我在windows-8中创建StreamWriter
对象时遇到了麻烦,通常我只是创建一个实例,只是传递一个字符串作为参数,但在Windows8中我得到一个错误,指示它应该接收一个流,但我注意到流是一个抽象类,有人知道如何编写xml文件的代码吗? BTW我使用.xml是因为我想保存序列化的对象,或者有人知道如何在Windows8中将序列化对象保存到文件中?
有什么想法吗?
当前正在使用Windows 8消费者预览版
代码:
StreamWriter sw = new StreamWriter("person.xml");
错误:
The best overloaded method match for 'System.IO.StreamWriter.StreamWriter(System.IO.Stream)' has some invalid arguments
发布于 2012-04-24 15:46:39
你可以使用下面这样的代码来代替StreamWriter:
StorageFolder folder = ApplicationData.Current.LocalFolder;
StorageFile file = await folder.CreateFileAsync();
using (IRandomAccessStream fileStream = await file.OpenAsync(FileAccessMode.ReadWrite))
{
using (IOutputStream outputStream = fileStream.GetOutputStreamAt(0))
{
using (DataWriter dataWriter = new DataWriter(outputStream))
{
//TODO: Replace "Bytes" with the type you want to write.
dataWriter.WriteBytes(bytes);
await dataWriter.StoreAsync();
dataWriter.DetachStream();
}
await outputStream.FlushAsync();
}
}
您可以查看WinRTXamlToolkit库中的StringIOExtensions类以供示例使用。
编辑*
虽然上面所有的方法都可以工作--它们是在WinRT中的FileIO
类可用之前编写的,这简化了上面解决方案所解决的大多数常见场景,因为您现在只需调用await FileIO.WriteTextAsync(file, contents)
将文本写入文件,并且也有类似的方法来读取、写入或附加字符串、字节、字符串列表或IBuffers
。
发布于 2013-04-09 22:48:52
您可以创建一个通用的静态方法,它可以通过我们的应用程序使用,如下所示
private async Task<T> ReadXml<T>(StorageFile xmldata)
{
XmlSerializer xmlser = new XmlSerializer(typeof(List<myclass>));
T data;
using (var strm = await xmldata.OpenStreamForReadAsync())
{
TextReader Reader = new StreamReader(strm);
data = (T)xmlser.Deserialize(Reader);
}
return data;
}
private async Task writeXml<T>(T Data, StorageFile file)
{
try
{
StringWriter sw = new StringWriter();
XmlSerializer xmlser = new XmlSerializer(typeof(T));
xmlser.Serialize(sw, Data);
using (IRandomAccessStream fileStream = await file.OpenAsync(FileAccessMode.ReadWrite))
{
using (IOutputStream outputStream = fileStream.GetOutputStreamAt(0))
{
using (DataWriter dataWriter = new DataWriter(outputStream))
{
dataWriter.WriteString(sw.ToString());
await dataWriter.StoreAsync();
dataWriter.DetachStream();
}
await outputStream.FlushAsync();
}
}
}
catch (Exception e)
{
throw new NotImplementedException(e.Message.ToString());
}
}
要编写xml,只需使用
StorageFile file = await ApplicationData.Current.LocalFolder.CreateFileAsync("data.xml",CreationCollisionOption.ReplaceExisting);
await writeXml(Data,file);
和读取xml的使用
StorageFile file = await ApplicationData.Current.LocalFolder.GetFileAsync("data.xml");
Data = await ReadXml<List<myclass>>(file);
https://stackoverflow.com/questions/10290820
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