找出两个字符串中有多少个单词相同答案

【问题标题】：Find how many words are the same in two strings找出两个字符串中有多少个单词相同
【发布时间】：2013-02-13 02:41:57
【问题描述】：

我有这个函数，我想比较两个字符串，然后返回存在多少个单词，但以下不起作用。我似乎总是得到 SameWordCount 的 0 和 MasterAddressWordCount 的 1

有什么想法吗？

// some more string cleaning
        mastermkAddressKey = mastermkAddressKey.Replace(",", " ").Replace(".", " ").Trim();
        mastermkAddressKey = Encoding.ASCII.GetString(Encoding.GetEncoding("Cyrillic").GetBytes(mastermkAddressKey));
        mastermkAddressKey = mastermkAddressKey.Replace("  ", " |").Replace("| ", "").Replace("|", "");
        mastermkAddressKey = QbaseStrings.RemoveDuplicateWords(mastermkAddressKey);

        duplicatemkAddressKey = duplicatemkAddressKey.Replace(",", " ").Replace(".", " ").Trim();
        duplicatemkAddressKey = Encoding.ASCII.GetString(Encoding.GetEncoding("Cyrillic").GetBytes(duplicatemkAddressKey));
        duplicatemkAddressKey = duplicatemkAddressKey.Replace("  ", " |").Replace("| ", "").Replace("|", "");
        duplicatemkAddressKey = QbaseStrings.RemoveDuplicateWords(duplicatemkAddressKey);

        string[] masterAddressSeparateWords = mastermkAddressKey.Split(new char[' '], StringSplitOptions.RemoveEmptyEntries);
        string[] duplicateAddressSeparateWords = duplicatemkAddressKey.Split(new char[' '], StringSplitOptions.RemoveEmptyEntries);

        int SameWordCount = 0;
        int MasterAddressWordCount = 0;

        foreach (string masterWord in masterAddressSeparateWords)
                {
                    foreach (string duplicateWord in duplicateAddressSeparateWords)
                    {
                        if (masterWord == duplicateWord) {SameWordCount++;}
                    }

                    MasterAddressWordCount++;
                }

        int WordDifference = MasterAddressWordCount - SameWordCount;

        if (WordDifference == 0) { return "sure"; }
        if (WordDifference > 0 && WordDifference < 3) { return SameWordCount.ToString() + " " + MasterAddressWordCount.ToString(); }
        if (WordDifference > 2 && WordDifference < 5) { return "possible"; }

【问题讨论】：

标签： c# sql-server user-defined-functions sqlclr

【解决方案1】：

你的问题是因为new char[' ']，你在这里的意思是new char[] {' '}。编译器（非常有帮助）在这里将' ' 转换为int，使其变为char[int]。这意味着：

new char[' ']

其实是一样的：

new char[32]

这最终成为一个大的无用 char[] 数组，而不是你所追求的单个空间。

您可以通过查看为以下对象生成的 IL 清楚地看到这一点：

var a = new char[' '];

这是：

IL_0001:  ldc.i4.s    20
IL_0003:  newarr      System.Char
IL_0008:  stloc.0     // a

20 是 32 的十六进制表示。

【讨论】：

【解决方案2】：

我通过更改以下几行解决了这个问题：

string[] masterAddressSeparateWords = mastermkAddressKey.Split(new char[' '], StringSplitOptions.RemoveEmptyEntries);
        string[] duplicateAddressSeparateWords = duplicatemkAddressKey.Split(new char[' '], StringSplitOptions.RemoveEmptyEntries);

收件人：

string[] masterAddressSeparateWords = mastermkAddressKey.Split(' ');
string[] duplicateAddressSeparateWords = duplicatemkAddressKey.Split(' ');

【讨论】：