经验分享 · 2023年3月13日

FileStream:The process cannot access the file because it is being used by another process

.Net 6.0, webapi项目

FileStream fs  = new FileStream(filePath, FileMode.Open) 访问站点内目录没有问题,访问其它文件夹出现上述错误,

改为 FileStream fs  = new FileStream(filePath, FileMode.Open, FileAccess.ReadWrite, FileShare.ReadWrite); 后正常。

先看下面一段代码(先以共享的方式打开文件读写,然后以只读的方式打开相同文件):

FileStream fs  = new FileStream(filePath, FileMode.Open, FileAccess.ReadWrite, FileShare.ReadWrite);
FileStream fs2 = new FileStream(filePath, FileMode.Open, FileAccess.Read) 或者 new FileStream(filePath, FileMode.Open, FileAccess.Read, FileShare.Read);

第一句成功执行,第二句呢?它抛出访问违规异常:The process cannot access the file ‘c:\Odma32.log’ because it is being used by another process!

查阅MSDN,无果。后经多次实验,包括使用Windows API CreateFile(…),最终发现正确的用法居然是:

FileStream fs  = new FileStream(filePath, FileMode.Open, FileAccess.ReadWrite, FileShare.ReadWrite);
FileStream fs2 = new FileStream(filePath, FileMode.Open, FileAccess.Read, FileShare.ReadWrite);

在打开fs2时,它必须制定FileShare为ReadWrite,因为fs以ReadWrite的方式打开的。

结论:FileShare不只是对随后的打开文件请求有影响,但事实是它对已经打开的文件句柄也有影响。MSDN中关于FileShare的解释不到位,应该CreateFile条目中的这一段也放进去:

You cannot request a sharing mode that conflicts with the access mode that is specified in an existing request that has an open handle. CreateFile would fail and the GetLastError function would return ERROR_SHARING_VIOLATION.