[英]In C# Dot Net, How to handle a exception when you want to de-serialize a xml file?, but by default the file doesn't exists
在 C# Dot Net,當你想反序列化一個 xml 文件時如何處理異常,但默認情況下該文件不存在。 因為您必須運行該程序才能創建一個。
以下是我需要幫助的地方。
public static Compare_Data[] Deserialize()
{
Compare_Data[] cm;
cm = null;
string path = @"C:\Users\XYZ\Desktop\BACKUP_DATA\log.xml";
XmlSerializer xs = new XmlSerializer(typeof(Compare_Data[]));
if (File.Exists(path))
{
using (FileStream fs = new FileStream(path, FileMode.Open))
{
// This will read the XML from the file and create the new instance of Compare_Data.
cm = (Compare_Data[])xs.Deserialize(fs);
return cm;
}
}
else
{
using (FileStream fs = new FileStream(path, FileMode.Create))
{
xs.Serialize(fs); /// what to add here ?
}
}
return null;
}
如果一般,您不希望您的方法有副作用。 在這種情況下,在else
分支中創建一個空日志文件可能是不必要的,當有數據要記錄時,應該由單獨的Serialize()
方法處理。
您的代碼可以像這樣簡化:
public static Compare_Data[] Deserialize()
{
const string path = @"C:\Users\XYZ\Desktop\BACKUP_DATA\log.xml";
if (!File.Exists(path))
{
// return null or an empty array, depending on how
// you want the calling code to handle this.
return null;
}
using (FileStream fs = new FileStream(path, FileMode.Open))
{
var xs = new XmlSerializer(typeof(Compare_Data[]));
return (Compare_Data[])xs.Deserialize(fs);
}
}
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