接受 Process.Start 中的多行字符串

Question

我正在从目录位置解析路径：

假设 InitialPath = @"C:\Users\username\Documents\Visual Studio 2015\Projects"

我有一个循环：

var list = new List<string>();

foreach(var folder in Directory.GetDirectories(InitialPath) {
    var folder = Path.GetFileName(folder);
    var file = Path.GetFileName(Directory.GetFiles(folder, "*.sln").Single());

    list.Add(InitialPath +  "\" + folder + "\" + file); //would then result something like "C:\Users\username\Documents\Visual Studio 2015\Projects\Folder1\Project1inFolder1.sln"
}

如果我尝试 list 的路径并将其分配给 richbox 作为其 .text 值，它 returns 是单行文本。

但是当我在 MessageBox 上显示它时，字符串被分成两行，如下所示：

我需要强迫它不要分成几行。我的意思是，无论字符串的长度如何，我都只需要它是一个单行字符串，因为 Process.Start() 不会接受该字符串，因为它会被切成多行。参考如下：

PS: sorry for not being able to explain my question eligibly, english is not my natural language

以防万一，这是我的代码片段：

using MaterialSkin;
using MaterialSkin.Controls;
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Security.Principal;

namespace The_Projects {
    public partial class MainForm : MaterialForm {
        public MainForm() {
            InitializeComponent();

            var materialSkinManager = MaterialSkinManager.Instance;
            materialSkinManager.AddFormToManage(this);
            materialSkinManager.Theme = MaterialSkinManager.Themes.LIGHT;
            materialSkinManager.ColorScheme = new ColorScheme(Primary.BlueGrey800, Primary.BlueGrey900, Primary.BlueGrey500, Accent.LightBlue200, TextShade.WHITE);
        }

        public class DirectoryInformation {
        private string _FolderName;
        private string _Solution;
        private DateTime _Created;
        private DateTime _Accessed;
        private DateTime _Modified;
        private string _SecIdentity;
        private string _NTAccount;
        private double _FileSize;
        private int _FileCount;

        public string FolderName {
            get { return _FolderName; }
            set { _FolderName = value; }
        }

        public string Solution {
            get { return _Solution; }
            set { _Solution = value; }
        }

        public DateTime Created {
            get { return _Created; }
            set { _Created = value; }
        }

        public DateTime Accessed {
            get { return _Accessed; }
            set { _Accessed = value; }
        }

        public DateTime Modified {
            get { return _Modified; }
            set { _Modified = value; }
        }

        public string SecIdentity {
            get { return _SecIdentity; }
            set { _SecIdentity = value; }
        }

        public string NTAccount {
            get { return _NTAccount; }
            set { _NTAccount = value; }
        }

        public double FileSize {
            get { return _FileSize; }
            set { _FileSize = value; }
        }

        public int FileCount {
            get { return _FileCount; }
            set { _FileCount = value; }
        }
    }

        public string InitialPath = @"X:\_\Document\Visual Studio 2015\Projects\";
        public string FolderPath = string.Empty;
        public string Solution = string.Empty;

        private void MainForm_Load(object sender, EventArgs e) {
            var projectList = new List<DirectoryInformation>();

            foreach(var dirs in Directory.GetDirectories(InitialPath)) {
                var ac = File.GetAccessControl(dirs);
                var di = new DirectoryInfo(dirs);

                var dirInf = new DirectoryInformation() {
                    FolderName = Path.GetFileName(dirs),
                    Solution = Path.GetFileName(Directory.GetFiles(dirs, "*.sln").Single()),
                    Created = Directory.GetCreationTime(dirs),
                    Accessed = Directory.GetLastAccessTime(dirs),
                    Modified = Directory.GetLastWriteTime(dirs),
                    SecIdentity = ac.GetOwner(typeof(SecurityIdentifier)).ToString(),
                    NTAccount = ac.GetOwner(typeof(SecurityIdentifier)).Translate(typeof(NTAccount)).ToString(),
                    FileSize = (double) di.EnumerateFiles("*.*", SearchOption.AllDirectories).Sum(x => x.Length) / 1024000,
                    FileCount = Directory.GetFiles(dirs, "*.*", SearchOption.AllDirectories).Count()
                };

                projectList.Add(dirInf);
            }

            lstProjectList.DataSource = projectList;
            lstProjectList.DisplayMember = "FolderName";
        }

        private void lstProjectList_SelectedIndexChanged(object sender, EventArgs e) {
            var project = lstProjectList.SelectedValue as DirectoryInformation;

            lblFolder.Text = project.FolderName;
            lblCreated.Text = project.Created.ToString();
            lblAccess.Text = project.Accessed.ToString();
            lblModified.Text = project.Modified.ToString();
            lblIdentifier.Text = project.SecIdentity;
            lblOwner.Text = project.NTAccount;
            lblSize.Text = project.FileSize.ToString("F2") + " MB";
            lblCount.Text = project.FileCount.ToString();

            FolderPath = InitialPath + project.FolderName;
            Solution = FolderPath + "\" + project.Solution;
        }

        private void btnOpenProject_Click(object sender, EventArgs e) {
            Process.Start(@"C:\Program Files (x86)\Microsoft Visual Studio 14.0\Common7\IDE\devenv.exe", Solution);
            //Clipboard.SetText(Solution);
        }

        private void btnOpenFolder_Click(object sender, EventArgs e) {
            Process.Start("explorer.exe", FolderPath);
        }
    }
}

Answer 1

这正是您的 MessageBox 包装文本的方式。您在这里有两个选择：

创建自定义表单class
创建一个仅用于显示消息的表单对话框

编辑：

改变这个：

 Process.Start(@"C:\Program Files (x86)\Microsoft Visual Studio 14.0\Common7\IDE\devenv.exe", Solution);

到

 Process.Start(@"C:\Program Files (x86)\Microsoft Visual Studio 14.0\Common7\IDE\devenv.exe", "\"" + Solution + "\"");

这里发生的是 Process.Start 方法的第二个参数被视为第一个参数给出的可执行文件的参数。所以 process.start 的作用相当于（但不完全相同）打开命令提示符并输入：

 "C:\Program Files (x86)\Microsoft Visual Studio 14.0\Common7\IDE\devenv.exe" X:\_\Document\Visual Studio 2015\.....

并且命令提示符将 (space) 视为参数分隔符，因此它将 X:\_\Document\Visual 视为一个参数，将 Studio 视为下一个参数，依此类推。当您在字符串周围使用 "\"" 时，您是在告诉 Process.Start 整个字符串（包括 spaces）是单个参数。

Answer 2

你的示例中有一些错误，对我来说这很有效，我得到了所有 *.sln 文件

//this is just to show that you can get short file name if you need FileInfo
        var list = new Dictionary<string, string>();

        var files = Directory.GetFiles(InitialPath, "*.sln", SearchOption.AllDirectories);
        foreach (var file in files)
        {
            FileInfo fileInfo = new FileInfo(file);
            list.Add(fileInfo.Name, file);
        }

        Process.Start(list.FirstOrDefault().Value);

在我的电脑上启动没有问题。但是，如果您想要 devenv.exe 从打开解决方案开始，您可以这样做

 Process.Start(@"C:\Program Files (x86)\Microsoft Visual Studio 14.0\Common7\IDE\devenv.exe", @"devenv/""" + fullFilePath + @"""");

您需要参数 devenv/ 中的命令，并且您必须将路径用双引号引起来 ("fullFilePath")。

接受 Process.Start 中的多行字符串

accept multi line string in Process.Start

c#

string

process