Question

我正在尝试生成文件名， log.txt的 log1.txt log2.txt 等

int filenum = 0;
public static void Maker()
{
  string basename = "log";
  string extention = ".txt";
  string finalname = "log.txt";
  for (int i = 0; i < 300; i++)
        {
            using (StreamWriter writer = new StreamWriter(workingDIR + finalname, true))
        {
           // addadder(filenum);
            StringBuilder builder = new StringBuilder();
            builder.Append(base);
            builder.Append(filenum);
            builder.Append(extension);
            finalname = builder.ToString();

        }   } 
}

有没有更好的方法来创建这些类型的文件？第一个文件中没有数字EI log.txt 感谢

Answer 1

您可以替换所有这些：

StringBuilder builder = new StringBuilder();
builder.Append(base);
builder.Append(filenum);
builder.Append(extension);
finalname = builder.ToString();

用这个：

finalname = string.Format("log{0}.txt", filenum);

或与此：

finalname = string.Format("{0}{1}{2}", basename, filenum, extension);

Answer 2

要构建文件名，您可以使用字符串格式：

finalname = string.Format("log{0}.txt", (i + 1));

然后构建文件名，你应该使用System.IO命名空间中的Path.Combine

 using(StreamWriter writer = new StreamWriter(Path.Combine(workingDir, finalname), true))
 {

Answer 3

string basename = "log";
string extention = ".txt";
for (int i = 0; i < 300; i++)
{
    using(var writer = new StreamWriter(Path.Combine(
        workingDir, string.Format("{0}{1:000}{2}" basename, i, extension), true))
    {
        // write file content using writer
    }
}

Answer 4

您可以使用：

const string basename = "log";
const string extention = ".txt";
IEnumerable<string> fileNames = Enumerable.Range(0, 300).
   Select(index => string.Format("{0}{1}.{2}", basename, index, extension));

生成文件名序列

Answer 5

这是你在找什么？

       string fileName = "log.txt";
       int count = 0;
       if (File.Exists(fileName))
       {
           fileName = fileName + "(" + count.ToString() + ").txt";
           count++;
       }
       else
       {
           //Add your logic here
           File.Create(fileName);
       }

Answer 6

public static void Maker()
{
  string basename = "log";
  string extention = ".txt";
  string finalname = "log.txt";
  for (int i = 0; i < 300; i++)
  {
       using (StreamWriter writer = new StreamWriter(workingDIR + finalname, true))
       {
            if(i == 0)
            {
                finalname = basename + extention;
            }
            else
            {
                finalname = basename + i + extention;
            }
       }
  }
}

生成顺序文件名的最佳方法

6 个答案: