本文介绍了使用xmldocument读取xml的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
<?xml version="1.0" encoding="utf-8" ?>
<testcase>
<date>4/12/13</date>
<name>Mrinal</name>
<subject>xmlTest</subject>
</testcase>
我正在尝试使用c#读取上述xml,但是我在try catch块中得到了null异常,任何机构都可以提出所需的更改.
I am trying to read the above xml using c#, But i get null exception in the try catch block can any body suggest the required change.
static void Main(string[] args)
{
XmlDocument xd = new XmlDocument();
xd.Load("C:/Users/mkumar/Documents/testcase.xml");
XmlNodeList nodelist = xd.SelectNodes("/testcase"); // get all <testcase> nodes
foreach (XmlNode node in nodelist) // for each <testcase> node
{
CommonLib.TestCase tc = new CommonLib.TestCase();
try
{
tc.name = node.Attributes.GetNamedItem("date").Value;
tc.date = node.Attributes.GetNamedItem("name").Value;
tc.sub = node.Attributes.GetNamedItem("subject").Value;
}
catch (Exception e)
{
MessageBox.Show("Error in reading XML", "xmlError", MessageBoxButtons.OK);
}
.........
.............
推荐答案
testcase
元素没有属性.您应该查看它的子节点:
The testcase
element has no attributes. You should be looking to it's child nodes:
tc.name = node.SelectSingleNode("name").InnerText;
tc.date = node.SelectSingleNode("date").InnerText;
tc.sub = node.SelectSingleNode("subject").InnerText;
您可以像这样处理所有节点:
You might process all nodes like this:
var testCases = nodelist
.Cast<XmlNode>()
.Select(x => new CommonLib.TestCase()
{
name = x.SelectSingleNode("name").InnerText,
date = x.SelectSingleNode("date").InnerText,
sub = x.SelectSingleNode("subject").InnerText
})
.ToList();
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