Question

在Jumpstart视频中（构建适用于Windows Phone 8 Jumpstart的应用）他们展示了给我错误的示例代码： “异步方法的返回类型必须是无效的，T的任务或任务

这是示例代码：

//open
private async string loadStringAsync()
{
   string theData = string.Empty;

   //Get a reference to the local folder
   StorageFolder localFolder = ApplicationData.Current.LocalFolder;
   StorageFile storageFile = await localFolder.GetFileAsync("MyFile.store");

   //Open it and read it
   Stream readStream = await storageFile.OpenStreamForReadAsync();
   using (StreamReader reader = new StreamReader(readStream))
   {
     theData = await reader.ReadToEndAsync();
   }
   return theData;
}

我想它需要一个简单的修复，但由于我是编程的新手并且该特定视频的注释被禁用，我不知道如何...

Answer 1

var a = loadStringAsync();

private async Task<string> loadStringAsync()
{
    string theData = string.Empty;

    //Get a reference to the local folder
    StorageFolder localFolder = ApplicationData.Current.LocalFolder;
    StorageFile storageFile = await localFolder.GetFileAsync("MyFile.store");

    //Open it and read it
    Stream readStream = await storageFile.OpenStreamForReadAsync();
    using (StreamReader reader = new StreamReader(readStream))
    {
       theData = await reader.ReadToEndAsync();
    }
    return theData;
}

从LocalStorage中读取

1 个答案: