应用程序将条目写入xml文档 - 作为数据库 - 将这些条目显示为datagridview中的行。每个“入口”节点都有其“entryno”属性。每次删除一个条目时，这个“entryno”编号序列就会被打破。我能够实现代码来为每个创建的新条目增加“entryno”，但在尝试上述操作时已经卡住了一段时间。
private void Deletbtn_Click(object sender, EventArgs e)
{
XmlNode node = doc.SelectSingleNode("//entry[@entryno='" + Dgv.CurrentRow.Cells[0].Value + "']");
try
{
if (node != null && Dgv.CurrentRow.Selected == true)
{
doc.DocumentElement.RemoveChild(node);
// Trying to refresh the sequence each time an node is deleted.
int Attrno = int.Parse(doc.SelectSingleNode("//entry[@entryno]").Value);
do
{
Attrno = 0;
Attrno++;
} while (Attrno < doc.ChildNodes.Count);
doc.SelectSingleNode("//entry[@entryno]").InnerText = Attrno.ToString();
doc.Save(Application.StartupPath + @"\SleepRecords.xml");
}
else
MessageBox.Show("Click the left side of the grid to select a row to delete.", "Select row", MessageBoxButtons.OK, MessageBoxIcon.Information);
}
catch (Exception ex)
{
MessageBox.Show(ex.Message, "Error", MessageBoxButtons.OK, MessageBoxIcon.Exclamation);
}
ReloadData();
Resetbtn_Click(Deletbtn, e);
makePdf();
}
仅供参考，这是我在每个创建的新条目（仅相关的代码段）上增加“entryno”属性的方法。
XmlNodeList nodes = doc.SelectNodes("//entry");
int max = 0;
foreach (XmlNode node in nodes)
{
int nodeAttr = int.Parse(node.Attributes["entryno"].Value);
max = nodeAttr;
}
max++;
存储条目数据的Xml Document。 datagridview从这里获取信息。
<?xml version="1.0" encoding="utf-8" standalone="yes"?>
<!-- These are the entries -->
<entries>
<entry entryno="1">
<weekday>Saturday</weekday>
<date>16/12/2017</date>
<time>21:42</time>
<action>Out of bed</action>
<mind>Ok</mind>
<body>Ok</body>
</entry>
<entry entryno="2">
<weekday>Sunday</weekday>
<date>17/12/2017</date>
<time>02:56</time>
<action>Awake in bed</action>
<mind>Ok</mind>
<body>Ok</body>
</entry>
<entry entryno="3">
<weekday>Sunday</weekday>
<date>17/12/2017</date>
<time>03:07</time>
<action>Awakening</action>
<mind>Ok</mind>
<body>Ok</body>
</entry>
<entry entryno="4">
<weekday>Sunday</weekday>
<date>17/12/2017</date>
<time>03:18</time>
<action>Awakening</action>
<mind>Ok</mind>
<body>Ok</body>
</entry>
<entry entryno="5">
<weekday>Sunday</weekday>
<date>17/12/2017</date>
<time>03:38</time>
<action>Out of bed</action>
<mind>Ok</mind>
<body>Ok</body>
</entry>
</entries>
谢谢你的帮助。
简单使用xml linq：
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;
namespace ConsoleApplication1
{
class Program
{
const string FILENAME = @"c:\temp\test.xml";
static void Main(string[] args)
{
XDocument doc = XDocument.Load(FILENAME);
int count = 1;
foreach (XElement entry in doc.Descendants("entry"))
{
entry.SetAttributeValue("entryno", count++);
}
}
}
}