c# - 如何从 FileStreamResult 获取内容

Question

我正在使用第三方库，其中一种方法返回 FileStreamResult。

public FileStreamResult GenerateFile(OutFormat format, dynamic params);

我的 Controller 中的一个 Action 调用了这个方法:

public ActionResult GenerateExcel()
{
    Utils.XCore core = new Utils.XCore(...); // where ... are contructor params
    // ... other codes here ...
    return core.GenerateFile(OutFormat.EXCEL, new { FileName = "Report" });
}

这没问题，但有时我想将多个 Excel 工作表合并为一个，就像这样:

public ActionResult GenerateExcel()
{
    Utils.XCore core = new Utils.XCore(...); // where ... are contructor params
    // ... other codes here ...
    var excel1 = core.GenerateFile(OutFormat.EXCEL, new { FileName = "rpt1" });
    var excel2 = core.GenerateFile(OutFormat.EXCEL, new { FileName = "rpt2" });
    var excel3 = core.GenerateFile(OutFormat.EXCEL, new { FileName = "rpt3" });

    var finalContent = combineFile(excel1, excel2, excel3);

    return new FileStreamResult(finalContent, "application/ms-excel")
       {
           FileDownloadName = "Report.xls"
       };
}

我现在的问题是我不知道如何从 FileStreamResult 获取内容。关于如何做的任何想法？即使是包含网络链接的评论也非常受欢迎。谢谢!

Answer 1

如果我正确理解你的问题，你想处理/获取来自 FileStreamResult 的内容。该类包含一个名为 FileStream 的属性，它是一个 Stream 对象。现在，可以使用以下修改后的代码将流对象保存为文件 from this site :

private void streamToFile(Stream fileStream, string filePath)
{
    using (FileStream writeStream = new FileStream(filePath, 
                                    FileMode.Create, 
                                    FileAccess.Write))
    {
        int length = 1024;
        Byte[] buffer = new Byte[length];
        int bytesRead = fileStream.Read(buffer, 0, length);
        while (bytesRead > 0)
        {
            writeStream.Write(buffer, 0, bytesRead);
            bytesRead = fileStream.Read(buffer, 0, length);
        }
        fileStream.Close();
        writeStream.Close();
    }
}

下面是使用方法:

var excel1 = core.GenerateFile(OutFormat.EXCEL, new { FileName = "rpt1" });
string filePath = "C:\\yourFileName.xls"; // path of your newly saved file 
using (Stream reportStream = excel1.FileStream)
{
     streamToFile(reportStream, filePath);
}