Question

我正在尝试使用相同的标签读取XML数据，但我无法获取所有数据。

示例XML代码：

<print_set>
    <line align="center" weight="normal" font="Courier New" height="20">-------------------------------</line>
    <line align="center" weight="normal" font="Courier New" height="20">MARRY BROWN FAMILY RESTAURANT</line>
    <line align="center" weight="normal" font="Courier New" height="20">Anna Nagar East</line>
    <line align="center" weight="normal" font="Courier New" height="20">AH 45, 4th Avenue</line>
    <line align="center" weight="normal" font="Courier New" height="20">5000023</line>
    <line align="center" weight="normal" font="Courier New" height="20">Shanthi Colony</line>
    <line align="center" weight="normal" font="Courier New" height="20">Chennai</line>
    <line align="center" weight="normal" font="Courier New" height="20">For Home Delivery : 46218777</line>
    <line align="center" weight="normal" font="Courier New" height="20">OrderOnline:marrybrownindia.com</line>
    <line align="center" weight="normal" font="Courier New" height="20">-------------------------------</line>
    <line align="left" weight="normal" font="Courier New" height="20">Employee:vinod</line>
    <line align="left" weight="normal" font="Courier New" height="20">Bill Number:Ma-70</line>
    <line align="left" weight="normal" font="Courier New" height="20">Date:26/07/2014    Time:12:14 PM</line>
    <line align="center" weight="normal" font="Courier New" height="20">-------------------------------</line>
</print_set>

我用来读它的代码：

XmlNodeList xnListCommon = xml.SelectNodes("/print_set");

foreach (XmlNode xn in xnListCommon)
{
    company_name = xn["line"].InnerText;
    font_style = xn["line"].Attributes["font"].Value;
    weight = xn["line"].Attributes["weight"].Value;
    alignment = xn["line"].Attributes["align"].Value;
    height = Convert.ToInt32(xn["line"].Attributes["height"].Value);
}

Answer 1

使用LINQ-to-XML，使用Descendants()非常简单。

从路径中，使用Load()：

var doc = XDocument.Load("path-to-file");

或者，从现有的XML字符串中，使用Parse()：

var doc = XDocument.Parse("your XML string");

然后，获取XML中的行：

var lines = doc.Root.Descendants("line");

如有必要，您可以保留元素名称。我不确定你到底做了什么，但你的循环看起来像这样：

foreach (var line in lines)
{
    company_name = line.Value;
    font_style = line.XAttribute("font").Value;
    weight = line.XAttribute("weight").Value;
    alignment = line.XAttribute("align").Value;
    height = Convert.ToInt32(line.XAttribute("height").Value);
}

使用C＃读取具有相同标签的xml数据

1 个答案: