我正在使用第三方库,其中一个方法返回FileStreamResult
。
public FileStreamResult GenerateFile(OutFormat format, dynamic params);
我的控制器中的操作调用此方法:
public ActionResult GenerateExcel()
{
Utils.XCore core = new Utils.XCore(...); // where ... are contructor params
// ... other codes here ...
return core.GenerateFile(OutFormat.EXCEL, new { FileName = "Report" });
}
这样会好的,但有时我想将多个Excel工作表合并为一个,如下所示:
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?甚至包含网站链接的评论也非常受欢迎。谢谢!
答案 0 :(得分:2)
如果我正确理解了您的问题,您希望处理/获取FileStreamResult
中的内容。该类包含一个名为FileStream
的属性,它是一个Stream
对象。现在,流对象现在可以使用以下修改后的代码from this site保存为文件:
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();
}
}
以下是如何使用:
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);
}