我有一个Dictionary<string, Tuple<string, List<string>>> email
,我想写入XML(Serialize),并从XML加载到Dictionary。
我这样写了XML:
public void WritetoXML(string path) {
var xElem = new XElement(
"emailAlerts",
email.Select(x => new XElement("email", new XAttribute("h1", x.Key),
new XAttribute("body", x.Value.Item1),
new XAttribute("ids", string.Join(",", x.Value.Item2))))
);
xElem.Save(path);
}
但是我坚持使用LoadXML,它接受XML路径并将其加载到此电子邮件类中的字典中
这是我到目前为止所做的:
public void LoadXML(string path) {
var xElem2 = XElement.Parse(path);
var demail = xElem2.Descendants("email").ToDictionary(x => (string)x.Attribute("h1"),
(x => (string)x.Attribute("body"),
x => (string)x.Attribute("body")));
}
背景信息我的XML应该是这样的
<emailAlerts>
<email h1='Test1' body='This is a test' ids='1,2,3,4,5,10,11,15'/>
</emailAlerts>
答案 0 :(得分:1)
试试这个:
var str = @"<emailAlerts>
<email h1='Test1' body='This is a test' ids='1,2,3,4,5,10,11,15'/>
</emailAlerts>";
var xml = XElement.Parse(str);
var result = xml.Descendants("email").ToDictionary(
p => p.Attribute("h1").Value,
p => new Tuple<string, List<string>>(
p.Attribute("body").Value,
p.Attribute("ids").Value.Split(',').ToList()
)
);
答案 1 :(得分:0)
您可以尝试这种方式:
public void LoadXML(string path)
{
var xElem2 = XElement.Load(path);
var demail = xElem2.Descendants("email")
.ToDictionary(
x => (string)x.Attribute("h1")
, x => Tuple.Create(
(string)x.Attribute("body")
, x.Attribute("ids").Value
.Split(',')
.ToList()
)
);
}
path
参数包含XML文件的路径,那么您应该使用XElement.Load(path)
而不是XElement.Parse(path)
。 Tuple.Create()
构建Tuple
实例通常比new Tuple()
样式