Question

所以我试图将变量存储到文本文件中，以便我的程序以后可以调用它们，但是我需要这些文件具有设置的名称，并且当我尝试使其永远无法正常工作时。

我尝试过在foreach循环中使用foreach循环，我尝试了很多事情，但是所有这些事情都只是弄乱了变量数据

string[] lines = { name, FirstBlock, MonIstem, WedIstem, ThridBlock, FourthBlock, "Design Time", SixthBlock, TueIstem, ThurIstem, EighthBlock, NinthBlock, "Design Time", FriIstem };
string[] names = { "name", "FirstBlock", "MonIstem", "WedIstem", "ThirdBlock", "FourthBlock", "Design Time", "SixthBlock", "TueIstem", "ThurIstem", "EightBlock", "NinthBlock", "Design Time", "FriIstem", };
foreach (string TXTname in names)
{
    Console.WriteLine($"Saving {TXTname}");
}
foreach (string line in lines)
{
    string getNameOfVar = nameof(line);
    using (FileStream bs = File.OpenWrite($@"C:\Users\gn193755\Documents\{TXTname}.txt"))
    {
        byte[] thing = new UTF8Encoding(true).GetBytes(line);
        bs.Write(thing, 0, thing.Length);
    }
}

没有错误消息，但是当我打开文本文件时，要么只有一个文本文件，要么它们上都带有相同的单词，或者它们只是乱码文本，所以我一直希望使用正确的文件名。正确的文字。

Answer 1

我太笨了，我做了更多研究，找到了.zip文件。

               var lines = new [] { name, FirstBlock, MonIstem, WedIstem, ThridBlock, FourthBlock, "Design Time", SixthBlock, TueIstem, ThurIstem, EighthBlock, NinthBlock, "Design Time", FriIstem };
            var names = new [] { "name", "FirstBlock", "MonIstem", "WedIstem", "ThirdBlock", "FourthBlock", "Design Time", "SixthBlock", "TueIstem", "ThurIstem", "EightBlock", "NinthBlock", "Design Time", "FriIstem", };
            var linesAndnames = lines.Zip(names, (l, n) => new { Line = l, Name = n });
            foreach (var ln  in linesAndnames)
            {
              var path = $@"C:\Users\gn193755\Documents\{ln.Name}.txt";
              File.WriteAllText(path, ln.Line);
            }
        }

Answer 2

您需要将第二个foreach放在第一个foreach中。
如果它们是字符串变量，作为实验，请使用以下内容创建文件

string getNameOfVar = nameof(line);
var path = $@"C:\Users\gn193755\Documents\{TXTname}.txt");
File.WriteAllText(path, line);

尝试此操作，以确保正确写入行字符串：

var path = $@"C:\Users\gn193755\Documents\output.txt");
File.AppendAllText(path, line);

Answer 3

从代码中可以看出，foreach循环是独立的。

string[] lines = { name, FirstBlock, MonIstem, WedIstem, ThridBlock, FourthBlock, "Design Time", SixthBlock, TueIstem, ThurIstem, EighthBlock, NinthBlock, "Design Time", FriIstem };
string[] names = { "name", "FirstBlock", "MonIstem", "WedIstem", "ThirdBlock", "FourthBlock", "Design Time", "SixthBlock", "TueIstem", "ThurIstem", "EightBlock", "NinthBlock", "Design Time", "FriIstem", };
        foreach (string TXTname in names)
        {
                Console.WriteLine($"Saving {TXTname}");

            foreach (string line in lines)
            {

                string getNameOfVar = nameof(line);
                using (FileStream bs = File.OpenWrite($@"C:\Users\gn193755\Documents\{TXTname}.txt"))
                {
                    byte[] thing = new UTF8Encoding(true).GetBytes(line);
                    bs.Write(thing, 0, thing.Length);
                }
            }
        }

尝试使用上面的代码

如何在没有foreach的情况下使“ foreach（名称中的字符串TXTname字符串）”为变量，或者我需要foreach？

3 个答案: