5

我正在使用此示例将一些变量保存到 xml 文件中：

如何将当前类设置为返回类型结果

这是我的设置文件代码：

using System;
using System.IO;
using System.Xml.Serialization;

namespace ssscc.Settings
{
  public class AppSettings
  {
    public string ReceiptLine1 { set; get; }
    public string ReceiptLine2 { set; get; }
    public string ReceiptLine3 { set; get; }
    public string ReceiptLine4 { set; get; }
    public string ReceiptLine5 { set; get; }
    public string ReceiptLine6 { set; get; }
    public bool ReceiptLine1Enabled { set; get; }
    public bool ReceiptLine2Enabled { set; get; }
    public bool ReceiptLine3Enabled { set; get; }
    public bool ReceiptLine4Enabled { set; get; }
    public bool ReceiptLine5Enabled { set; get; }
    public bool ReceiptLine6Enabled { set; get; }

    public string GatewayUserName { set; get; }
    public string GatewayPassword { set; get; }
    public string GatewayId { set; get; }

    private static string GetSettingsFile()
    {
      var exePath = System.Windows.Forms.Application.StartupPath;
      var sharedDirectory = Path.Combine(exePath, "shared");
      var settingsDirectory = Path.Combine(sharedDirectory, "settings");
      var settingsFile = Path.Combine(settingsDirectory, "ssscc.xml");

      if (!Directory.Exists(sharedDirectory))
      {
        Directory.CreateDirectory(sharedDirectory);
      }

      if (!Directory.Exists(settingsDirectory))
      {
        Directory.CreateDirectory(settingsDirectory);
      }

      return settingsFile;
    }

    internal void SaveSettings()
    {
      var serializer = new XmlSerializer(typeof(AppSettings));
      using (var stream = File.OpenWrite(GetSettingsFile()))
        serializer.Serialize((Stream)stream, this);
    }

    internal static AppSettings GetInstance()
    {
      try
      {

      if (!File.Exists(GetSettingsFile()))
        return null;

      var serializer = new XmlSerializer(typeof(AppSettings));
      using (var stream = File.OpenRead(GetSettingsFile()))
      {
        return (AppSettings)serializer.Deserialize(stream);
      }

      }
      catch (Exception ex)
      {
        Console.WriteLine(ex.Message);
        throw;
      }
    }

  }
}

当我保存数据时，初始保存正常，并在文件末尾显示：

<?xml version="1.0"?>
<AppSettings xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <ReceiptLine1 />
  <ReceiptLine2 />
  <ReceiptLine3 />
  <ReceiptLine4 />
  <ReceiptLine5 />
  <ReceiptLine6 />
  <ReceiptLine1Enabled>false</ReceiptLine1Enabled>
  <ReceiptLine2Enabled>true</ReceiptLine2Enabled>
  <ReceiptLine3Enabled>false</ReceiptLine3Enabled>
  <ReceiptLine4Enabled>false</ReceiptLine4Enabled>
  <ReceiptLine5Enabled>false</ReceiptLine5Enabled>
  <ReceiptLine6Enabled>false</ReceiptLine6Enabled>
  <GatewayUserName>asdfasdf</GatewayUserName>
  <GatewayPassword>asdf</GatewayPassword>
  <GatewayId>sdf</GatewayId>
</AppSettings>

当我更新文件并再次保存时，我得到了这样的结果：

<?xml version="1.0"?>
<AppSettings xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <ReceiptLine1 />
  <ReceiptLine2 />
  <ReceiptLine3 />
  <ReceiptLine4 />
  <ReceiptLine5 />
  <ReceiptLine6 />
  <ReceiptLine1Enabled>false</ReceiptLine1Enabled>
  <ReceiptLine2Enabled>true</ReceiptLine2Enabled>
  <ReceiptLine3Enabled>false</ReceiptLine3Enabled>
  <ReceiptLine4Enabled>false</ReceiptLine4Enabled>
  <ReceiptLine5Enabled>false</ReceiptLine5Enabled>
  <ReceiptLine6Enabled>false</ReceiptLine6Enabled>
  <GatewayUserName>asdfasdf</GatewayUserName>
  <GatewayPassword>asdf</GatewayPassword>
  <GatewayId>sdf</GatewayId>
</AppSettings>>

它>>在最后看到两个。

任何人都明白为什么它>>在我的 xml 文件末尾保存两个？

我的代码错误如下：

4

1 回答 1

11

这是因为您正在使用File.OpenWrite：

对于现有文件，它不会将新文本附加到现有文本。相反，它会用新字符覆盖现有字符。如果您用较短的字符串（例如“Second run”）覆盖较长的字符串（例如“This is a test of the OpenWrite method”），则文件将包含字符串的混合（“OpenWrite 方法的Second runtest ”）。

虽然从您的示例中不清楚，但我怀疑新内容比旧内容短一个字节，因此您会看到原始文件中的右尖括号。

我怀疑你应该File.Create改用。

于 2013-02-01T20:36:05.790 回答