我想发送一个文本文件和我的应用程序,并在执行我的应用程序时读取它。我该如何阅读该特定文件?我已将文件设置为内容并在较新时复制。
答案 0 :(得分:4)
private async void ProjectFile()
{
var _Path = @"Metro.Helpers.Tests\MyFolder\MyFolder.txt";
var _Folder = Windows.ApplicationModel.Package.Current.InstalledLocation;
var _File = await _Folder.GetFileAsync(_Path);
var _ReadThis = await Windows.Storage.FileIO.ReadTextAsync(_File);
}
读取适合我的项目文件的代码。希望它也有助于其他人:)
答案 1 :(得分:2)
您的问题的答案在很大程度上取决于您想要阅读文件的 以及它包含的内容。
您可以在File Access Sample app on MSDN中找到一些文件访问示例,其中一个示例是:
if (file != null)
{
using (IRandomAccessStream readStream = await file.OpenAsync(FileAccessMode.Read))
{
using (DataReader dataReader = new DataReader(readStream))
{
UInt64 size = readStream.Size;
if (size <= UInt32.MaxValue)
{
UInt32 numBytesLoaded = await dataReader.LoadAsync((UInt32)size);
string fileContent = dataReader.ReadString(numBytesLoaded);
OutputTextBlock.Text = "The following text was read from '" + file.Name + "' using a stream:" + Environment.NewLine + Environment.NewLine + fileContent;
}
else
{
OutputTextBlock.Text = "File " + file.Name + " is too big for LoadAsync to load in a single chunk. Files larger than 4GB need to be broken into multiple chunks to be loaded by LoadAsync.";
}
}
}
}