write("{"Products":[{"@ProductCode":" //colection item
variable1.data1","@CategoryName":"//colection item
variable1.data2"}, {"@ProductCode":"//colection item
variable2.data1","@CategoryName":"//colection item
variable11.data1"},
{"@ProductCode":"//colection item variable13.data1","@CategoryName":"//colection item variable13.data3"}]}");
// in side the file stream Writer
答案 2 :(得分:0)
试试这个
public string getData(ref XmlDocument doc) {
JObject productobj = new JObject();
var productsenum = from p in doc.GetElementsByTagName("product").Cast<XmlElement>()
select p;
JArray products = new JArray();
foreach (XmlElement p in productsenum) {
JObject pobj = new JObject();
pobj["ProductCode"] = p.GetAttribute("ProductCode");
pobj["CategoryName"]= p.GetAttribute("CategoryName");
products.Add(pobj);
}
JObject product = new JObject();
product["Product"] = products;
productobj["Products"] = product;
return productobj.ToString();
}