替换字符串中的许多可变长度占位符

Question

晚上好，

我正在尝试查找所有出现的以“{”开头并以“}”结尾的字符串，然后将它们替换为不同的字符串。为了帮助说明，下面是一个简短的例子。

string exampleString = "My name is {Name}. I was born as {Race}. I am {Height} tall. My hair is {HairLength}.";

在上面的字符串中，我想删除 {Name}、{Race}、{Height} 和 {HairLength}，并用其他内容替换它们。我现在在使用子字符串时遇到了问题，我想，因为我似乎无法在正确的位置拆分字符串，只能删除括号中的单词。

我想我已经为此精疲力尽，并且犯了一些愚蠢的错误，但我现在看不到它们。下面是我用来拆分字符串的（混乱的）代码。

public static string ForPlayer ( string originalString )
{

    string richText = originalString;

    PropertyInfo propertyInfo;
    int openingInstance = 0;
    int closingInstance = 0;

    int length = 0;
    foreach ( char c in originalString.ToCharArray ())
    {

        if ( c == '{' )
        {

            openingInstance = length;
        }

        if (  c == '}' && openingInstance > closingInstance )
        {

            int lastBrace = closingInstance;
            closingInstance = length;
            int lengthToNextBrace = originalString.IndexOf ( "{", closingInstance ) - closingInstance - 1;

            string richInstance = originalString.Substring ( openingInstance + 1, closingInstance - openingInstance - 1 );
            propertyInfo = Player.player.GetType ().GetProperty ( richInstance );
            if ( propertyInfo != null )
            {

                string beginning;
                string end;

                if ( lastBrace == 0 )
                {

                    beginning = originalString.Substring ( 0, openingInstance );
                } else
                {

                    beginning = originalString.Substring ( closingInstance + 1, openingInstance - lastBrace );
                }

                if ( lengthToNextBrace > -1 )
                {

                    end = originalString.Substring ( closingInstance + 1, lengthToNextBrace );
                } else
                {

                    end = originalString.Substring ( closingInstance + 1, originalString.Length - closingInstance - 1 );
                }

                richText = beginning + propertyInfo.GetValue ( Player.player, null ) + end;
                UnityEngine.Debug.Log ( beginning + propertyInfo.GetValue ( Player.player, null ) + end );
            }
        }

        length += 1;
    }

    return richText;
}

我得到：

DEBUG: "My name is default. I was born as "

DEBUG: ". I am {Height} tdefault. I am "

DEBUG: " tall. Mdefault tall. My hair is "

ERROR: "ArgumentOutOfRangeException: startIndex + length > this.length"

我愿意：

My name is defaultName. I was born as 

defaultRace. I am 

defaultHeight tall. My hair is 

defaultHairLength.

===编辑===

谢谢大家的解答！不幸的是，我不知道我想要替换的字符串是什么，而且有很多可能性，如果可能的话，我宁愿使用我目前通过反射找到它的方法。

Answer 1

var replacement=new Dictionary<string,string>{
    {"Name","defaultName"},
    {"Race","defaultRace"},
    {"Height","defaultHeight"},
    {"HairLength","defaultHairLength"}
};
string exampleString="My name is {Name}. I was born as {Race}. I am {Height} tall. My hair is {HairLength}.";

foreach(var kv in replacement)
{
  exampleString = exampleString.Replace("{" + kv.Key + "}", kv.Value);
}

从 PetSerAl 窃取第一部分

Answer 2

这是我使用正则表达式的方法。

using System.Text.RegularExpressions;
...
class Program
{
    static int occurence = 0;
    static string[] defValues = new string[] { "DefName", "DefRace", "DefHeight", "DefHair" };
    static string ReplaceWithDefault(Match m)
    {
        if (occurence < defValues.Length)
            return defValues[occurence++];
        else
            return "NO_DEFAULT_VALUE_FOUND";
    }

    static void Main(string[] args)
    {
        string exampleString = "My name is {Name}. I was born as {Race}. I am {Height} tall. My hair is {HairLength}.";
        string replaced = Regex.Replace(exampleString, "\{[^\}]+\}", new MatchEvaluator(ReplaceWithDefault));
        occurence = 0;
        Console.WriteLine(exampleString);
        Console.WriteLine(replaced);
    }
}

Answer 3

试试这个：

public static string ForPlayer(string originalString)
{
    return
        Regex
            .Matches(originalString, "{(.*?)}")
            .Cast<Match>()
            .Select(x => x.Groups[1].Value)
            .Select(x => new
            {
                From = x,
                To = Player.player
                    .GetType()
                    .GetProperty(x)
                    .GetValue(Player.player)
                    .ToString()
            })
            .Aggregate(originalString, (a, x) => a.Replace("{" + x.From + "}", x.To));
}

我使用以下测试 class:

在您的样本输入 "My name is {Name}. I was born as {Race}. I am {Height} tall. My hair is {HairLength}." 上测试了这个

public class Player
{
    public string Name {get; set; }
    public string Race {get; set; }
    public string Height {get; set; }
    public int HairLength {get; set; }
}

var player = new Player()
{
    Name = "Fred", Race = "English", Height = "Tall", HairLength = 33
};

得到这个结果：

"My name is Fred. I was born as English. I am Tall tall. My hair is 33."

这个更好：

public static string ForPlayer(string originalString)
{
    return
        Regex
            .Replace(originalString, "{(.*?)}", m =>
                Player.player
                    .GetType()
                    .GetProperty(m.Groups[1].Value)
                    .GetValue(Player.player)
                    .ToString());
}

Answer 4

如果占位符实际上是属性对象的名称，那么您可以利用 FormatWith 来完成这项工作。

string exampleString = "My name is {Name}. I was born as {Race}. I am {Height} tall. My hair is {HairLength}.";
string replacedString = exampleString.FormatWith(Player.player);

Answer 5

这是一个干净、简单的版本。它使用 IndexOf 方法查找大括号而不是遍历字符。它还使用 StringBuilder 来尽可能避免复制字符串。

public static string ForPlayer(string originalString, object player)
{
    var type = player.GetType();
    var sb = new StringBuilder(originalString.Length);
    var lastEnd = 0;    // after the last close brace
    var start = originalString.IndexOf('{');    // start brace

    while (start != -1) // go until we run out of open braces
    {
        var end = originalString.IndexOf('}', start + 1);   // end brace
        if (end == -1)  // if there's a start brace but no end, just quit
            break;
        // copy from the end of the last string to the start of the new one
        sb.Append(originalString, lastEnd, start - lastEnd);
        // get the name of the property to look up
        var propName = originalString.Substring(start + 1, end - start - 1);
        // add in the property value
        sb.Append(type.GetProperty(propName).GetValue(player, null));
        lastEnd = end + 1;  // move the pointer to the end of the last string
        start = originalString.IndexOf('{', lastEnd);   // find the next start
    }

    // copy the end of the string
    sb.Append(originalString, lastEnd, originalString.Length - lastEnd);
    return sb.ToString();
}

替换字符串中的许多可变长度占位符

Replace many variable-length placeholders in string

c#

replace