如何从文件中读取某些数据并将其显示出来？

Question

假设我有一个文件，它包含以下内容：

Game Name: Candy Crush 
Game Type: Match Three 
Game Difficulty Rating: 8 

Game Name: Maplestory 
Game Type: RPG 
Game Difficulty: 6 

Game Name: Runescape 
Game Type: RPG 
Game Difficulty: 6 

Game Name: GTA 
Game Type: Video 
Game Difficulty: 7 

如何发布仅一个消息框RPG游戏？我想显示列出的每个RPG游戏的所有数据。

我刚才试过这段代码，但它似乎是显示所有它。

private void button3_Click(object sender, EventArgs e) 
{ 
    var file = File.ReadAllText("Games.txt"); 
    var enter = Environment.NewLine; 

    var gamestrings = games.Select(x => string.Format("{0}\r\n{1}\r\n{2}", x.Name, x.Type, x.Difficulty)); 
    MessageBox.Show(string.Join(enter + enter, gamestrings)); 

    var games = file.Split(new[] { enter + enter }, StringSplitOptions.RemoveEmptyEntries).Select(x => x.Split(new[] { enter }, StringSplitOptions.RemoveEmptyEntries)).Select(e => new { Name = e[0], Type = e[1], Difficulty = e[2] }).Where(x => x.Type.Contains("RPG")); 
} 

给我的错误：

Cannot use local variable 'games' before it is declared 

A local variable named 'e' cannot be declared in this scope because it would give a  different meaning to 'e', which is already used in a 'parent or current' scope to denote something else 

Answer 1

首先，删除StreamReader的东西，并留下这个：

var file = File.ReadAllText(@"c:\Games.txt"); 

（请确保你在里面输入正确的文件路径！）

其次，由于你的数据文件有两个Enter S离开，让我们的这种优势。让我们将字符串分割成份这2分进入，然后除以重新每个输入有每场比赛的细节到一个单一的记录，为此，我们将这个变量方便地定义：

var enter = Environment.NewLine; 

三，LINQ到拯救！！

您可以在lambda语法之间进行选择：

var games = file.Split(new[] {enter + enter}, StringSplitOptions.RemoveEmptyEntries) 
       .Select(x => x.Split(new[] {enter}, StringSplitOptions.RemoveEmptyEntries)) 
       .Select(e => new {Name = e[0], Type = e[1], Difficulty = e[2]}) 
       .Where(x => x.Type.Contains("RPG")); 

或查询语法：

var games = from g in file.Split(new[] {enter + enter}, StringSplitOptions.RemoveEmptyEntries) 
      let gamestrings = g.Split(new[] {enter}, StringSplitOptions.RemoveEmptyEntries) 
      let gamerecord = new {Name = gamestrings[0], Type = gamestrings[1], Difficulty = gamestrings[2]} 
      where gamerecord.Type.Contains("RPG") 
      select gamerecord; 

看到了吗？现在您有包含3个属性匿名类型的强类型列表：

• 名称
• 类型
• 难度

这些，你可以，好像他们是普通班操作和访问他们属性如下：

var gamestrings = games.Select(x => string.Format("{0}\r\n{1}\r\n{2}", x.Name, x.Type, x.Difficulty)); 
MessageBox.Show(string.Join(enter + enter, gamestrings)); 

Answer 2

你有一对夫妇与代码和问题的问题。

FileStream inputFileStream = new FileStream("Games.txt", FileMode.Open, FileAccess.Read); 
StreamReader reader = new StreamReader(inputFileStream); 

这将打开一个流以从文件Games.txt中读取内容。
您密切与流：

inputFileStream.Close(); 
reader.Close(); 

这应该以相反的顺序来完成。但是，没有必要关闭FileStream和StreamReader。关闭FileStream也将关闭StreamReader。

一个更好的办法是使用using关键字：

using (var inputFileStream = new FileStream("Games.txt", FileMode.Open, FileAccess.Read)) 
using (var reader = new StreamReader(inputFileStream)) 
{ 
    ... 
} 

在这样做，所以你永远不会忘记关门。

但是所有这些代码在你的程序中是没有用的。你完全不使用StreamReader。
相反，你从Games.txt阅读内容：

string contents = File.ReadAllText("Games.txt"); 

然后告诉它：

MessageBox.Show(contents); 

你真正想要做的是符合像读行：

using (var inputFileStream = new FileStream("Games.txt", FileMode.Open, FileAccess.Read)) 
using (var reader = new StreamReader(inputFileStream)) 
{ 
    while (reader.Peek() >= 0) 
    { 
    var gameName = reader.ReadLine(); 
    var gameType = reader.ReadLine(); 
    var gameDifficulty = reader.ReadLine(); 
    var emptyLine = reader.ReadLine(); 

    // Here you'll insert the logic to filter out what you want to show. 
    } 
} 

Answer 3

以下是一种方法：

void Main() 
{ 
string[] lines = System.IO.File.ReadAllLines(@"C:\temp\test.txt"); 
var test = new List<string>(); 
foreach (var line in lines) 
{ 
    if (!string.IsNullOrWhiteSpace(line)) 
    { 
     var values = line.Split(':'); 
     test.Add(values[1].Trim()); 
    } 
} 

List<Game> games = new List<Game>(); 
for (int i = 0;i < test.Count();i = i + 3) 
{ 
    Game game = new Game(); 
    game.Name = test[i]; 
    game.Type = test[i+1]; 
    game.Difficulty = test[i+2]; 

    games.Add(game); 
} 
var rpgGames = games.Where(w => w.Type == "RPG"); 
} 

class Game { 
public string Name {get;set;} 
public string Type {get;set;} 
public string Difficulty {get;set;} 
} 

Answer 4

你也可以试试这个：

1. 创建一个包含属性的游戏类：

   class Game 
   { 
       public string Name { get; set; } 
       public string Type { get; set; } 
       public string Difficulty { get; set; } 
   } 
   

2. 然后在你的方法，添加以下代码：

   var lines = File.ReadLines("test.txt"); 
   var games = new List<Game>(); 
   
   var currGame = new Game(); 
   foreach (string line in lines) 
   { 
       var parts = line.Split(':'); 
       if (parts.Length == 1) 
       { 
        games.Add(currGame); 
        currGame = new Game(); 
       } 
       else if (parts[0].EndsWith("Name")) 
        currGame.Name = parts[1].Trim(); 
       else if (parts[0].EndsWith("Type")) 
        currGame.Type = parts[1].Trim(); 
       else 
        currGame.Difficulty = parts[1].Trim(); 
   } 
   
   var output = string.Join(System.Environment.NewLine, games.Where(x => x.Type == "RPG") 
       .Select(x => string.Format("{0} [{1}] => Difficulty: {2}", x.Name, x.Type, x.Difficulty)) 
       .ToArray()); 
   Console.Write(output);