Question

例如，我有一个输入："Test your Internet connection bandwidth. Test your Internet connection bandwidth."（重复两次），我想搜索字符串互联网和带宽。

string keyword = tbSearch.Text //That holds value: "internet bandwidth"
string input = "Test your Internet connection bandwidth. Test your Internet connection bandwidth.";

Regex r = new Regex(keyword.Replace(' ', '|'), RegexOptions.IgnoreCase);
if (r.Matches(input).Count == siteKeyword.Split(' ').Length)
{
    //Do something
}

这不起作用，因为它找到2“互联网”和2“带宽”，所以它计数4但关键字长度是2.那么我能做什么？

Answer 1

var pattern = keyword.Split()
        .Aggregate(new StringBuilder(),
                   (sb, s) => sb.AppendFormat(@"(?=.*\b{0}\b)", Regex.Escape(s)),
                   sb => sb.ToString());

if (Regex.IsMatch(input, pattern, RegexOptions.IgnoreCase))
{
    // contains all keywords
}

第一部分是从关键字生成模式。如果有两个关键字"internet bandwidth"，则生成的正则表达式模式将如下所示：

"(?=.*\binternet\b)(?=.*\bbandwidth\b)"

它将匹配以下输入：

"Test your Internet connection bandwidth."
"Test your Internet connection bandwidth. Test your Internet bandwidth."

以下输入将不匹配（不包含所有单词）：

"Test your Internet2 connection bandwidth bandwidth."
"Test your connection bandwidth."

另一个选项（分别验证每个关键字）：

var allWordsContained = keyword.Split().All(word => 
    Regex.IsMatch(input, String.Format(@"\b{0}\b", Regex.Escape(word)), RegexOptions.IgnoreCase));

Answer 2

不确定你要做什么，但你可以尝试这样的事情：

public bool allWordsContained(string input, string keyword)
{
    bool result = true;
    string[] words = keyword.Split(' ');

    foreach (var word in words)
    {
        if (!input.Contains(word))
            result = false;
    }

    return result;
}

public bool atLeastOneWordContained(string input, string keyword)
{
    bool result = false;
    string[] words = keyword.Split(' ');

    foreach (var word in words)
    {
        if (input.Contains(word))
            result = true;
    }

    return result;
}

Answer 3

这是解决方案。线索是获取结果列表并制作Distinct（）......

    string keyword = "internet bandwidth";
    string input = "Test your Internet connection bandwidth. Test your Internet connection bandwidth.";

    Regex r = new Regex(keyword.Replace(' ', '|'), RegexOptions.IgnoreCase);
    MatchCollection mc = r.Matches(input);
    List<string> res = new List<string>();

    for (int i = 0; i < mc.Count;i++ )
    {
        res.Add(mc[i].Value);
    }

    if (res.Distinct().Count() == keyword.Split(' ').Length)
    {
        //Do something
    }

Answer 4

Regex r = new Regex(keyword.Replace(' ', '|'), RegexOptions.IgnoreCase);
int distinctKeywordsFound = r.Matches(input)
                             .Cast<Match>()
                             .Select(m => m.Value)
                             .Distinct()
                             .Count();
if (distinctKeywordsFound == siteKeyword.Split(' ').Length)
{
    //Do something
}

以任何顺序使用正则表达式查找两个字符串

4 个答案: