C# 用随机生成的字符串替换每个单词

StackOverFlow的人们好，下面我写的错误代码有问题。我的问题是如何用随机生成的字符串替换下面的每个单词

我的问题是下面的代码，现在它正在随机替换所有的单词，但是它们都是相同的，我想要的是用随机生成的字符串替换它们，而不是所有的单词都是相同的

我已经在参考资料中多次使用这些字符串

我在参考资料中的话是：basenet、R02、R01、R03、R0、vistubularcut、naturalipad、braskausa、moonsteelir、dubilomo

我的代码：

公共字符串randogen1int a1，随机{ StringBuilder result1=新StringBuilderR1； 字符串=textBox1.Text； 对于int i=0；i 我根据ASCII表选择了范围： 如果还需要随机长度，只需在调用方法中创建一个新的随机实例

编辑：

根据评论，这里有一个更好的方法

私有静态随机rdn=新随机； 私有字符串随机字符串长度 { 返回新的stringcharrdn。下一个'A'，'Z'，长度； } 最后，您可以简单地编写：replace1[index]=RandomString4；

String.Replacestring，String将用第二个参数字替换第一个参数字的每个实例。如果要用不同的字符串替换同一个字的每个实例，则必须将该字符串的字分隔开，对其进行迭代，每次找到要替换的字时，repl用一个随机字符串将其置为a

下面是一个示例，假设您已经有了获取随机单词的方法：

string Source = "The quick brown fox jumps over the lazy dog.\r\nThe quick brown fox jumps over the lazy dog.";
string[] ReplaceWords = { "quick", "fox", "lazy", "dog" };

string[] SourceWords = Source.Split(' ');
string result = "";
for (int i = 0; i < SourceWords.Length; i++)
{
    if (ReplaceWords.Contains(SourceWords[i]))
        result += " " + GetRandomWord();
    else
        result += " " + SourceWords[i];
}
result = result.Trim() //Remove white spaces at the begining and end of the string.

现在，让我们创建列表：

string[] Nouns = {"dog", "cat", "fox", "horse", "bird"};
string[] Adjectives {"lazy", "sleepy", "quick", "big", "small"};

现在我们的方法是：

string GetRandomWord(Context c)
{
    Random R = new Random();
    switch (c)
    {
        case Context.Noun:
            return Nouns[R.Next(0, Nouns.Length)];
            break;
        case Context.Adjective:
            return Adjectives[R.Next(0, Adjectives.Length)];
            break;
    }
}

现在，对文本替换代码进行轻微修改：

string Source = "The quick brown fox jumps over the lazy dog.\r\nThe quick brown fox jumps over the lazy dog.";
string[] ReplaceAdjectives = { "quick", "lazy" };
string[] ReplaceNouns = { "fox", "dog" };

string[] SourceWords = Source.Split(' ');
string result = "";
for (int i = 0; i < SourceWords.Length; i++)
{
    if (ReplaceAdjectives.Contains(SourceWords[i]))
        result += " " + GetRandomWord(Context.Adjective);
    else if (ReplaceNouns.Contains(SourceWords[i]))
        result += " " + GetRandomWord(Context.Noun);
    else
        result += " " + SourceWords[i];
}
result = result.Trim() //Remove white spaces at the begining and end of the string.

然后我们相应地修改代码：

string Source = "The quick brown fox jumps over the lazy dog.\r\nThe quick brown fox jumps over the lazy dog.";
string[] ReplaceAdjectives = { "quick", "lazy" };
string[] ReplaceNouns = { "fox", "dog" };

bool GuaranteeMatch = true;

string[] SourceWords = Source.Split(' ');
string result = "";
Random R = new Random();
for (int i = 0; i < SourceWords.Length; i++)
{
    if (GuaranteeMatch)
    {
        int I = R.Next(0, Adjectives.Length) //Adjectives and Nouns have the same Length. This is a requirement for this method to work.
        if (ReplaceAdjectives.Contains(SourceWords[i]))
            result += " " + GetMatchingWord(Context.Adjective, I);
        else if (ReplaceNouns.Contains(SourceWords[i]))
            result += " " + GetMatchingWord(Context.Noun, I);
        else
            result += " " + SourceWords[i];
    }
    else
    {
        if (ReplaceAdjectives.Contains(SourceWords[i]))
            result += " " + GetRandomWord(Context.Adjective);
        else if (ReplaceNouns.Contains(SourceWords[i]))
            result += " " + GetRandomWord(Context.Noun);
        else
            result += " " + SourceWords[i];
    }
}
result = result.Trim() //Remove white spaces at the begining and end of the string.

现在，如果GuaranteMatch为真，我们将始终得到如下结果： 结果=棕色的大马跳过昏昏欲睡的猫。\r\n懒惰的棕色狗跳过小鸟

---

它甚至可以返回原来的句子，因为被替换的单词也存在于要替换的单词列表中。

你想要的东西列表不是一个问题。你的问题是什么？这不是调试错误代码的服务；如果你有一个明确的、具体的问题并有答案，就去问这个问题选项。如果您想了解如何调试错误代码的建议，请参阅@Ashlou foreach是C中该类型循环的合适关键字。您的编辑将其更改为for each，这是不正确的。编辑其他人的代码时请小心，尤其是在问题中。@EricLippert好吧，我的问题是如何用随机泛型替换这些单词ted字符串。您希望用作替换的字符串是随机的，还是希望随机选择数组中的一个字符串作为替换字符串？@BMinster基本上我希望替换所有字符串，并且希望所有字符串都是随机的。请注意，您可能希望将随机实例设为私有静态；这样，如果方法被快速连续多次调用。您不会得到相同的输出。请注意，如果长度较长，则效率会非常低。最好是创建一个具有适当长度的字符数组，或使用字符串生成器。请注意，修改其输入的方法更难调试。并且，您的长度变量不再按照它所说的那样执行；现在它意味着要添加的字符，而不是字符串长度。考虑使用本地剩余而不是修改长度。注意，65和90是魔术数字。如果您使用的是“A”而不是65，代码变得更清晰。“EricLippert谢谢，做了一个编辑来改进我的答案。

string Source = "The quick brown fox jumps over the lazy dog.\r\nThe quick brown fox jumps over the lazy dog.";
string[] ReplaceAdjectives = { "quick", "lazy" };
string[] ReplaceNouns = { "fox", "dog" };

string[] SourceWords = Source.Split(' ');
string result = "";
for (int i = 0; i < SourceWords.Length; i++)
{
    if (ReplaceAdjectives.Contains(SourceWords[i]))
        result += " " + GetRandomWord(Context.Adjective);
    else if (ReplaceNouns.Contains(SourceWords[i]))
        result += " " + GetRandomWord(Context.Noun);
    else
        result += " " + SourceWords[i];
}
result = result.Trim() //Remove white spaces at the begining and end of the string.

string GetMatchingWord(Context c, int i)
{
    switch (c)
    {
        case Context.Noun:
            return Nouns[i];
            break;
        case Context.Adjective:
            return Adjectives[i];
            break;
    }
}

string Source = "The quick brown fox jumps over the lazy dog.\r\nThe quick brown fox jumps over the lazy dog.";
string[] ReplaceAdjectives = { "quick", "lazy" };
string[] ReplaceNouns = { "fox", "dog" };

bool GuaranteeMatch = true;

string[] SourceWords = Source.Split(' ');
string result = "";
Random R = new Random();
for (int i = 0; i < SourceWords.Length; i++)
{
    if (GuaranteeMatch)
    {
        int I = R.Next(0, Adjectives.Length) //Adjectives and Nouns have the same Length. This is a requirement for this method to work.
        if (ReplaceAdjectives.Contains(SourceWords[i]))
            result += " " + GetMatchingWord(Context.Adjective, I);
        else if (ReplaceNouns.Contains(SourceWords[i]))
            result += " " + GetMatchingWord(Context.Noun, I);
        else
            result += " " + SourceWords[i];
    }
    else
    {
        if (ReplaceAdjectives.Contains(SourceWords[i]))
            result += " " + GetRandomWord(Context.Adjective);
        else if (ReplaceNouns.Contains(SourceWords[i]))
            result += " " + GetRandomWord(Context.Noun);
        else
            result += " " + SourceWords[i];
    }
}
result = result.Trim() //Remove white spaces at the begining and end of the string.