在读写文件C#时共享违反IOException [英] Sharing violation IOException while reading and writing to file C#

问题描述

这是我的代码:

public static TextWriter twLog = null;
private int fileNo = 1;
private string line = null;

TextReader tr = new StreamReader("file_no.txt");
TextWriter tw = new StreamWriter("file_no.txt");
line = tr.ReadLine();
if(line != null){
    fileNo = int.Parse(line);
    twLog = new StreamWriter("log_" + line + ".txt");
}else{
    twLog = new StreamWriter("log_" + fileNo.toString() + ".txt");  
}
System.IO.File.WriteAllText("file_no.txt",string.Empty);
tw.WriteLine((fileNo++).ToString());
tr.Close();
tw.Close();
twLog.Close();

它抛出此错误:

IOException:在路径C:\ Users \ Water Simulation \ file_no.txt上共享冲突

IOException: Sharing violation on path C:\Users\Water Simulation\file_no.txt

我要执行的操作只是打开一个文件名为log_x.txt的文件，并从file_no.txt文件中获取"x".如果file_no.txt文件为空，则使日志文件的名称为log_1.txt并输入"fileNo + 1到file_no.txt.新程序启动后，新的日志文件名称必须为log_2.txt.但是我遇到此错误，我不明白我在做什么错.谢谢您的帮助.

What i'm trying to do is just open a file with log_x.txt name and take the "x" from file_no.txt file.If file_no.txt file is empty make log file's name log_1.txt and write "fileNo + 1" to file_no.txt.After a new program starts the new log file name must be log_2.txt.But i'm getting this error and i couldn't understand what am i doing wrong.Thanks for help.

推荐答案

好吧，您正在尝试打开文件file_no.txt来读取，来使用单独的流进行写入.这可能不起作用，因为文件将被读取流锁定，因此无法创建写入流，并且您会收到异常.

Well, you're trying to open the file file_no.txt for reading and for writing using separate streams. This may not work as the file will be locked by the reading stream, so the writing stream can't be created and you get the exception.

一种解决方案是先读取文件，关闭流，然后在增加fileNo后写入文件.这样，文件一次只能打开一次.

One solution would be to read the file first, close the stream and then write the file after increasing the fileNo. That way the file is only opened once at a time.

另一种方法是像这样创建用于读写访问的文件流:

Another way would be to create a file stream for both read and write access like that:

FileStream fileStream = new FileStream(@"file_no.txt", 
                                       FileMode.OpenOrCreate, 
                                       FileAccess.ReadWrite, 
                                       FileShare.None);

对此问题的接受的答案似乎有即使我假设您不想允许共享读取，也是一个很好的解决方案.

The accepted answer to this question seems to have a good solution also, even though I assume you do not want to allow shared reads.

可能的替代解决方案
我了解您想在程序启动时创建唯一的日志文件.这样做的另一种方法是:

Possible alternate solution
I understand you want to create unique log files when your program starts. Another way to do so would be this:

int logFileNo = 1;
string fileName = String.Format("log_{0}.txt", logFileNo);

while (File.Exists(fileName))
{
    logFileNo++;
    fileName = String.Format("log_{0}.txt", logFileNo);
}

这会增加数量，直到找到不存在日志文件的文件编号为止.缺点:如果有log_1.txt和log_5.txt，则下一个文件不是log_6.txt，而是log_2.txt.

This increases the number until it finds a file number where the log file doesn't exist. Drawback: If you have log_1.txt and log_5.txt, the next file won't be log_6.txt but log_2.txt.

要解决此问题，您可以使用掩码log_*.txt枚举目录中的所有文件，并通过执行一些字符串操作找到最大数量.

To overcome this, you could enumerate all the files in your directory with mask log_*.txt and find the greatest number by performing some string manipulation.

可能性无限:-D

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