如何在C＃中打开文件 [英] How to open file in C#

问题描述

如何用字符串值打开文件



我尝试过：

< pre> string c = string.Empty; 
 string s = string.Empty; 
 string ex = string.Empty; 

例如c =sg，s =g，ex = FA1



然后如何链接c，s，ex

 xxx = @C：\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ ，ex.xlsx in c drive 
解决方案
 
请自行尝试一下： c＃路径合并 -  Google搜索 [ ^ ] 
 
 
非常感谢Richart 
 
i尝试
 
 
 string path1 =C：\\how \\+ c +\\+ s +\\+ ex +。xlsx; 
 xxx = path1; 
 
 
 
我用它来构建文件路径。它的用法如下：
 var pathtest =。xslx。GetFilePath（C：，how，c，s，ex）; 
 
 
使用System; 
使用System.IO; 
使用System.Text; 
 
 namespace YourNameSpace 
 {
 public static class FileExtensions 
 {
 private static StringBuilder sb = new StringBuilder（）; 
 
 private static char pathdelimiter ='\\'; 
 
 public static string GetFilePath（this string extension，params string [] pathElements）
 {
 sb.Clear（）; 
 
 string path = null; 
 
 int last = pathElements.Length  -  1; 
 
 for（int i = 0; i< last; i ++）
 {
 sb.Append（pathElements [i]）; 
 sb.Append（pathdelimiter）; 
} 
 
 sb.Append（pathElements [last]）; 
 sb.Append（扩展名）; 
 
 try 
 {
 path = Path.GetFullPath（sb.ToString（））; 
} 
 catch（例外e）
 {
开关（e.GetType（）。姓名）
 {
 caseNotSupportedException：
抛出新的ArgumentException（@扩展名或路径元素滥用冒号，e.InnerException）; 
 
 caseArgumentNullException：
抛出新的ArgumentNullException（@扩展名或路径元素为空，e.InnerException）; 
 
 caseInvalidCastException：
抛出新的ArgumentException（@扩展名或路径元素包含无效字符，e.InnerException）; 
 
 caseSecurityException：
抛出新的ArgumentException（@无法访问该位置，e.InnerException）; 
 
 casePathTooLongException：
抛出新的ArgumentException（@路径太长，e.InnerException）; 
} 
} 
 
返回路径; 
} 
} 
} 
 
 
 
how to open file with string values



What I have tried:



<pre>string c = string.Empty;
        string s = string.Empty;
        string ex = string.Empty;




for example c="sg" ,s ="g",ex="FA1"



then how to link c,s,ex 



xxx = @"C:\how\c\s\ex.xlsx";


the above one shows error because it opens how,c,c,ex.xlsx in c drive
解决方案
Please try a little research for yourself: c# path combine - Google Search[^]

thanks you so much Richart

i tried 

 string path1 = "C:\\how\\" +c + "\\" + s + "\\" + ex + ".xlsx";
xxx = path1;

I use this for building file-paths. It's used like this:
var pathtest = ".xslx".GetFilePath("C:", "how", "c", "s", "ex");
using System;
using System.IO;
using System.Text;

namespace YourNameSpace
{
    public static class FileExtensions
    {
        private static StringBuilder sb = new StringBuilder();

        private static char pathdelimiter = '\\';

        public static string GetFilePath(this string extension, params string[] pathElements)
        {
            sb.Clear();

            string path = null;

            int last = pathElements.Length - 1;

            for (int i = 0; i < last; i++)
            {
                sb.Append(pathElements[i]);
                sb.Append(pathdelimiter);
            }

            sb.Append(pathElements[last]);
            sb.Append(extension);

            try
            { 
                path = Path.GetFullPath(sb.ToString());
            }
            catch (Exception e)
            {
                switch (e.GetType().Name)
                {
                    case "NotSupportedException":
                        throw new ArgumentException(@"extension or path elements misuse a colon", e.InnerException);

                    case "ArgumentNullException":
                            throw new ArgumentNullException(@"extension or path elements are null", e.InnerException);

                    case "InvalidCastException": 
                        throw new ArgumentException(@"extension or path elements contain invalid characters", e.InnerException);

                    case "SecurityException":
                        throw new ArgumentException(@"No access to that location", e.InnerException);

                    case "PathTooLongException":
                        throw new ArgumentException(@"path is too longn", e.InnerException);
                }
            }

            return path;
        }
    }
}


                        
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