我写了这段代码。
string filepath = (string)command.ExecuteScalar();
string gfile = Server.MapPath("//" + filepath);
connection.Close();
string path = newFile(aid);
string AttributeDeclaration = "@ATTRIBUTE";
string AttributeDeclaration2 = "@attribute";
string Relation = "@RELATION";
string Relation2 = "@relation";
string Data = "@DATA";
string Data2 = "@data";
string line;
using (FileStream fsReader = new FileStream(gfile, FileMode.Open, FileAccess.ReadWrite, FileShare.ReadWrite))
{
using (StreamReader sr = new StreamReader(fsReader))
{
while ((line = sr.ReadLine()) != null)
{
if (line.StartsWith(Relation) | line.StartsWith(Relation2))
{
using (FileStream fs = new FileStream(path, FileMode.Open, FileAccess.Read, FileShare.ReadWrite))
{
fs.Close();
}
try
{
using (StreamWriter writer = new StreamWriter(path, true))
{
writer.WriteLine(line);
writer.Flush();
writer.Close();
writer.Dispose();
}
}
catch (IOException)
{
}
else if (line.StartsWith(AttributeDeclaration) | line.StartsWith(AttributeDeclaration2))
{
var data = line.Split(new Char[] { ' ', '\t' }, 3);
string attri = (data[0]);
string name = (data[1]);
string type = (data[2]);
Save(aid, attri, name, type);
}
}
sr.Close();
}
}
}
}
private string newFile(int aid)
{
string folderName = @"C:\Users\valentina\Downloads\Desktop\web respository"; //@"C:\Users\valentina\Documents\Visual Studio 2010\Projects\WebRepository3\WebRepository3\files";
string pathString = System.IO.Path.Combine(folderName, "CreateFiles");
string fileName = System.IO.Path.GetRandomFileName();
pathString = System.IO.Path.Combine(pathString, fileName);
string newfilePath = System.IO.Path.ChangeExtension(pathString, ".txt");
System.IO.File.Create(newfilePath);
SqlConnection con = new SqlConnection("Data Source=VALENTINA-PC\\SQLEXPRESS;Initial Catalog=repository_db;Integrated Security=True");
SqlCommand command = con.CreateCommand();
command.CommandText =
@"INSERT INTO new_file
(set_id, path)
VALUES
(@set_id, @path)";
con.Open();
command.Parameters.Add("@set_id", SqlDbType.Int).Value = aid;
command.Parameters.Add("@path", SqlDbType.NVarChar, 1000).Value = newfilePath;
command.ExecuteNonQuery();
con.Close();
return (newfilePath);
}
但我有错误进程无法访问该文件,因为它正被另一个进程使用。错误在于使用(FileStream fs = new FileStream(path,FileMode.Open,FileAccess.Read,FileShare.ReadWrite))。如果我删除它,错误行将在StreamWriter下面。我用ReadAllLInes尝试了一切,没有关闭...... Cam任何人都帮我解决了这个问题。
代码必须是从txt,文件读取,然后是从其他txt文件中的关系写入开始的行。
答案 0 :(得分:0)
尝试使用DirectoryInfo创建目录,然后使用File:
添加文本 string filePath ="Your_file_path.txt";
DirectoryInfo FileCreator = new DirectoryInfo(filePath);
File.AppendAllText(filePath, "some conetnt");
File.AppendAllText(filePath, "make sure you spell content wrong");