- 我正在尝试这个**和它的工作正常。我想用户输入文件路径使用逗号分隔。我不明白如何采取单一的单一路径和添加到单一的zip文件。
using System;
using System.IO;
using System.IO.Compression;
namespace ConvertMultipleFilesIntoZip
{
internal class Program
{
static void Main(string[] args)
{
try
{
Console.WriteLine("..........Convert multiple files into a zip file...........");
Console.WriteLine();
Console.WriteLine();
string zipPath = @"d:\result" + DateTime.Now.ToString("yyyyMMddHHmmssffff") + ".zip";
using (ZipArchive archive = ZipFile.Open(zipPath, ZipArchiveMode.Create))
{
Console.WriteLine("Enter the 1st file path, Which you want to add: ");
string sourcePath1 = Console.ReadLine();
Console.WriteLine("Enter the 2st file path, Which you want to add: ");
string sourcePath2 = Console.ReadLine();
Console.WriteLine("Enter the 3st file path, Which you want to add: ");
string sourcePath3 = Console.ReadLine();
Console.WriteLine("Enter the 4st file path, Which you want to add: ");
string sourcePath4 = Console.ReadLine();
archive.CreateEntryFromFile(sourcePath1, "file1.txt");
archive.CreateEntryFromFile(sourcePath2, "file2.txt");
archive.CreateEntryFromFile(sourcePath3, "file1.txt");
archive.CreateEntryFromFile(sourcePath4, "sanju1.nupkg");
/*archive.CreateEntryFromFile(@"d:\file1.txt", "file1.txt");
archive.CreateEntryFromFile(@"c:\Intel\file2.txt", "file2.txt");
archive.CreateEntryFromFile(@"d:\file1.txt", "file1.txt");
archive.CreateEntryFromFile(@"d:\sanju1.nupkg", "sanju1.nupkg");*/
}
Console.WriteLine("Zip file path is " + zipPath + " .");
}
catch (FileNotFoundException dirEx)
{
Console.WriteLine("File does not exist: " + dirEx.Message);
}
}
}
}
- 问题**：给定文件路径与逗号分隔，我想拆分路径，并添加到一个压缩文件中的每个文件添加。
1条答案
按热度按时间6ojccjat1#
要用逗号分隔文件路径并将每个文件添加到zip文件，可以使用以下方法：
首先，提示用户输入文件路径并将其存储在字符串变量中：
使用Split方法将文件路径字符串拆分为字符串数组：
循环遍历文件路径数组，并将每个文件添加到zip文件中：
这将用逗号分隔文件路径，从每个路径中删除任何前导或尾随空格，并使用连续的文件名（例如“file1.txt”、“file2.txt”等）将每个文件添加到zip文件中。